Chemistry 12 - Notes on Unit 2 -...

Chemistry 12 - Notes on Unit 2 - Equilibrium

You might remember from the last unit, when we were dealing with Potential Energy Diagrams, that the term Reverse Reaction came up many times. It's important to know that many chemical reactions are reversible. That is:

Reactants → Products or Reactants ← Products

Reactants form Products Products form Reactants

For example, under certain conditions, one mole of the colourless gas N2O4 will decompose to form two moles of brown NO2 gas: N2O4 2 NO2 colourless brown Under other conditions, you can take 2 moles of brown NO2 gas and change it into one mole of N2O4 gas: N2O4 2 NO2 colourless brown In other words, this reaction, as written may go forward or in reverse, depending on the conditions.NOW, here's something to think about! If we were to put some N2O4 in a flask, the N2O4 molecules would collide with each other and some of them would break apart to form NO2 .

This process, of course is indicated by the forward reaction: N2O4 2 NO2 Once this has happened for awhile, there is a build up of NO2 molecules in the same flask. (After all, this is what the reaction is making.)

Now these NO2 molecules don't just sit there! They are of course moving around and colliding with things. Once in awhile, two NO2 molecules will collide with each other and guess what? They join to form a molecule of N2O4 !

This process, as you might have guessed is indicated by the reverse reaction: N2O4 2 NO2 Two things you'll have to realize is that as long as there is N2O4 present, the forward reaction will keep on happening and as long as there is NO2 present, the reverse

reaction will keep on happening!

Also, you must keep in mind that all these molecules are mixed in the same container! At one particular time a molecule of N2O4 might be breaking up, and at the same time two molecules of NO2 might be joining to form another molecule of N2O4! So here's an important thing to understand: In any reversible reaction, the forward reaction and the reverse reaction are going on at the same time! This is sometimes shown with a double arrow

:

N2O4 2 NO2

The double arrow means that both the forward and reverse reaction are happening at the same time.

Just a little comment here. The word "happening" has a similar meaning to the word " dynamic ". Just remember that!

*************************************************** Now what we're going to do is look at how the rate of the forward reaction changes if we put some pure N2O4 in a flask: If we put some pure N2O4 in a flask (No NO2 yet!), there will be a high concentration of N2O4 . That is, there will be lots of N2O4 molecules to collide with each other. So at the beginning of our little experiment, (which we will call " time " 0 ") the rate of the forward reaction is quite fast.

N2O4 2 NO2 So if we were to make a graph of the rate of the forward reaction vs. time, the graph might start out something like this:

1098765432100.0

0.1

0.2

0.3

0.4

0.5

0.6

TIME (min)

RATE (forward)

OK, now you might ask: "Why does the rate of the forward reaction go down ? Well, if you recall Unit 1, as the forward reaction proceeds: N2O4 2 NO2 the N2O4 is used up and so it's concentration goes down. Also, you must remember that if the concentration of a reactant goes down, there is less chances of collisions and the rate of the reaction decreases. As the reaction continues, the slower rate will use up N2O4 more slowly, so the [N2O4 ] will not decrease so quickly and therefore the rate will not decrease quite as quickly. (Read the last sentence over a couple of times and make sure it makes sense to you!) For those "graph wise" people, you will probably guess that this means that the slope of the line on the graph gets more gradual. The rest of the graph might look something like this:

864200.0

0.1

0.2

0.3

0.4

0.5

0.6

TIME (min)

RATE (forward)

NOW, it's time we consider the rate of the reverse reaction. As you might recall, when we have a container full of pure N2O4 , initially there is no NO2 in the container. Since there is no NO2 , there are no NO2 molecules to collide with each other, and the rate of the reverse reaction is zero. But of course, as time goes on, NO2 is formed from theforward reaction (N2O4 2 NO2) so in a short time, some NO2 molecules can start colliding and the reverse reaction will begin: ( N2O4 2 NO2 ) As MORE NO2 is formed by the forward reaction, the rate of the reverse reaction gradually increases

. Now, for you "graph buffs", the graph of the Rate of the reverse reaction vs. Time might look like this:

1098765432100.0

0.1

0.2

0.3

0.4

0.5

0.6

RATE (rev erse)

TIME (min)

OK. Now, lets look at the graph for the forward rate and the reverse rate together:

1098765432100.0

0.1

0.2

0.3

0.4

0.5

0.6

RATE (f orward)

RATE (rev erse)

TIME (min) NOW, focus your attention on the graph at " Time = 4 minutes " . You will notice that at this point: the rate of the forward reaction = the rate of the reverse reaction At this point, NO2 is being used up at the same rate that it is being formed: N2O4 2 NO2 Because this is so, you should be able to convince yourself that the [NO2] is no longer changing! Because the reverse rate is equal to the forward rate, N2O4 is being formed at the same rate it is being used up. So, also the [N2O4] is no longer changing either! Can you predict what will happen to the graph after 4.0 minutes? YOU GUESSED IT! The rates of the forward reaction and the reverse reaction, no longer change because the [N2O4] is constant and the[NO2] is also constant. The graph will look like this:

1210864200.0

0.1

0.2

0.3

0.4

0.5

0.6

RATE (f orward)

RATE (rev erse)

TIME (min) The situation happening from 4.0 minutes on in this graph has a special name and a special significance. At this point, the system (meaning the container, the N2O4 and the NO2 ) is said to be at equilibrium. To describe it even more precisely, we can say that we have reached a state of dynamic equilibrium

..

Here are some things that you must understand about dynamic equilibrium: 1. The reaction has not

stopped!

2. The forward and the reverse reaction continue

reverse reactions are said to be " balanced ") eg. for the reaction:

to take place, but their rates are equal so there are no changes in concentrations of reactants or products. (The forward and

N2O4 2 NO2 for each N2O4 molecule that breaks up to form two NO2 molecules, two other NO2 molecules combine to form another N2O4 molecule. All this is happening on the microscopic level, so we don't see individual molecules reacting.

3. As far as we can see from the "outside", there appears to be nothing happening. All observable properties are constant. These include the concentrations of all reactants and products, the total pressure, colour, temperature etc. 4. If no changes were made in conditions and nothing is added or taken away, a system

at equilibrium would remain that way forever, the forward and reverse reactions "ticking away", but balanced so that no observable changes happen. Here are a couple of other things to consider before we summarize everything: 1. Changing the temperature can alter the rates of the reactions at equilibrium. This could "throw off" the balance. So, for a system at equilibrium, the temperature must remain constant and uniform throughout the system.

2. Letting material into or out of the system will affect rates so a system at equilibrium is a closed system

.

3. Again, consider the equilibrium reaction: N2O4 2 NO2 In the example that we did to construct the graphs, we had started with pure N2O4 and no NO2. The forward reaction rate was high at the start, but the reverse reaction rate eventually "caught up", the rates became equal and equilibrium was established. Can you guess what would happen if we had started with pure NO2 instead (no N2O4 )? To make a long story short, the reverse reaction rate would start out high, but the forward rate would eventually " catch up ". When the rates became equal, again equilibrium would be established. We can summarize all this by saying that the equilibrium can be approached from the left (starting with reactants) or from the right

(starting with products)

Just a little term before we summarize: The word macroscopicobservable. (The opposite is microscopic, which means too small to see eg. molecular level). Some

means large scale or visible or

macroscopic properties are total pressure, colour, concentrations, temperature, density etc. Alright, let's summarize:

Characteristics of a System at Equilibrium 1. The rate of the forward reaction = The rate of the reverse reaction 2. Microscopic processes (the forward and reverse reaction) continue in a balance which yields no macroscopic changes. (so nothing appears to be happening.) 3. The system is closed and the temperature is constant and uniform throughout. 4. The equilibrium can be approached from the left (starting with reactants) or from the right (starting with products).

In our previous examples you will recall that when we started with N2O4 or with NO2, things happened and equilibrium was eventually established. In general:

If sufficient activation energy is available, systems not

at equilibrium will tend to move toward

a position of equilibrium.

Enthalpy You have probably met with the concept of enthalpy in Unit 1 and in Chemistry 11. Looking it up in the glossary of the textbook defines it as: " The heat content of a system. " Another way to think of enthalpy is as "Chemical Potential Energy". Any change in the Potential Energy of a system means the same thing as the " Enthalpy Change ". The symbol for Enthalpy is " H ". Therefore the "change in Enthalpy" of a chemical reaction is called "∆H".

As far as we're concerned in Chemistry 12, a Potential Energy Diagram (like we looked at in the last unit) is the same thing as an "Enthalpy Diagram". Let's look at an example:

Enthalpy

∆H

Progress of Reaction

The "Enthalpy Change"

Reactants

Products

The reaction shown on this graph is Exothermic. This means that heat is released or given off. What was given off as heat energy was lost as Enthalpy from the reactants. The net energy change between the products and the reactants is called the Enthalpy Change ( ∆H). As you can see, the enthalpy change ( ∆H) in this reaction is negativ e . (This is always the case when the Products are lower on the Potential Energy (Enthalpy) Graph.)

An exothermic reaction

So, we can make a statement here: In an Exothermic Reaction (∆H is negative), the Enthalpy

is decreasing.

In an Endothermic Reaction (∆H is positive), the Enthalpy

is increasing.

Of course, you might remember another way to show an endothermic or exothermic reaction. In this case, the "Heat Term" is written right in the equation. (This is called a "thermochemical equation".) If the heat term is on the left side, it means heat is being used up and it's endothermic. If the heat term is on the right side, heat is being released and it's exothermic.

Look at the following examples: 1. A + B C + D ∆H = -24 kJ is exothermic so enthalpy is decreasing

.

2. X + Y Z ∆H = 87 kJ is endothermic so enthalpy is increasing

.

3. E + D F + 45 kJ is exothermic so enthalpy is decreasing

.

4. G + J + 36 kJ L + M is endothermic so enthalpy is increasing

.

Make sure you are very familiar with the facts given above before you go on the the next section! Ask if you have any problems with it. Given an equation with the ∆H shown or a thermochemical equation with the heat term on left or right, you will be expected to identify it as endothermic or exothermic and to determine whether the enthalpy is increasing or decreasing

as the reaction is proceeding in the forward direction.

Chemical systems will tend toward a state of minimum enthalpy activation energy is available and

if sufficient no other factors are considered

.

Another way of stating this might be: A chemical reaction will favour the side (reactants or products) with minimum enthalpy

if no other factors are considered.

Thus for an exothermic reaction, if no other factors are considered:

Enthalpy

∆H


The "Enthalpy Change"

Reactants

Products

An exothermic reaction

Here, the Products have lower enthalpy than the Reactants so the reaction tends to "favour the products". In other words, if the reactants are mixed, they will tend to form products spontaneously (without outside assistance) rather than remain as reactants.

The products will be favoured because the products have minimum enthalpy. In other words, there is a natural tendency here for reactants to spontaneously form products Let's look at a diagram for an endothermic reaction:

∆H is positive

An endothermi c reaction

Reactants

Products

Enthalpy


In the case of an endothermic reaction, the enthalpy of the Reactants is lower than the enthalpy of the Products. Since chemical systems favour a state of minimum enthalpy, the Reactants are favoured in this case. In other words if the reactants are mixed, they will tend to remain as reactants rather than forming products.

As you can see by looking at the examples above, there are two ways of looking at what happens to the enthalpy: If the reaction is exothermic, the products formation of products (move toward the

have minimum enthalpy and the right) is favourable

.

If the reaction is endothermic, the reactants the formation of products (move toward the right) is

have minimum enthalpy and un

this case the formation of favourable. In

reactants (move toward the left) is favourable

.

*******************************************************

Now, consider the simple melting of water: H2O(s) + heat H2O(l) (the subscript (s) stands for solid) (the subscript (l) stands for liquid) If we were to look at only the enthalpy in this process, you can see that the reactant ( H2O(s)) would have minimum enthalpy and would be favoured. Can you see what this statement would mean? It would mean

that all of the water in the universe should exist only as a solid

! (It would not be favourable for water to exist as a liquid!) We would all be frozen solid!!!!

Obviously there is something wrong with this reasoning! We know that there is liquid water in the universe, so what gives? The answer to this problem lies in looking at another factor that governs equilibrium. That factor is called entropy

(or randomness or disorder)

*******************************************************

Entropy

Entropy simply means disorder

, or lack of order.

Student's Chemistry binders are a good example. At the beginning of the semester, papers are neat and ordered. By the end of the semester, pages are torn and falling out. The thing has been overloaded and the rings are usually bent or broken. Pages are loose and often the plastic coating is in shreds. The entropy of the binder has increased over time! This situation is not unusual, even in chemical reactions! In Grade 8, you probably learned about the arrangement of molecules in solids, liquids and gases. Well, look on the next page and you'll find it again. (yawn!):

The particles in a solid are very close together and very ordered . A solid has very low entropy . (very little disorder)

The particles in a liquid are fairly close together and they are not as ordered as they are in the solid. A liquid has higher entropy than a solid , but less entropy than a gas .

The particles of a gas are very far apart and they are moving randomly (all different directions - any old way!). They have very little order. A gas has very high entropy .

So we can summarize by saying that: Entropy of a Solid < Entropy of a Liquid < Entropy of a Gas

Chemists and successful Chemistry Students (THAT'S YOU!) can look at a chemical equation with subscripts showing the phases and tell which has maximum entropy, the reactants or the products. In other words, they can look at an equation and tell whether entropy is increasing or decreasing as the reaction proceeds to the right. In the following examples, the entropy is increasing

(or the products have greater entropy):

1. There is a gas (or gases) on the right, when there are no gases on the left of the equation: CaCO3(s) + 2 HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) a gas is formed on the right. 2. When there are gases on both sides, the products have greater entropy when there are more moles of gas on the right (add up coefficients of gases on left and right.): 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

There are (4 + 5) = 9 moles of gas on the left

There are (4 + 6) = 10 moles of gas on the right.

Another way to look at the last example is to say that: " The side with the greater number of moles of gas has the greatest entropy. " 3. When a solid dissolves in water, the products ( the aqueous solution of ions ) have greater entropy. This makes sense because:

+ -+ -

--

- +

+

-

-+

+

+

+

-

OHH

OH

H

OH

H OH H

+

OHH

OH

H

OHH

OH

H

- OHH

OH

H

OH

H OH H

+

OHH

OH

H

OHH

OH

H

-

SOLID

Dissolving

An Aqueous Solution

The ions in a Solid are very ordered. They have low entropy .

When dissolved in water, the ions are separated and surrounded by water molecules. The ions are much more spread out and disordered. The entropy of an aqueous solution is higher than that of a solid .

If you have any questions about these, check with your teacher!

*********************************************** Now, we looked at the process: H2O(s) + heat H2O(l) Remember we decided that all the H2O in the universe should remain as a solid because H2O(s) has lower enthalpy than H2O(l) and nature favours a state of minimum enthalpy

.

Well, now we can explain why there is some liquid water in the universe (lots of it): H2O(l) has higher entropy

than H2O(s)

If left alone for a long time, systems tend to get more disordered. (Like this classroom, your bedroom or your binders!) There is a natural tendency in nature toward maximum disorder or maximum entropy! Chemical systems will tend toward a state of maximum entropy


Another was of stating this might be:

A chemical reaction will favour the side (reactants or products) with maximum entropy


Remember, the other factor which controlled reactions was enthalpyAlso remember that:

. (chemical potential energy).

Chemical systems will tend toward a state of minimum enthalpy activation energy is available and

if sufficient no other factors are considered

.

or (see next page)...... A chemical reaction will favour the side (reactants or products) with minimum enthalpy


Be careful that you don't get the terms enthalpy (chemical potential energy) and entropy (disorder) confused. They do look and sound somewhat the same! Sorry! Remember you figure out which has the most enthalpy (reactants or products) by looking at the H or the heat term. Also, remember that you can figure out which has the more entropy (reactants or products) by looking at the subscripts which represent the phases. Also, we can combine the rules about "natural tendencies" to come up with this: In nature, there is a tendency toward minimum enthalpy and maximum entropy

.

Now, let's consider this process again: H2O(s) + heat H2O(l) The two tendencies are said to "oppose each other

" in this case:

The tendency toward minimum enthalpy would favour the reactant !( since you have to add heat energy to H2O(s) to get H2O(l) , H2O(s) has minimum enthalpy) In this case the tendency toward maximum entropy would tend to favour the product. (A liquid has more entropy (disorder) than a solid) We say that:

When the two tendencies oppose each other products), the reaction will

(one favours reactants, the other favours reach a state of equilibrium

.

That is, there will be some reactants and some products present. The relative amounts of each depends on conditions like temperature, pressure, concentration etc. Since this is the case with : H2O(s) + heat H2O(l) , there is some solid water and some liquid water in the universe. (In other words, there is a state of equilibrium) Which one is present in the greater amount is determined largely by the temperature.

*********************************************** Now, lets consider another simple process: A glass bottle is knocked down from a high shelf onto a concrete floor and the glass shatters: Bottle on a high shelf Thousands of pieces of glass on a concrete floor The bottle falls down and not up! This happens because there is a natural tendency toward minimum gravitational potential energy (like minimum enthalpy in chemistry) In other words the tendency toward minimum gravitational potential energy favours the products (the low bottle rather than the high) (To stretch this analogy further, we could consider that the person who knocked the bottle off of the shelf was simply supplying the " activation energy ") Remember that the bottle broke into thousands of pieces when it hit the concrete. The broken pieces of glass have more disorder (entropy) than the bottle, so in this process, the tendency toward maximum entropy also favours the products!

Bottle on a high shelf → Thousands of pieces of glass on a concrete floor

Products have minimum "enthalpy", so the tendency toward minimum enthalpy favours the products.

Products have maximum entropy, so the tendency toward maximum entropy favours the products.

There is no "equilibrium" here when the process is finished. That bottle has completely fallen down and it is all broken. (This bottle is no longer on the shelf and it is no longer an "unbroken bottle") We can summarize what happened here:

Processes in which both entropy favour the products, will

the tendency toward minimum enthalpy and toward maximum go to completion

.

(ie. All reactants will be converted into products. There will be no reactants left once the process is finished!) Here's an example of a chemical reaction in which this happens: 2K(s) + 2H2O(l) 2KOH(aq) + H2(g) + heat This process is exothermic (the heat term is on the right) so the products have lower enthalpy. The tendency toward minimum enthalpy favours the products

.

There is a mole of gas on the right (H2(g)) and no gases in the reactants Therefore, the products have greater entropy. The tendency toward maximum entropy favours the products

.

Since both tendencies favour the products, this reaction will go to completion

.

That is, all of the reactants (assuming you have the correct mole ratios eg. 2 moles of K to 2 moles of H2O) will be converted to products. If one reactant is in excess, the limiting reactant will be completely consumed. So, if you put a little bit of potassium in a beaker of water, the reaction will keep going until all of the potassium is used up. There will be no potassium left once the reaction is complete. In other words, the reverse reaction does not occur!

***********************************************

Let's consider one more process: 2KOH(aq) + H2(g) + heat 2K(s) + 2H2O(l) In this case, the tendency toward minimum enthalpy favours the reactants

, and the

tendency toward maximum entropy also favours the reactants

.

Processes in which both entropy favour the reactants, will

the tendency toward minimum enthalpy and toward maximum not occur at all!

.

(ie. None of the reactants will be converted into products. There will be no products formed!)

NOTE: This would be like thousands of pieces of glass spontaneously sticking together, forming a bottle and jumping up onto a high shelf! This does not occur at all. (At least I've never seen it happen!)

***********************************************

To summarize: When the two tendencies oppose each other products), the reaction will

(one favours reactants, the other favours reach a state of equilibrium

.

Processes in which both entropy favour the

the tendency toward minimum enthalpy and toward maximum products, will go to completion

.

Processes in which both entropy favour the

the tendency toward minimum enthalpy and toward maximum reactants, will not occur at all!

.

Remember the following? When a chemical system is at equilibrium, the rate of the forward reaction

is equal to the rate of the reverse

reaction.

As long as no changes are made to conditions of a system at equilibrium, this situation would

just go on forever with no changes in macroscopic properties. This might be nice, but it would be boring! Also, humans like to change and manipulate

systems to their advantage. (For example, to produce more of a useful product) In order to disturb an equilibrium all we have to do is change the rate of either the forward or

reverse reaction so that they are no longer equal. Let's do it and see what happens:

Effect of Temperature

Consider the equilibrium system: N2O4(g) + heat 2 NO2(g) colourless brown At equilibrium, NO2 is being formed at the same rate as it is being used up, so its

concentration is constant. The system is a medium brown colour at room temperature. Let's look at a potential energy diagram for this reaction (Notice that it is endothermic)


PE (kJ)

0

N O2 4

2NO 2

N2O4(g) + heat 2 NO2(g) 1. Now, consider the forward reaction and the reverse reaction. Which reaction do you think would be most affected by an increase in temperature? Forward reaction i

Increasing the temperature will speed up the forward reaction more than the reverse:

One way to look at it is: N2O4(g) + heat 2 NO2(g)

In an endothermic reaction, the forward reaction needs heat, so it's rate will be increased more by an increase in temperature.

So, increasing the temperature, the forward reaction is faster than the reverse reaction for

awhile: (This can be shown by making the forward arrow longer) N2O4(g) + heat 2 NO2(g) So guess what happens to [NO2] while the forward reaction is faster than the reverse reaction?

N2O4(g) + heat 2 NO2(g) The NO2 is formed faster than it is used up, so it's concentration increases. The N2O4 is used up faster than it is formed, so it's concentration decreases. This might be shown as follows: N2O4(g) + heat 2 NO2(g) Since there are now more molecules of NO2 to run into each other, The rate of the reverse reaction will also speed up. Because there is less N2O4 after awhile, the forward reaction will slow down. So as you might guess, after awhile, the rate of the reverse reaction will again equal the rate

of the forward reaction and again we have equilibrium

!(a new equilibrium!)

But remember this

:

The forward rate was faster than the reverse

rate for awhile.(increasing [NO2] and decreasing [N2O4])

But the reverse rate was never faster than the forward rate even though it finally caught up. It may take a little deep thought about this, but : If NO2 was being formed faster than being used up for awhile but never used up faster than it

was being formed, it's concentration will be higher when the new With similar thought, you might also see that the [N2O4] will be lower in the new equilibrium.

equilibrium is established.

So, to summarize

:

1. Original equilibrium: N2O4(g) + heat 2 NO2(g) 2. Temperature is increased and the endothermic (forward in this case) reaction rate increases: N2O4(g) + heat 2 NO2(g) 3. A new equilibrium is established in which there is more NO2 and less N2O4.

N2O4(g) + heat 2 NO2(g)

N2O4(g) + heat 2 NO2(g) colourless brown When we have more product(s) than we had before and less reactants we say that: The equilibrium has shifted to the (the stuff on the right has increased and the stuff on the left has decreased)

right

So what will happen to the colour in flask containing the equilibrium mixture when it is put into boiling water and heated? You're right, of course, it will get darker ............(because of a higher [NO2] which is brown.)

************************************************* If the flask were placed in ice water, the endothermic (forward in this case) reaction would

slow down. You would get a situation like this: (The rate of the forward reaction is slower than the rate of

the reverse reaction.)

The forward reaction is endothermic. When heat is removed, it slows down.


Now, the reverse reaction is faster than the forward reaction, so N O is being formed faster than it is being used up, and NO is being used up faster than it forms.

2 4

2

While the reverse reaction is faster than the forward, the [N2O4] will build up and the [NO2] will decrease...


N2O4(g) + heat 2 NO2(g) So what happens now, since [N2O4] is higher, the rate of the forward reaction will gradually

increase and after a certain time, it will again be equal to the rate of the reverse reaction. At this point, we have a new equilibrium

!

In this equilibrium [N2O4] will be higher than it was originally (for awhile it was being formed faster than it was being used up), and the [NO2] will be lower than it was originally.(for awhile it was being used up faster than it was formed.)

We say that the equilibrium has shifted to the left

. (or shifted toward the reactant side)


It's at equilibrium again because the rates are equal, even though the amounts are different than in the original equilibrium

To summarize the effect of temperature

:

When the temperature is increased, the endothermic reaction will speed up and the

equilibrium will shift toward the side without the heat term

.

(A new equilibrium is established in which there is a higher concentration of substances on the side without the heat term and a lower concentration of substances on the side with the heat term.)

When the temperature is decreased, the endothermic reaction will slow down and the

equilibrium will shift toward the side with the heat term

.

(A new equilibrium is established in which there is a lower concentration of substances on the side without the heat term and a higher concentration of substances on the side with the heat term.)

**************************************************

Effect of Concentration and Partial Pressure Next, we will consider what happens when we change the concentration of a reactant or the

partial pressure of a reactant. First, we'd better explain the term "partial pressure". When you have a gas mixture, the pressure exerted by one gas in the mixture is called the

partial pressure of that gas. The more of that gas you have, the greater it's partial pressure. For example: In a certain gas mixture containing NO and CO2 gases: Partial Pressure of NO = 40 kPa (kilopascals-a unit of pressure) Partial Pressure of CO2 = 60 kPa Total Pressure = 40 + 60 = 100 kPa If you add some NO, it's partial pressure will go up. For example: Partial Pressure of NO = 50 kPa (kilopascals-a unit of pressure) Partial Pressure of CO2 = 60 kPa Total Pressure = 50 + 60 = 110 kPa Since a higher partial pressure results from putting more of a certain gas in the same volume,

it is really just another way of saying concentration. That are different quantities in different units (Concentration is in moles/L, Partial pressure is in kPa) But when one goes up, the other goes up. From now on, when we mention "Partial Pressure" changes, they will have exactly the same

effect as Concentration changes. Consider the following system at equilibrium: CO2(g) + NO(g) CO(g) + NO2(g) Of course, at equilibrium: the rate of the forward reaction = the rate of the reverse reaction

Let's, all of a sudden, add some CO2 to the container that contains all of these. (What we are doing is increasing the [CO2] or the partial pressure of CO2.)

CO2(g) + NO(g) CO(g) + NO2(g) Since the [CO2], a reactant is now higher, there will be more chances of collision between

CO2 and NO, so the forward reaction will speed up. CO2(g) + NO(g) CO(g) + NO2(g) This will cause the [CO] and the [NO2] to increase and the [CO2] and [NO] to decrease CO2(g) + NO(g) CO(g) + NO2(g) Because [CO] and the [NO2] have increased, the rate of the reverse reaction will speed up. When the rate of the reverse reaction is again = the rate of the forward reaction, we will

again have equilibrium. (A new equilibrium!)

CO2(g) + NO(g) CO(g) + NO2(g) In this new equilibrium, [CO] and [NO2] will be higher than they were originally and [CO2] and [NO] will be lower than they were after we added the CO2. In this case, the equilibrium is said to have shifted to the right. (or shifted to the product side.) NOTE: Remember, we added some CO2. In the new equilibrium [CO2] is less than after we

added it, but it doesn't quite go down to the level it was before we added any. [NO] will be quite low because it goes down and we didn't add any.

*************************************************** Effects of Changing the Volume of the Container or Total Pressure One thing to remember from Chemistry 11 here is that: As the volume of a fixed number of moles of gas is decreased (the gas is compressed), the pressure will increase. Let's consider the equilibrium: 2NO2(g) N2O4(g) brown colourless Let's say that for some strange reason, we had 2 molecules of NO2 only in a certain volume:

NO 2

NO 2

There NO molecules will move around randomly, hitting the sides of the container and exerting pressure.

2

Now, suppose we were to completely convert those two molecules of NO2 into N2O4: 2NO2(g) N2O4(g) We would now have only 1 molecule (a molecule of N2O4) in the same volume. This means there are only half as many molecules hitting the sides of the container, and therefore: The pressure will be only half of what is was with the 2 molecules of NO2:

N O2 4

Having only 1 molecule means the system only has half the pressure it would have with 2 molecules

So, to summarize: The greater the number of moles (or molecules) of gas in a particular volume, the greater the pressure.

******************************************************** Now, let's get back to the original equilibrium mixture, where we have some NO2 and some N2O4:

2NO2(g) N2O4(g) brown colourless

Let's say we have this system in a syringe and we quickly decrease the volume by pushing the plunger in. Recall that decreasing the volume is exactly the same thing as increasing the pressure. Initially, the colour will go darker because everything (including the brown NO2) is compressed. However, when you increase the pressure on something, there is a natural tendency for the

system to do anything it can in order to offset that increase. (For example, when you squish a balloon in one place, the air will be forced to another place

and the balloon will bulge somewhere else, or the balloon will pop to decrease the pressure!)

When we increase the pressure on the 2NO2(g) N2O4(g) system, it can offset that

increase by: converting more NO2 into N2O4 !

2NO2(g) N2O4(g) brown colourless

in other words: shifting to the side with less moles of gas. (as shown by the coefficients) This "shift to the right" will use up some brown NO2 converting it to colourless N2O4, and the

colour of the system will gradually get lighter again. So, in summary: When the total pressure is increased (volume is decreased) in an equilibrium system with

gases, the equilibrium will shift toward the side with less moles of gas in the equation. or, as you might guess: When the total pressure is decreased (volume is increased) in an equilibrium system with

gases, the equilibrium will shift toward the side with more moles of gas in the equation.

***************************************************** Effect of Catalysts Consider the equilibrium system: N2O4(g) + heat 2 NO2(g) colourless brown Adding a catalyst to this system would decrease the activation energy by providing a

different route (or mechanism) for the reaction: (see the Potential Energy Diagram on the next page.)

PE (kJ)

0

N O2 4

2NO 2

Uncatalyzed Reaction

Catalyzed Reaction

Using a catalyst to provide a route will decrease the activation energy for the forward reaction and for the reverse reaction by the same amount. This means the forward reaction will speed up, but so will the reverse reaction. In fact, the rates of both the forward and reverse reactions will still remain equal to each other

(even though they are both faster.) Therefore, the equilibrium will not shift! Adding a catalyst to a system not at equilibrium will simply speed things up so that

equilibrium will be attained faster. It does not alter any of the concentrations etc. at equilibrium!

To summarize, equilibrium is affected by: 1. Temperature - If the temp. is increased, the equilibrium will shift toward the side without

the heat term. -If the temp. is decreased the equilibrium will shift toward the side with the heat term.

2. Concentration - If the [a reactant] is increased, the equilibrium will shift toward the right (the product side) - If the [a product] is increased, the equilibrium will shift toward the left (the reactant side) 3. Partial Pressure of Gases - the same effects as concentration. 4. Total Volume and Total Pressure - If pressure is increased (volume decreased), the equilibrium will shift to the side with less moles of gas. - If the pressure is decreased (volume increased), the equilibrium will shift toward the side with more moles of gas. And lastly, remember: Catalysts - Have no effect on equilibrium. They may help a system reach equilibrium faster, that's all! Now, Let's look at LeChatelier's Principle. Hopefully this will make it all easier as it gives us a simply way to look at the previous two concepts together: LeChatelier's Principle: If a closed system at equilibrium is subjected to a change, processes will occur that tend to counteract that change. - "Processes" usually mean that the equilibrium will shift to the left or right. - "Counteract" means that if you do something to a system at equilibrium, the system will shift in such a way as to try to "undo" what you did. - Remember, equilibrium only occurs in a closed system.

Examples of Counteracting: - If you add heat to a system, it will shift in a way that it tends to "use up" the added heat. - If you remove heat from a system, it will shift in a way that it tends to "produce heat" - If you increase the concentration of a certain substance in an equilibrium mixture, the system will shift so as to reduce the concentration of that substance. - If you decrease the concentration of a certain substance in an equilibrium mixture, the system will shift so as to increase the concentration of that substance. - If you increase the total pressure on an equilibrium system involving gases, the system will shift in a way that will reduce the total pressure. - If you decrease the total pressure on an equilibrium system involving gases, the system will shift in a way that will increase the total pressure.

*********************************************************** Effect of Changes in Temperature NOTE: When using LeChatelier's Principle, it is easier to have all reactions with "heat" in "Thermochemical Form". eg.) The reaction: A + B C ∆H = -56 kJ is exothermic (∆H is negative), so heat is given off or written on the right. Thermochemical form: A + B C + 56 kJ The reaction: D + E F + G ∆H = 43 kJ is endothermic (∆H is positive), so heat is absorbed, or written on the left Thermochemical form: D + E + 43 kJ F + G Okay, let's find out how LeChatelier's Principle applies here: Let's say we have an endothermic reaction:

A + B + heat C + D If we increase the temperature of this system, we are adding heat. In order to counteract our change, the equilibrium will move in such a way as to use up heat. Since heat is on the left, the forward reaction uses up heat, so it will predominate and the equilibrium will shift toward the right.

Which means a new equilibrium will be established in which there is more C and D and less A and B than in the original equilibrium. To summarize: When the temperature is increased, the equilibrium will shift away from the side with the heat term. Now, if the temperature was decreased, the equilibrium would shift in such a way that would produce heat (to counteract the change). To do this, it would shift toward the side with the heat term. (in other words, produce heat) To summarize: When the temperature is decreased, the equilibrium will shift toward the side with the heat term. Effect of Changes in Concentration or Partial Pressure Consider the equilibrium equation:

H2(g) + I2(g) 2HI(g) If we add some H2 to a flask containing this mixture at equilibrium, [H2] will immediately increase. In order to counteract this change, the equilibrium will shift to the right in order to "use up" some of the extra H2. (In other words to decrease the [H2]). Consider the equilibrium equation: H2(g) + I2(g) 2HI(g) Let's say now that we somehow take away some I2. [I2] will immediately decrease. In order to counteract this change, the equilibrium will shift to the left in order to increase [I2] again . If we were to add some HI, the [HI] would immediately increase. In order to counteract this change, the equilibrium would shift to the Left. We can summarize the effects of changing concentrations by saying: If the concentration of a substance in an equilibrium system is increased by us, the equilibrium will shift toward the other side of the equation, in order to counteract the change. or

If the concentration of a substance in an equilibrium system is decreased by us, the equilibrium will shift toward the side of the equation with that substance, in order to counteract the change. Changing the Partial Pressure of a gas in an equilibrium system has the same effect as changing the concentration of that gas. For example: Given the equilibrium equation: H2(g) + Br2(g) 2HBr(g) If the partial pressure of H2 is increased, the equilibrium will shift to the right. If the partial pressure of H2 is decreased, the equilibrium will shift to the left. If the partial pressure of HBr is increased, the equilibrium will shift to the left. If the partial pressure of HBr is decreased, the equilibrium will shift to the right.

*********************************************************** Effect of Changes in Total Pressure or Volume for Gaseous Systems Recall from the last tutorial that: The more moles of gas in a certain volume, the higher the pressure. Also, recall LeChatelier's Principle: LeChatelier's Principle: If a closed system at equilibrium is subjected to a change, processes will occur that tend to counteract that change. Thus: If the total pressure of a system at equilibrium is increased, the equilibrium will shift toward the side with less moles of gas (as shown by coefficients) in order to reduce the total pressure. For example: Given the equilibrium equation: N2(g) + 3H2(g) 2NH3(g)

1 + 3 = 4 moles of gas on this side.

2moles of gas on this side.

If the total pressure on this system is increased, the equilibrium would shift to the right (the side with fewer moles of gas). This counteracts the imposed change by reducing the pressure. To review: A shift to the right in this case would mean that once the new equilibrium is established, the [NH3] would be higher than before, the [N2] and the [H2] would be lower than before. If the total pressure on this system is decreased, the equilibrium would shift to the left (the side with more moles of gas). This counteracts the imposed change by increasing the pressure. To see what happens when we change the total volume of the container we must remember that: Increasing the volume of a closed system with gases will decrease the pressure. (the molecules have more room so they exert less pressure on the sides of the container) So, given changes in volume of the container, just remember that the changes in pressure are just the opposite. For example: Given the equilibrium equation: N2(g) + 3H2(g) 2NH3(g)

1 + 3 = 4 moles of gas on this side.

2moles of gas on this side.

If the total volume of the container is increased, this means that the total pressure is decreased. The equilibrium will then shift to the side with more moles (to the left in this case), in order to counteract the change and try to increase the pressure again. If the total volume of the container is decreased, this means that the total pressure is increased. The equilibrium will then shift to the side with less moles (to the right in this case), in order to counteract the change and try to decrease the pressure again.

What is Keq ? The "K" in Keq stands for "Constant". The "eq" means that the reaction is at equilibrium. Very roughly, Keq tells you the ratio of Products/Reactants for a given reaction at

equilibrium at a certain temperature.

K eq =[Products] [Reactants]

It's not quite this simple when we deal with real substances. Let's take an example. It has been found for the reaction: 2HI(g ) H2(g) + I

2(g)

that if you take the [H2], the [I2] and the [HI] in an equilibrium mixture of these at 423 °C, the expression:

[H ] 2 [I ]22[HI]

= 0.0183

The K expression

eq

The value of Keq

The value of this ratio stays at 0.0183 regardless of what we might try to do with the

concentrations. The only thing that changes the value of Keq for a given reaction is the temperature! Writing Keq Expressions In the example just above this, the equation was:

2HI(g) H2(g) + I2(g) and the Keq expression was:

[H ] 2 [I ]22[HI]

K =eq

Notice a couple of things here. The concentrations of the products are on the top (numerator) and the concentration of the reactant is on the bottom. (denominator).

Also, notice that the coefficient "2" in the "2HI" in the equation ends up as an exponent for

[HI] in the Keq expression. Thus we have 2[HI] in the denominator.

Notice that in the Equilibrium Constant Expression (Keq ), whatever is written on the right of the arrow in the equation (products) is on top and whatever is written on the left of the arrow in

the equation (reactants) is on the bottom. This is always the case in a Keq expression, regardless of which reaction (forward or reverse)

predominates at a certain time.

*********************************************************

The Keq Expressions for Solids and Liquids Consider the following reaction:

CaCO3(s) CaO(s) + CO2(g)

You might expect the Keq expression to be something like:

Keq = [CaO(s)] [CO2(g)] __________________ [CaCO3(s)] But when you consider a solid, the number of moles per litre or molecules in a certain volume

is constant. The molecules everywhere in the solid are about the same distance apart and are the same size: See the diagram on the next page...

In a Solid, equal volumes anywhere within the solid have an equal number of molecules.. Therefore we say that the concentration of a solid is constant.

Going back to our example: Consider the following reaction:


You might expect the Keq expression to be something like: Keq = [CaO(s)] [CO2(g)] __________________ [CaCO3(s)] Since CaO and CaCO3 are solids, we can assume that their concentrations are constant. We can therefore rewrite the Keq expression as follows:

Keq = (a constant) [CO2(g)] ____________________ (a constant) Now, if we rearrange:

Keq (a constant) = [CO2(g)] (a constant) You'll notice that now, on the left side, we have an expression which consists of only

constants. Chemists simply combine all these constants on the left and call it the equilibrium constant

(Keq ) In other words, the concentrations of the solids are incorporated into the value for Keq.

Therefore, the Keq expression for the equation:


is simply: Keq = [CO2] The bottom line is: When we write the Keq expression for a reaction with solids, we simply leave out the solids. Liquids also have a fairly constant concentration. They don't expand or contract that much

even with changes in temperature. The same argument that was used for solids can also be used for liquids. Thus, we can expand

the last statement: When we write the Keq expression for a reaction with solids or liquids, we simply leave out the solids and the liquids. Gases and aqueous solutions do undergo changes in concentration so they are always

included in the Keq expression.

*********************************************************** Value of Keq and the Extent of Reaction Remember that Keq is a fraction (or ratio). The products are on the top and the reactants are

on the bottom. Remember that in a fraction: The larger the numerator the larger the value of the fraction. The larger the denominator the smaller the value of the fraction. At 200 °C, the Keq for the reaction:

N2(g) + 3H2(g) 2NH3(g) is known to be 626.

Keq is equal to the ratio: [NH3]2

[N2] [H2]3

__________________

Since this ratio is very large (626) at 200°C, we can say that [NH3]2 (the numerator) must be

quite large and [N2] [H2]3 (the denominator) must be small:

N2(g) + 3H2(g) 2NH3(g)

In other words, a large value for Keq means that at equilibrium, there is lots of product and very little reactant left.

Even another way to say this is: The larger the value for Keq the closer to completion the reaction is at equilibrium. (NOTE: "Completion" means reactants have been completely converted to products.) A very small value for Keq means that there is very little product and lots of reactant at

equilibrium. In other words, a very small value for Keq means that the reaction has not occurred to a very

great extent once equilibrium is reached. Consider the following reaction: A + B C + D Keq = 1.0

The Keq expression is: [C] [D] Keq = ________ = 1.0 In this case the ratio of [C] [D] to [A] [B] is 1.0. [A] [B] This means that there is about the same amount of products as reactants. At equilibrium, this reaction has proceeded to "about half way" to completion. Keq and Temperature You probably couldn't help but notice that in some of the examples above when the Keq was

given, the temperature was also mentioned eg.) "At 200 °C, the Keq for the reaction:

N2(g) + 3H2(g) 2NH3(g) is known to be 626."

You probably wondered, " What's the 200 °C got to do with it and what should I do with the "200"?

Well, here's something you need to know: When the temperature changes, the value of Keq also changes. Let's see how this works:

Consider the following endothermic reaction: A + B + heat C The Keq expression for this is: [C]

[A] [B] Keq = __________

Now, let's say that we increase the temperature of this system. By LeChatelier's Principle,

adding heat to an endothermic reaction will make it shift to the right:

Shift to the Right

A + B + heat C

Heat is Added

Because it shifts to the right, a new equilibrium is established which has a higher [C] and a lower [A] and [B]. Therefore the Keq will have a larger numerator and a smaller denominator: [C] This will make the value of Keq Keq = __________ larger than it was before. [A] [B] So we can summarize by saying: When the temperature is increased in an endothermic reaction, the equilibrium will shift to the right and the value of Keq will increase. For an endothermic reaction, decreasing the temperature would make the equilibrium shift to

the left. This would cause [C] to decrease and the [A] and [B] to increase:

Shift to the Left

A + B + heat C

Heat is Removed

Now, in the Keq expression, the numerator would be smaller and the denominator would be

larger: [C] This will make the value of Keq Keq = ____________ smaller than it was before.

[A] [B]

So we can say: When the temperature is decreased in an endothermic reaction, the equilibrium will shift to the left and the value of Keq will decrease. Changes in Concentration and Keq Now, as you know, changing the concentration of one of the reactants or products will cause

the reaction to shift right or left. But this does not change the value for Keq as long as the temperature remains constant.

How can this be? Let's have a look: Consider the reaction:

A + B C + D Keq = 4.0 The Keq expression is: [C] [D] Keq = ___________ = 4.0 [A] [B] Let's say we quickly add some C to the system at equilibrium. Of course [C] would increase, and temporarily equilibrium would be destroyed. Since [C] is so large, the ratio: [C] [D] ___________ would be > 4.0 (the high [C] makes the numerator large) [A] [B] But, of course, things don't stay like this. When [C] has been increased, the equilibrium will

shift to the left (by LeChatelier's Principle)

Shift to the Left

A + B C + D

In the shift to the left [A] and [B] will get a little larger and the big [C] will get smaller and [D] will get smaller. This will decrease the value of the numerator and increase the value of the denominator

until the ratio: [C] [D] ___________ is again = 4.0 [A] [B]

As long as the temperature is not changed, the equilibrium will always shift just enough to

keep the ratio equal to the value of the equilibrium constant (Keq)!

********************************************************* Effect of Catalysts on the Value of Keq As we saw in the Tutorial on LeChatelier's Principle: Addition of a catalyst speeds up the forward reaction and the reverse reaction by the same amount. Therefore, it does not cause any shift of the equilibrium. Because there is no shift, the value of the Keq will also remain unchanged. Addition of a catalyst to a system at equilibrium does not change the value of Keq! Effect of Pressure or Volume on the Value of Keq Like changes in concentration, changes in the total pressure or volume can cause an

equilibrium to shift left or right. (If there is a different number of moles of gas on each side.) For example: Given the reaction:

N2(g) + 3H2(g) 2NH3(g) Keq = 626

So the ratio: [NH3]2 ___________ = 626 [N2] [H2]3

Let's say the volume of the container is decreased. This increases the total pressure of the

system. Increasing the pressure will increase the concentrations of all three species the same amount. Since there are more moles of gas (N2(g) + 3H2(g)) on the left side, there is more "stuff"

increased in the denominator of the ratio, so the value of the ratio will temporarily go down: So the ratio: [NH3]2 ___________ will be < 626

[N2] [H2]3

There are two substances in the denominator that are increased, so the whole denominator goes up more than the numerator.

But, by LeChatelier's Principle, increased pressure will cause the equilibrium to shift to the right:

N2(g) + 3H2(g) 2NH3(g) 4 moles of gas 2 moles of gas

This will bring [NH3] up, so the numerator of the ratio ( [NH3]2 ) will increase. Thus, the value of the ratio increases again. And guess what? It increases until it just reaches 626 again. The ratio is now equal to Keq and equilibrium has

again been achieved! So the ratio: [NH3]2

___________ is again equal to 626

[N2] [H2]3 So, to summarize: A change in total volume or total pressure does not change the value of the equilibrium constant Keq. The equilibrium will shift to keep the ratio equal to Keq.

There are four main types of Keq questions

Type 1 - Calculating Keq Given Equilibrium Concentrations For the purposes of this Tutorial it is useful to clarify the following in your mind: A chemical system can be thought of as being either: 1. At Equilibrium or 2. Not At Equilibrium (Initial) A system which is not at equilibrium will move spontaneously to a position of being at equilibrium.

In Type 1 calculations, all species in the system are at equilibrium already, so there will be no

changes in concentration. The value for Keq is calculated simply by "plugging" the values for equilibrium

concentrations into the Keq expression and calculating. Whenever a question says something like " ...the equilibrium concentrations of the following

are..." or something to that effect, it is a Type 1 Calculation.

Let's do an example: Given the equilibrium system: PCl5(g) PCl3(g) + Cl

2(g)

The system is analyzed at a certain temperature and the equilibrium concentrations are as follows:

[PCl5] = 0.32 M, [PCl3] = 0.40 M and the [Cl2

] = 0.40 M.

Calculate the Keq for this reaction at the temperature this was carried out. SOLUTION: Step 1 - Use the balanced equation to write the Keq expression: PCl5(g) PCl3(g) + Cl2(g)

so

][

]][[

5

23

PClClPCl

Keq =

Step 2 - "Plug in" the values for the equilibrium concentrations of the species:

]32.0[

]40.0][40.0[=Keq

Step 3 - Calculate the value of Keq .

50.0]32.0[

]40.0][40.0[==Keq

Answer: The Keq = 0.50 for this reaction. Notice the answer is in 2 SD’s like the lowest # of SD’s in the data. Notice that there are no units given in the answer. Even though Keq technically has

units, they are fairly meaningless and they are just dropped. So don't include any units when you state the value for Keq .

************************************************************

A variation of Type 1 problems is when you are given the Keq and all the equilibrium

concentrations except one and you are asked to calculate that one. The solution for this type of problem is simply writing out the Keq expression, filling in what

you know and solving for the unknown. Read through this example:

At 200°C, the Keq for the reaction: N2(g) + 3H2(g) 2NH3(g)

is 625

If the [N2] = 0.030 M, and the [NH3] = 0.12 M, at equilibrium, calculate the equilibrium [H2

].

Solution: All concentrations given are at equilibrium so: Write out the Keq expression:

][][

][

23

2

23

NHNH

Keq =

][][

][

23

2

23

NHNH

Keq =

Plug in what is known:

]030.0[][

]12.0[625 32

2

H=

Cross-multiplying: (625) [H2]3 (0.030) = (0.12)

2

Solving for [H2]

3

)030.0(625

)12.0(][2

32 =H

)1068.7(000768.075.18

0144.0][ 432

−== xH

Take the cube root of both sides: MMxxH 092.0102.91068.7][ 23 4

2 === −− notice 2 sd’s like data

*********************************************************** Type 2 - Given Initial Concentrations of all Species and equilibrium

concentration of one species and asked to calculate the equilibrium concentrations of all species or the Keq (Also called “ICE” problems)

Remember:

A chemical system can be thought of as being either: 1. At Equilibrium or 2. Not At Equilibrium A system which is not at equilibrium will move spontaneously to a position of being at

equilibrium. In this type of problem, we start out with "INITIAL" concentrations of all the species. "Initial" usually means "NOT AT EQUILIBRIUM YET" or “WHAT YOU START WITH”.

We abbreviate Initial Concentration as [I] where “I” stands for “Initial” and not “Iodine”. The [ ]’s stand for “Molar Concentration”

When the system is not at equilibrium, the "reaction will shift" left or right until it reaches

equilibrium. In this type of problem, there will be one species which we will know the concentration of

initially and at equilibrium. We can find the change in the concentration (which we abbreviate as “[C]” where the “C” stands for the words “Change in” and [ ]’s stand for Concentration) of this species and by using mole ratios in the balanced equation, find the changes in concentration “[C]” of the other species. From this we can calculate the equilibrium concentration (which we abbreviate as “[E]”) of all the species.

There is a lot to keep track of here, so this is best done using a little table (called an “ICE”

table) Remember, there are 3 stages: Initial, Change and Equilibrium ( hence the name “ICE”

problem) Here’s a little problem on the next page…. Given the reaction:

N2(g) + 3H2(g) 2NH

3(g)

Some H2 and N2 are added to a container so that initially, the [N2

[H] = 0.32 M and

2

At a certain temperature and pressure, the equilibrium [H] = 0.66 M.

2

] is found to be 0.30 M.

a) Find the equilibrium [N2] and [NH3

].

b) Calculate Keq

at this temperature and pressure.

Let's look at the two "time-frames": INITIALLY or [I] - We are given [N2] and [H2]. Since we are not told anything about NH3,

we assume that initially, [NH3

] = 0.

AT EQULIBRIUM or [E] - We are given [H2

] once equilibrium is reached. We need to find the other two concentrations at equilibrium.

Note that we know [H2] initially, and at equilibrium, so we can easily find the change in or [C] of the [H2

] as the system approaches equilibrium.

We start by making a table as follows: (NOTE: We sometimes call this an (ICE) table.) (Notice that the Species in the top row are always written in the same order as they appear in the balanced

equation. This prevents confusion and minimizes errors when transferring this information.) N2 + 3H2 2NHInitial conc. [I]

3

(change in conc.) [C] Equilibrium conc [E]

Now we fill the table in with all the information we are given in the question. Study this for a

couple of minutes and make sure you're convinced you know exactly where everything goes in the chart and why. Ask if you don't understand at this point!

Some H2 and N2 are added to a container so that initially, the [N2

[H] = 0.32 M and

2

At a certain temperature and pressure, the equilibrium [H] = 0.66 M.

2

] is found to be 0.30 M.

N2 + 3H2 2NHInitial conc. [I]

3 0.32 0.66 0

(change in conc.) [C] Equilibrium conc [E] 0.30

If you look at the chart, you will see that we know [H2] initially and at equilibrium. We can

see that [H2] has decreased. We show this by making [C] negative for [H2

].

To calculate how much it has gone down, we subtract 0.30 from 0.66 ( 0.36). So [C]

(Change in Concentration) for [H2

] = -0.36. We can insert this in the proper place in the table:


3 0.32 0.66 0

(change in conc.) [C] -0.36 Equilibrium conc [E] 0.30

Now to find how the other concentrations have changed, we use the equation and the mole

ratios. To keep things consistent, I like to always put "[C]’s." on top of the equation. (The

concentrations are in the same ratio as the moles) To determine whether a [C] is negative or positive, we use what I call the "Teeter-Totter" rule. If you are on the same side of the teeter-totter as someone going down, you will go down also, and the person on the other side will go up. If you are on the same side of the teeter-totter as someone going up, you will go up also, and the person on the other side will go down.

The " " in the middle of the equation is like the pivot of the teeter-totter. If you don't care for teeter-totters or silly little analogies like this, just remember, if a reaction

shifts right, all reactants go down and all products go up. If it shifts left, all products go down and all reactants go up.

Looking back at our table, we see that [H2

] went down by 0.36 moles/L. ( [C] = -0.36 )

We find [C] for N2

-0.12

like this: x 1/3 -0.36

1 N2(g) + 3 H2(g) 2NH

3(g)

Since [H2] and [N2] are going down, the reaction must be shifting to the right and [NH3

We can also find it by using mole ratios:

] must be going up. Hence it will have a positive [C] (change in concentration)

-0.12 -0.36 x 2/3 +0.24 1 N2(g) + 3 H2(g) 2NH

3(g)

We can now insert these [C]’s into the table:


3 0.32 0.66 0

(change in conc.) [C] -0.12 -0.36 +0.24 Equilibrium conc [E] 0.30

Now we use the changes in concentration ([C]’s) and the initial concentrations to find the equilibrium concentrations of each species. See the table on the next page:


3 0.32 0.66 0

(change in conc.) [C] -0.12 -0.36 +0.24 Equilibrium conc [E] 0.32 - 0.12= 0.20 0.30 0 + 0.24 = 0.24

Now we can answer question "a". The equilibrium [N2] = 0.20 M and [NH3

] = 0.24 M

The "b" part of the question asked us to calculate the value of Keq for this reaction at these conditions.

First we write the expression for Keq : N2(g) + 3 H2(g) 2NH

3(g)

322

23

]][[][

HNNH

Keq =

To calculate Keq , we plug in the values for the equilibrium concentrations of all the

species. These are in the last row of the table above (the [E]’s (Equilibrium conc.))

)'2(117.10)30.0)(20.0(

)24.0(]][[

][3

2

322

23 sSDHN

NHKeq ====

************************************************************* In another variation of Type 2 Problems, we are sometimes given the initial moles when we

have something other than a 1.0 L container. In this case, you must find initial concentrations [I]. by using the familiar formula:

L

molMMolarity =)(

Let's do an example: Consider the equilibrium system: A + 3 B 2 C 0.20 moles of A and 0.60 moles of B are placed in a 2.0 L container. When equilibrium is

reached, the [A] is found to be 0.08 M. Calculate the equilibrium [B] and the equilibrium [C]

ML

molAInitial 10.00.2

20.0][ ==

ML

molBInitial 30.00.2

60.0][ ==

Notice that in this case the equilibrium concentration (not moles!) of A is given. This can go

right in the table under equilibrium concentration [E] of A. A + 3 B 2 C Initial conc.(M=mol/L) [I] 0.20 ÷ 2 = 0.10 0.60 ÷ 2 = 0.30 0 (change in conc.) [C] Equilibrium conc. [E] 0.08

These are now ready to be plugged into the [I] row of the ICE table

We can see that [A] goes from 0.10 M down to 0.08 M (see column "A" in Table) So the (change in conc. or [C]) for "A" is -0.02 Use this and the mole ratios to calculate the [C]’s for B and C. (If [A] goes down, [B] must also go down and [C] must go up (Shifting to the Right))

-0.02 x 3/1 -0.06 A + 3 B 2 C

-0.02 x 2/1 + 0.04 A + 3 B 2 C These can now be inserted into the "[C]" row in the table, and the equilibrium concentrations

can be determined:

A + 3 B 2 C Initial conc.(M=mol/L)

[I] 0.20 ÷ 2 = 0.10 0.60 ÷ 2 = 0.30 0

(change in conc.) [C] -0.02 -0.06 +0.04 Equilibrium conc. [E] 0.08 0.30 - 0.06 = 0.24 0 + 0.04 = 0.04

In one more variation of "Type 2", they may give you initial moles and equilibrium moles (not equilibrium concentration) in something other than a 1.0 L container. In this case, you would have to calculate the [E] (Equilibrium concentration) from the

equilibrium moles and the volume of the container using: L

molM =

Here's an example: Given the equilibrium: X + 2Y 2 Z When 2.0 moles of X and 3.5 moles of Y are placed in a 5.0 L container at 25°C, an

equilibrium is established in which there are 2.5 moles of Z. Calculate [X], [Y] and [Z] at equilibrium. Notice that moles of Z (not [Z]) is given at equilibrium. We can find the Equilibrium [Z]

using the formula: M = moles/ L. This can then be placed in the table and the rest of the calculations can be done:

ML

molZmEquilibriu 50.050.05.2][ ==

X + 2Y 2 Z [I] 2.0 ÷ 5.0 = 0.40 3.5 ÷ 5.0 = 0.70 0 [C] [E] 2.5 ÷ 5.0 = 0.50

X + 2Y 2 Z [I] 2.0 ÷ 5.0 = 0.40 3.5 ÷ 5.0 = 0.70 0 [C] [E] 2.5 ÷ 5.0 = 0.50

Change in concentration [C], for Z must be = +0.50 (since it started at 0 and ended up as 0.50 M) Using mole ratios, we can find the [C]’s for X and Y: (make sure you understand how this is

done!) -0.25 -0.50 +0.50 X + 2Y 2 Z These are then plugged into the [C] row on the table and the rest of the calculations are

performed:

X + 2Y 2 Z [I] 2.0 ÷ 5.0 = 0.40 3.5 ÷ 5.0 = 0.70 0 [C] -0.25 -0.50 +0.50 [E] 0.40 – 0.25 = 0.15 0.70 – 0.50 = 0.20 2.5 ÷ 5.0 = 0.50

We now have the answers to the question: [X] = 0.15 M, [Y] = 0.20 M, and [Z] = 0.50 M at equilibrium. NOTE: If you were ever asked for the Equilibrium Moles of a substance (like X in this

example), you would just take the equilibrium concentration and the litres given and use: moles = M x Litres eg.) moles X = 0.15 M x 5.0 L = 0.75 moles of X

********************************************************** Type 3 - Determining What a Reaction Will Do, Given Initial

Concentrations of all Species and Keq . The ratio of [Products] to [Reactants] with the initial concentrations is called the

Trial Keq. By the "ratio", we mean the initial concentrations plugged in to the Keq expression. For example: For the reaction A(g) + 2B(g) 3C

(g)

2

3

]][[][BA

CTrialKeq =

plug INITIAL concentrations into this expression to get Trial Keq

(NOTE: This is the same expression that is used for the Keq . The only difference is that the Keq has equilibrium concentrations, while the Trial Keq has initial concentrations.)

The value of the Trial Keq (initial concentrations) is then compared with the value for Keq (at equilibrium). If Trial Keq < Keq the reaction will shift to the right. (a higher [products] is needed to bring the ratio up so that it does equal Keq ) If Trial Keq > Keq the reaction will shift to the left. (a lower [products] and a higher [reactants] is needed to bring down the ratio of products to reactants to make it equal the Keq .) If Trial Keq = Keq , the reaction will not shift at all. (It is already at equilibrium and no macroscopic properties, like concentration will change unless the equilibrium is disturbed.)

Let's do an example: For the equilibrium reaction:

CO(g) + H2O(g) CO2(g) + H

2(g)

The value of Keq = 10.0 at a temperature of 600 °C. A reaction mixture is analyzed and found to contain 0.80M CO, 0.050M H2O,

0.50M CO2 and 0.40M H2

.

Determine which direction (left, right or not at all) the reaction will have to shift in order to reach equilibrium.

Here is how we would do this: It doesn't say that the concentrations given are at equilibrium, so we call them initial

concentrations. We write the equilibrium equation and the expression for Keq :

CO(g) + H2O(g) CO2(g) + H

2(g)

]0][{]][[

2

22

HCOHCOKeq =

If we plug in INITIAL CONCENTRATIONS, the value for this is called the Trial Keq :

0.5)050.0)(80.0()40.0)(50.0(

]0][[]][[

2

22 ===HCOHCOTrialKeq

Now in the question, they gave the actual Keq as 10.0 (See the 3rd

line of this question.)

So, in this case the Trial Keq < Keq so this means that the reaction will have to shift to the right in order to reach equilibrium. (In other words, the ratio (5.0) is lower than it has to be at

equilibrium (10.0). In order to make the ratio bigger, the reaction has to shift to the right to give a higher [products] and a lower [reactants].)

Type 4 - Given Initial Concentrations Only and Keq and asked to Calculate

the Equilibrium Concentrations of the Species. This type of problem gives you the initial concentrations but it does NOT give you any

concentrations or moles at equilibrium. It's best to follow through this example. (BEWARE: It is long! I have explained all of the

steps and even gone through the algebra (Yes algebra!) step by step.) Example: At a particular temperature, for the reaction:

H2 + I2 2HI The Keq = 55.6 If the initial [H2] = 0.200 M and [I2

] = 0.200 M, what is the equilibrium [HI]?

The first thing we need to determine is, Which way does it have to shift in order to reach equilibrium? To do this, we look at the value for the ratio:

]][[

][

22

2

IHHI

when the initial values are put in (the Trial Keq) and then we compare with what the ratio has

to be at equilibrium (which is, the value for Keq !) Initially, we have no (0) HI, and [H2] and [I2

] are both 0.200 M, so the Trial Keq is:

0)200.0)(200.0(

)0(]][[

][ 2

22

2

==IH

HI

At equilibrium, this ratio must be equal to the Keq which is 55.6. (Given at beginning of question.) So, obviously, the ratio must go up (from 0 to 55.6) as the system approaches equilibrium. In order for the ratio of product (HI) to reactant (H2 & I2

) to go up, the reaction must shift to the right!

H2 + I2 2HI will shift RIGHT in order to reach equilibrium.

This means [H2] and [I2

] will go down, while [HI] will go up as equilibrium is approached.

OK. Now we've answered the first question, we can continue with the question: The table for type 4 is similar but with one difference. You need the 3 important ones, namely

Initial Concentration [I], ∆ Conc. [C], and Equilibrium Concentration [E] This time, however, we will leave three rows for Equilibrium Concentration. You will see

why as we go through the problem.(I'm going to re-state the question here so you don't have to look back!)

At a particular temperature, for the reaction: H2 + I2 2HI The Keq = 55.6

If the initial [H2] = 0.200 M and [I2

] = 0.200 M, what is the equilibrium [HI]?

H2 + I2 2HI [I] 0.200 0.200 0 [C] [E] [E] [E]

Since you are not given anything at equilibrium, how do you know what [C] is going to be for

any of them? Well, you don't! But what you DO know is that the reaction will shift to the right in order to reach equilibrium.

(Remember we calculated that using a Trial Keq!) So [H2] and [I2

] will go down and [HI] will go up!

H2 + I2 2HI

What we don't know is HOW MUCH these will go up or down! So what we do is let the amount that [H2] goes down = x ( So [C] for H2

is -x) (negative because it is going down!)

We can then determine [C]’s for the other species using mole ratios: -x -x +2x H2 + I2 2HI

(Of course [HI] goes up by 2x because of the coefficient "2" DON’T FORGET THIS!!!!) These can now be put into the table and the equilibrium concentrations can be figured out in

terms of "x":

H2 + I2 2HI

x 2/1

[I] 0.200 0.200 0 [C] -x -x +2x [E] 0.200 - x 0.200 - x 0 + 2x = 2x [E] [E]

So, how does this help us? Well, remember these are equilibrium concentrations and we have

something that relates equilibrium concentrations.... Yes, friends, the Keq expression! We know the value of Keq (55.6 in this case), so we plug the values from the table into the

Keq expression and solve for x! H2 + I2 2HI

6.55]][[

][

22

2

==IH

HIKeq

Plugging in values for equilibrium concentrations from the [E] row of the ICE Table:

6.55)200.0)(200.0(

)2( 2

=−−

=xx

xKeq

This looks an awful lot like a quadratic, but relax! (we don't use those in Chem. 12)

6.55)200.0)(200.0(

)2( 2

=−−

=xx

xKeq

or

6.55)200.0(

)2(2

2

=−

=x

xKeq

We can simplify this by taking the square root of both sides of the equation:

6.55)200.0(

)2(2

2

=− x

x

4565.7)200.0(

2=

− xx

We can now continue with our algebra to solve for "x". By cross multiplying: 2x = 7.4565 (0.200 - x)

Working out the right side: 2x = 1.4913 - 7.4565 x Adding 7.4565 x to both sides: 2x = 1.4913 - 7.4565 x Adding 7.4565 x to both sides: 2x + 7.4565 x = 1.4913 or (2 + 7.4565) x = 1.4913 or 9.4565 x = 1.4913 or

1577.04565.94913.1

==x

So we finally have that x = 0.1577, but what the heck was "x" (It was so long ago, you might not remember!)

"x" was the amount that the [H2

Now to find out the equilibrium concentrations, we plug the value for "x" back into the table: ] went down during the shift.

x = 0.1577

H2 + I2 2HI [I] 0.200 0.200 0 [C] -x -x +2x [E] 0.200 - x 0.200 - x 0 + 2x = 2x [E] 0.200 - 0.1577 0.200 - 0.1577 2(0.1577) [E] 0.0423 0.0423 0.315

Now we can answer the final question: "What is the equilibrium [HI]" We see from the table

that it is 0.315 M. Equilibrium [HI] = 0.315 M Final NOTE: We've done a lot of work here and there is always the possibility of a mistake.

We can actually check to see if our answer is correct. We can do this by plugging the final "Equilibrium Concentrations" into the Keq expression. If we've done everything right, the value should come out very close to the given value of Keq

, which is 55.6 in this case. Let's try it!

Notice our answer is 3SD’s (like our data) but we use more than 3sd’s when working through the algebra. Don’t round much until the very end!

45.55)0423.0()315.0(

]][[][

2

2

22

2

===IH

HIKeq (close enough)

Chemistry 12 - Notes on Unit 2 -...

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