Energy comes in many forms:

1. kinetic energy is the energy of motion.2. potential energy is the energy stored in

stressed objects, like a bent bow or compressed spring, or water at the top of a waterfall.

3. chemical energy is the energy stored in chemical bonds, which is released when a chemical reaction, like burning, occurs.

4. nuclear energy is the energy stored within the nucleus of an atom:• Fission• Fusion

Heat

• Heat is thermal energy. • Heat is the product of motion. • All matter has heat in it, since all

matter is made up of particles which move.

• Something has no heat only if all motion has stopped inside it. This point is called absolute zero.

• Absolute zero is the lowest temperature that can be reached:– it is equal to - 273 ° Celsius– it is also equal to 0 Kelvins (K)

http://www.answers.com/topic/thermally-agitated-molecule-gif

Heat

• It is not possible to determine the amount of heat an object has, since it is tied up with the various types of energy.

• Heat always flows from a warmer to a colder object. It continues to flow until both objects are at the same temperature.

• Different masses or volumes of a substance require different amounts of heat to produce a similar rise in temperature.

Heat

• Different substances of the same mass require different amounts of heat to produce a similar rise in temperature: – Given equal inputs of heat energy, antifreeze will

rise much more rapidly in temperature than water will.

– Put another way, if water and antifreeze are heated, water requires more heat energy per unit mass to produce a similar temperature rise.

– Put a third way, in equal masses of water and antifreeze at the same temperature, the water contains more heat.

Specific Heat Capacity• Also called Specific Heat, or

Heat Capacity.• Is the amount of heat required

to heat 1 g or 1 kg of a substance by 1°C (or 1 K)

• Measured in kiloJoules per kilogram degree Celsius (kJ/kg·C).

• All substances have their own specific heat capacity; it is a characteristic physical property.

Measuring Changes in Heat• Involves 3 things:

1. Heat Capacity – is a constant for each substance or phase of a substance.

2. Mass – the more mass, the more heat.

3. Temperature Change – can be positive or negative.

Measuring Changes in HeatAmount of = Specific x Mass x

Change in Heat Heat

Temperature

Q = C x M x ΔT

Q = C x M x (Tf – Ti)

Q = C x M x ΔT

• This is an equation of four variables. If you have three of the variables you should be able to calculate the fourth. Remember a few things, though: 1. Make sure the units are correct. Include metric

conversion. 2. Heat can go into or come out of a substance.

– If heat goes in temperature goes up the change is positive.

– If heat comes out temperature goes down the change is negative.

3. Use the specific heat table for the C values (see handout).

Q = C x M x ΔT

1. Calculate the heat required to change the temperature of 1.5 kg water from 40.0°C to 80.0°C.

• Since Q = C x M x (Tf – Ti) and the substance is water, therefore

• C = 4.18 kJ/kgC• M = 1.5 kg• Tf = 80.0 C• Ti = 40.0 C• Q = (4.18 kJ/kgC)(1.5 kg)(80.0 C - 40.0 C)• = 250 kJ (2 significant digits)

Q = C x M x ΔT

2.Calculate the heat required to change the temperature of 0.300 kg sucrose from 80.0°C to 19.0°C.

• C = 1.25 kJ/kgC• M = 0.300 kg• Tf = 19.0 C• Ti = 80.0 C• Q = (1.25 kJ/kgC)(0.300 kg)(19.0 C - 80.0 C)• = -22.9 kJ

(a negative value, since the temperature dropped)

Q = C x M x ΔT

3. Calculate the heat required to increase the temperature of 650 g copper by 12.6°C.

•

• C = 0.385 kJ/kgC• M = 650 g x (1 kg / 1000 g) = 0.65 kg • ΔT = 12.6 C (only temp change is given)

• Q = (0.385 kJ/kgC)(0.65 kg)(12.6 C)• = 3.2 kJ (2 significant digits)

ΔT = Q C x M

4.Calculate the temperature change that occurs when 2500 kJ of heat is added to 4.45 kg of asbestos.

• C = 0.820 kJ/kgC• M = 4.45 kg• Q = 2500 kJ

• ΔT = 2500 kJ (0.820 kJ/kgC)(4.45 kg)

• = 690 C

ΔT = Q C x M

5. Calculate the temperature change that occurs when -652 J of heat is added to 241 g of mercury.

• C = 0.140 kJ/kgC• M = 241 g x (1kg / 1000 g) = 0.241 kg• Q = -652 J x (1 kJ / 1000 J) = -0.652 kJ

• ΔT = -0.652 kJ(0.140 kJ/kgC)(0.241 kg)

• = -19.3 C (the temperature dropped)

M = Q C x ΔT

6.Calculate the mass of wood that would undergo a temperature change of 26.5°C when 1360 kJ of heat is added.

• C = 1.76 kJ/kgC• ΔT = 26.5 C• Q = 1360 kJ• M = 1360 kJ

(1.760 kJ/kgC)(26.5 C)• = 29.2 kg

Phase Changes

0

2 0

4 0

6 0

8 0

1 0 0

1 2 0

1 4 0

-2 0

-4 0

-6 0

H ea tin g C u rv e o f W a te r

L IQU ID

GAS

MELT ING

FREEZ ING

CONDENSING

BO IL ING

Phase Changes

• Heat goes into the breaking of physical bonds between the molecules of the substance, and putting distance between the molecules:

– When a substance melts it goes from a rigid, hard solid to a liquid. Bonds are broken so that molecules have the freedom of movement.

– When a liquid becomes a gas the molecules get much further apart.

Phase Changes

• Heat of Fusion (Hf) - This is the heat required to melt or freeze a substance. Expressed in kilojoules per kilogram or kilojoules per mole. If a substance is freezing the Hf value is expressed as a negative.

• Heat of Vapourization (Hv) - This is the heat required to vapourize or boil a substance. Expressed in the same units as heat of fusion. If a substance is condensing the Hv value is expressed as a negative.

Calculating the Energy of Phase Changes

• Heat Required to Melt/Freeze =  Heat of Fusion   x   Mass of a Substance Substance                   

• Q = HfM (for water Hf is 334 kJ/kg)

• Heat Required to Boil/Condense   =   Heat of x   Mass ofa Substance Vapourization   Substance

 • Q = HvM (for water Hv is 2260 kJ/kg)

1. Calculate the heat required to melt 1.5 kg ice

• Since the phase change is melting the formula is

Q = Hf M

• where Hf = 334 kJ/kg

• and M = 1.5 kg

• Q = (336 kJ/kg)(1.5 kg)

= 5.0 x 102 kJ

2. Calculate the heat required to freeze 4.50 x 102 kg water.

• Since the phase change is freezing the formula is

Q = Hf M

• where Hf = - 334 kJ/kg (heat is taken out)

• and M = 4.50 x 102 kg

• Q = (- 336 kJ/kg)(4.50 x 102 kg)

= - 1.50 x 105 kJ

3. Calculate the heat required to boil 245 g water.

• Since the phase change is boiling the formula is

Q = Hv M

• where Hv = 2260 kJ/kg and M = 245 g x (1 kg/1000 g) = 0.245 kg

• Q = (2260 kJ/kg)(0.245 kg)

= 554 kJ

4. Calculate the heat required to condense 1.2 t steam.

• Since the phase change is condensing the formula is

Q = Hv M

• where Hv = - 2260 kJ/kg and M = 1.2 t x (1000 kg/ 1 t) = 1200 kg

• Q = (- 2260 kJ/kg)(1200 kg)

= 2.7 x 106 kJ

5. Calculate the heat required to melt 162 mol ice.

• Since the phase change is melting the formula is

Q = Hf M• where Hf = 334 kJ/kg • and M = is not given, but is 162 mol• The molar mass of water is 18.02 g/mol (from H2O) and• Mass = Molar mass x Number of moles

= (18.02 g/mol)(162 mol) = 2920 g x (1 kg / 1000 g) = 2.92 kg

• Q = (334 kJ/kg)(2.92 kg) = 975 kJ

Combined Heat Problems

0

2 0

4 0

6 0

8 0

1 0 0

1 2 0

1 4 0

-2 0

-4 0

-6 0

H ea tin g C u rv e o f W a te r

L IQU ID

GAS

MELT ING

FREEZ ING

CONDENSING

BO IL ING

how much heat is required to take 2.00 kg of ice at -20.0°C and turn it into steam at +110.0°C ?

• 5 steps are required:– heat ice from -20.00°C to 0°C– melt ice– heat water from 0°C to 100°C– boil water– heat steam from 100°C to 110.0°C

• any change in temperature uses the formula Q = C M ΔT

• phase changes use the formula Q = HfM or HvM

how much heat is required to take 2.00 kg of ice at -20.0°C and turn it into steam at +110.0°C ?

• heat ice from -20.00°C to 0°CQ = C M ΔT

= (2.06 kJ/kg·°C)(2.00 kg)(0°C – (-20.0°C))

= 82.4 kJ• melt ice

Q = Hf M= (334 kJ/kg)(2.00 kg)= 668 kJ

• heat water from 0°C to 100°CQ = C M ΔT

= (4.18 kJ/kg·°C)(2.00 kg)(100°C – 0°C)= 836 kJ

• boil waterQ = Hv M= (2260 kJ/kg)(2.00 kg)= 4520 kJ

• heat water from 100°C to 110°CQ = C M ΔT= (1.86 kJ/kg·°C)(2.00 kg)(110°C – 100°C)= 37.2 kJ

• Total = 84.2 + 668 + 836 +4520 kJ + 37.2 kJ

= 6150 kJ

Enthalpy

• refers to the energy change in chemical reactions.

• the symbol for enthalpy is H.• it is impossible to determine the heat

or energy contained in an object or chemical, but we can determine the change in energy in a chemical reaction.

• change in enthalpy is ΔH

Enthalpy Change

• reactions that give off energy are exothermic; their ΔH is negative.

• reactions that take in energy are endothermic; their ΔH is positive.

Heat of Formation

• is the enthalpy change when a compound is formed from its elements.

Heat of FormationElements Heat of reaction

(kJ/mol of product) H2 (g) + 1/2 O2 (g) H2O(g) - 241.8H2 (g) + 1/2 O2 (g) H2O(l) - 285.8S(s) + O2 (g) SO2 (g) - 296.8H2 (g) + S(s) + 2 O2 (g) H2SO4 (l) - 812S(s) + 3/2 O2 (g) SO3 (g) - 395.71/2 N2 (g) + 1/2 O2 (g) NO(g) + 90.371/2 N2 (g) + O2 (g) NO2 (g) + 33.851/2 N2 (g) + 3/2 H2 (g) NH3 (g) - 46.19C(s) + 1/2 O2 (g) CO(g) - 110.5C(s) + O2 (g) CO2 (g) - 393.5C(s) + 2 H2 (g) CH4 (g) - 74.862 C(s) + 3 H2 (g) C2H6 (g) - 83.8 3 C(s) + 4 H2 (g) C3H8 (g) - 104

Using Heats of Formation

1. C (s) + 1/2 O2 (g) CO(g)

Δ H = - 110.5 kJ

2. CO(g) + 1/2 O2 (g) CO2 (g) ΔH = - 283.0 kJ

3. C (s) + O2 (g) CO2 (g) Δ H = - 393.5 kJ

Using Heats of Formation

• C (s) + 1/2 O2 (g) CO(g) Δ H = - 110.5 kJ

• CO(g) + 1/2 O2 (g) CO2 (g)

ΔH = - 284.0 kJ

• C (s) + O2 (g) CO2 (g) Δ H = - 393.5 kJ

Hess’ Law• Hess's law of constant heat

summation:

• The enthalpy change for any reaction depends only on the products and reactants and is independent of the pathway or the number of steps between the reactant and product.

2 Al (s)   +   3 CuO (s)     3 Cu(s)   +   Al2O3

(s) • Identify the compounds in the equation; find heat of

formation equations from the table that contain them (don’t worry about the elements. They take care of themselves):

• 2 Al (s)   +   1½ O2 (g)     Al2O3 (s)          ΔH = - 1676.0 kJ/mol • Cu (s)   +   ½ O2 (g)     CuO (s)               ΔH = - 155.0 kJ/mol

• Flip the CuO equation to make the compound a reactant as well:

• Cu (s)   +   ½ O2 (g)     CuO (s)               ΔH = - 155.0 kJ/mol

• Becomes

• CuO (s)     Cu (s)   +   ½ O2 (g)               ΔH = + 155.0 kJ/mol

2 Al (s)   +   3 CuO (s)     3 Cu(s)   +   Al2O3

(s)

• Multiply the CuO equation by 3 to have the same number of molecules as in the original equation. The enthalpy change is multiplied by the same factor:

• 2 Al (s)   +   1½ O2 (g)     Al2O3 (s)     ΔH = - 1676.0 kJ/mol

• 3 x      CuO (s)     Cu (s)   +   ½ O2 (g)           ΔH = 3 x (+155.0

kJ/mol)

• This gives• • 2 Al (s)   +   1½ O2 (g)     Al2O3 (s)    ΔH = - 1676.0 kJ/mol • 3 CuO (s)    3 Cu (s)   +  1½ O2 (g)   ΔH = + 465.0 kJ/mol

2 Al (s)   +   3 CuO (s)     3 Cu(s)   +   Al2O3

(s)• Add the two equations together. You treat the equations

just like you would in math; add the material to the left of the arrow together and the material on the right of the arrow together. The enthalpy changes are also added:

• 2 Al (s)   +   1½ O2 (g)     Al2O3 (s)                ΔH = - 1676.0 kJ/mol • 3 CuO (s)    3 Cu (s)   +  1½ O2 (g)               ΔH = + 465.0 kJ/mol• 2 Al (s) +  1½ O2 (g) +  3 CuO (s)  Al2O3 (s) + 3 Cu (s) +  1½ O2(g)

•   ΔH = - 1211 kJ

• Cancel out like terms:• • 2 Al (s) +  3 CuO (s)  Al2O3 (s) + 3 Cu (s)     ΔH = - 1211 kJ

• You know you are right if the net equation you end up with is the same as the original equation you started with.

2 Al (s)   +   3 CuO (s)     3 Cu(s)   +   Al2O3

(s)• WHAT YOU ACTUALLY NEED TO SHOW:

• 2 Al (s)   +   1½ O2 (g)     Al2O3 (s)       ΔH = - 1676.0 kJ/mol • 3 x 3 CuO (s) 3 Cu (s)  + (3)½ O2 (g)   ΔH = 3 x (+155.0

kJ/mol)

• 2 Al (s) +  3 CuO (s)  Al2O3 (s) + 3 Cu (s)   ΔH = - 1211 kJ

2 C2H6 (g)   +   7 O2 (g)     4 CO2 (g)   +   6 H2O

(l)

• 2 x      C2H6 (g)     2 C (s)   +   3 H2 (g)           

ΔH = 2 x(+ 84.0 kJ/mol)

• 4 x      C (s)   +   O2 (g)     CO2 (g)                 

ΔH = 4 x (- 394.0 kJ/mol)

• 6 x      H2 (g)   +   ½ O2 (g)     H2O (l)            

ΔH = 6 x (- 286.0 kJ/mol)

• 2 C2H6 (g)   +   7 O2 (g)     4 CO2 (g)   +   6 H2O (l)

ΔH = - 3124 kJ

SiO2 (s)   +   C (s)     CO2 (g)   +   Si (s) Where the heat of formation of

SiO2 (s)   =   - 861 kJ/mol

• Heat of formation is formation of a compound from its elements. SiO2 is made of silicon (Si (s)) and oxygen (O2

(g)):

• 1x SiO2 (s) Si (s) + O2 (g)

ΔH = +861 kJ/mol

• 1x C (s) + O2 (g) CO2 (g)

Δ H = - 394 kJ/mol

• SiO2 (s)   +   C (s)     CO2 (g)   +   Si (s)

Δ H = + 467 kJ

Enthalpy and Entropy

• Enthalpy is one of the driving forces in the universe; reactions that release energy (negative ΔH) are favoured.

• Endothermic reactions do occur spontaneously; that is because of a second force, Entropy.

• Entropy is disorder• The more disorder, the more entropy.

Examples of Increasing Entropy• tossed salad

Examples of Increasing Entropy• Broken glass

Examples of Increasing Entropy• Melting ice

Examples of Increasing Entropy• Bedroom: Before

Examples of Increasing Entropy• Bedroom: After

Recognizing Entropy in Reactions:• Phase changes:

– solids have lowest entropy– gases have highest entropy

– C12H22O11 (s) C12H22O11 (l)

– increasing entropy

– C2H5OH (g) C2H5OH (l)

– decreasing entropy

Recognizing Entropy in Reactions:

• Mixtures have higher entropy than pure substances:– C12H22O11 (s) C12H22O11 (aq)

– increasing entropy

– Na1+(aq) + Cl1-

(aq) NaCl (s)

– decreasing entropy

Recognizing Entropy in Reactions:

• Side of reaction equation with more particles has higher entropy:

– 2 SO2 (g) + O2(g) 2 SO3 (g)

– decreasing entropy

– C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)

– increasing entropy

Spontaneous Processes

• Spontaneous processes are those that can proceed without any outside intervention.

• The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously

Spontaneous Processes

Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.

Spontaneous Processes• Processes that are spontaneous at one

temperature may be nonspontaneous at other temperatures.

• Above 0C it is spontaneous for ice to melt.• Below 0C the reverse process is spontaneous.

Combining Enthalpy and Entropy• If enthalpy and entropy are both

forces that affect the spontaneity of reactions how can we determine whether a reaction will be spontaneous or not ?

• GIBB’S FREE ENERGY

GIBB’S FREE ENERGY

• Gibb’s free energy (ΔG) combines enthalpy and entropy terms to tell if there is extra energy left over to cause a reaction to happen.

• ΔG = ΔH - T ΔS

• where T is the temperature in Kelvin (always positive)

ΔG = ΔH - T ΔS• a reaction is spontaneous if the ΔG function is

negative. That means there are 4 possibilities of ΔH and ΔS:

• ΔH ΔS Effect on the Reaction

• positive positive ΔG will be negative only if TΔS is very large; this reaction will be spontaneous only at high temperature.

• positive negative ΔG cannot be negative under any

conditions. This reaction will not proceed spontaneously.

ΔG = ΔH - T ΔS

• ΔH ΔS Effect on the Reaction

• negative positive ΔG will be negative under all conditions. This reaction will proceed spontaneously at all temperatures.

• negative negative ΔG will be negative only if TΔS is very small; this reaction will be spontaneous

only at low temperature.

1. H2O (l)     H2O (s)            ΔH =   - 6 kJ

• ΔH is negative. The reaction is exothermic.

• ΔS is negative. This is a phase change and the solid is more ordered than the liquid, so entropy is decreasing.

• ΔG   =   ΔH   -   TΔS

• this reaction will be spontaneous only if TΔS is as small as possible, so as to be smaller than the enthalpy change.

• This reaction will only proceed at low temperatures.

2. CaCO3 (s)   +   181 kJ CaO (s)   +   CO2 (g)

• ΔH is positive. The energy term is on the reactant side, which means it is going into the reaction, so it is endothermic.

• ΔS is positive. This is a decomposition reaction producing more molecules and one product is a gas.

• ΔG   =   ΔH   -   TΔS

• this reaction will be spontaneous only if TΔS is as large as possible, so as to be bigger than the enthalpy change.

• This reaction will only proceed at high temperatures.

3. CS2 (g)   +   3 O2 (g) CO2 (g)   +   2 SO2 (g)

ΔH =   + 1110 kJ• ΔH is positive. The reaction is endothermic.

• ΔS is negative. All the substances are gases, but the reactants have more particles than the products.

• ΔG   =   ΔH   -   TΔS

• this reaction is not spontaneous at any temperature because Gibbs Free Energy cannot be made to be positive.

4. NaCl (s) Na1+ (aq)   +   Cl1- (aq)

+   energy • ΔH is negative. The reaction is exothermic,

since it appears on the product side of the equation.

• ΔS is positive. Since the salt is dissolving the entropy is increasing.

• ΔG   =   ΔH   -   TΔS • since both factors are favourable for a

spontaneous reaction this reaction will be spontaneous at any temperature.

Kinetics

• Free energy tells us if a reaction is likely to be spontaneous, but tells us nothing about how fast it will be.

• Kinetics studies the rate at which a chemical process occurs.

• Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).

The Collision Model

• In a chemical reaction, bonds are broken and new bonds are formed.

• Molecules can only react if they collide with each other.

The Collision Model

Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

Activation Energy• In other words, there is a minimum amount of

energy required for reaction: the activation energy, Ea.

• Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

Factors That Affect Reaction Rates

• Physical State of the Reactants– In order to react, molecules must come in

contact with each other.– The more homogeneous the mixture of

reactants, the faster the molecules can react.

– nature of the reactants - the more complex the reaction, the slower the rate. • Reactions that only involve a change in charge

will proceed more rapidly than a reaction that involves breaking and reforming chemical bonds

Factors That Affect Reaction Rates

• Concentration of Reactants– As the concentration of reactants

increases, so does the likelihood that reactant molecules will collide.

Factors that Affect Reaction Rates• Surface Area

– the more surface area the more molecules exposed to react (more collisions).

– Gases react more quickly than solids

– powders react more quickly than larger lumps.

Factors That Affect Reaction Rates

• Temperature– At higher temperatures, reactant

molecules have more kinetic energy, move faster, and collide more often and with greater energy.

– The minimum energy needed for the reaction to occur is threshold energy.

Effect of Temperature on Rate

• Temperature is defined as a measure of the average kinetic energy of the molecules in a sample.

• At any temperature there is a wide distribution of kinetic energies.

Effect of Temperature on Rate

• As the temperature increases, the curve flattens and broadens.

• Thus at higher temperatures, a larger population of molecules has higher energy.

Effect of Temperature on Rate• If the dotted line represents the activation

energy, as the temperature increases, so does the fraction of molecules that can overcome the threshold energy barrier.

• As a result, the reaction rate increases.

Factors That Affect Reaction Rates

• Presence of a Catalyst– Catalysts speed up reactions by

changing the mechanism of the reaction.

– Catalysts are not consumed during the course of the reaction.

Catalysts• Catalysts increase the rate of a reaction by

decreasing the activation energy of the reaction.

• Catalysts change the mechanism by which the process occurs.

Catalysts

One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.

Catalysts

• A second mechanism is to become part of the activated complex.

• Enzymes perform this function.

Enzymes• Enzymes are catalysts in

biological systems.• The substrate fits into the

active site of the enzyme much like a key fits into a lock.

Inhibitors

• are the opposite of catalysts; they increase the activation energy and reduce the rate of a chemical reaction.

• they often act by forming a complex with one of the reactants, preventing it from having successful collisions with other reactants.

Potential Energy DiagramsIt is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile.

Potential Energy Diagrams

• It shows the energy of the reactants and products (and, therefore, E).

• The high point on the diagram is the transition state.

• The species present at the transition state is called the activated complex.

• The energy gap between the reactants and the activated complex is the activation energy barrier.

Drawing Potential Energy Diagrams• reactant energy = 150 kJ• threshold energy = 400 kJ• product energy = 250 kJ• activation energy =• ΔH = • exo- or endo-themic ?

Drawing Potential Energy Diagrams• reactant energy = 450

kJ• threshold energy = • product energy = • activation energy = 50

kJ• ΔH = - 200 kJ• exo- or endo-themic ?

Drawing Potential Energy Diagrams• reactant energy = • threshold energy = • product energy = 350

kJ• activation energy =

250 kJ• ΔH = +150 kJ• exo- or endo-themic ?

Drawing Potential Energy Diagrams

• reactant energy = • threshold energy = 300 kJ• product energy = • activation energy = 200

kJ• ΔH = -50 kJ• exo- or endo-themic ?

Reaction Rates

Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

Reaction Rates

One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration.

Concentration and Rate

Comparing Experiments 1 and 2, when [NH41+]

doubles, the initial rate doubles.

NH41+

(aq) + NO21-

(aq) N2 (g) + 2 H2O (l)

Concentration and Rate

Likewise, comparing Experiments 5 and 6, when [NO2

1-] doubles, the initial rate doubles.

NH41+

(aq) + NO21-

(aq) N2 (g) + 2 H2O (l)

Reaction Order

• is a description of the relationship between the reactants and their effect on the rate of a reaction

Reaction Order

• Zeroth Order – a change in concentration of a reactant has no effect on the rate of the reaction.

• In the rate equation the concentration is raised to the zero power.

Reaction Order

• First Order – a doubling in concentration of a reactant doubles the rate of the reaction.

• In the rate equation the concentration is raised to the first power.

Reaction Order

• Second Order – a doubling in concentration of a reactant causes the rate of the reaction to increase 4 times.

• In the rate equation the concentration is raised to the second power.

Rate Laws

• The overall reaction order can be found by adding the exponents on the reactants in the rate law.

Concentration and Rate

When [NH41+] doubles, the rate doubles.

When [NO21-] doubles, the rate doubles.

NH41+

(aq) + NO21-

(aq) N2 (g) + 2 H2O (l)

Writing Rate Law Equations• This means

Rate [NH41+]

Rate [NO21-]

Rate [NH1+] [NO21-]

or

Rate = k [NH41+] [NO2

1-]• This equation is called the rate law,

and k is the rate constant.• This reaction is second order overall.

Writing Rate Law Equations

• 2 ICl + H2 I2 + 2 HCl

• if [ICl] doubles, the rate doubles (first order)

• if [H2] quadruples, rate quadruples (first order)

• Rate = k [ICl]1 [H2]1

• Rate is second order overall

Experiment

[ICl] (mol/L)

[H2] (mol/L)

Initial Rate(mol/(L·s))

1 0.10 0.01 0.002

2 0.20 0.01 0.004

3 0.10 0.04 0.008

Calculating Rate Law Constant

• Rate = k [ICl]1 [H2]1

• k = Rate . [ICl][H2]

• k = (0.002 mol/(L·s)) (0.10 mol/L)(0.01 mol/L)

• k = 2

• Rate = 2 [ICl][H2]

Experiment [ICl] (mol/L)

[H2] (mol/L) Initial Rate(mol/(L·s))

1 0.10 0.01 0.002

2 0.20 0.01 0.004

3 0.10 0.04 0.008

Writing Rate Law Equations

• A + B C

• if [A] doubles, the rate quadruples (second order)

• if [B] doubles, rate unchanged (zeroth order)• Rate = k [A]2 [B]0 • Rate is second order overall

Experiment

[A] (mol/L) [B] (mol/L) Initial Rate(mol/(L·s))

1 0.002 0.05 2.0 x 10-5

2 0.004 0.05 8.0 x 10-5

3 0.002 0.10 2.0 x 10-5

Calculating Rate Law Constant

• Rate = k [A]2 [B]0 • k = Rate .

[A]2

• k = (2.0 x 10-5 mol/(L·s)) (0.002 mol/L)2

• k = 5• Rate = 5 [A]2

Experiment

[A] (mol/L) [B] (mol/L) Initial Rate(mol/(L·s))

1 0.002 0.05 2.0 x 10-5

2 0.004 0.05 8.0 x 10-5

3 0.002 0.10 2.0 x 10-5

Writing Rate Law Equations

• H2O2 (aq) + 2 HI (aq) 2 H2O (l) + I2 (aq)

• if [H2O2] doubles, the rate doubles (first order)

• if [HI] doubles, rate doubles (first order)

• Rate = k [H2O2]1 [HI]1

• Rate is second order overall

Experiment

[H2O2] (mol/L)

[HI] (mol/L)

Initial Rate(mol/(L·s))

1 0.05 0.05 0.002

2 0.05 0.10 0.004

3 0.10 0.05 0.004

Writing Rate Law Equations

• Rate = k [H2O2]1 [HI]1

• k = Rate . [H2O2][HI]

• k = (0.002 mol/(L·s)) (0.05 mol/L)(0.05 mol/L)

• k = 0.8• Rate = 0.8 [H2O2][HI]

Experiment [H2O2] (mol/L)

[HI] (mol/L) Initial Rate(mol/(L·s))

1 0.05 0.05 0.002

2 0.05 0.10 0.004

3 0.10 0.05 0.004