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Transcript of Chemical Equilibrium Some more complicated applications Text 692019 and your message to 37607.
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Chemical Equilibrium
Some more complicated applications
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The ICE chart is a powerful tool for many different equilibrium problems
But you can’t always make a simplifying assumption, and that means that you may need to do a little algebra
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A Quadratic EquationA quadratic equation is a 2nd order
polynomial of the general form:
a x2 + b x + c = 0
Where a, b, and c represent number coefficients and x is the variable
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The Quadratic Formula
a x2 + b x + c = 0
All quadratic equations have a solution for x that is given by:
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Equilibrium & the Quadratic Formula
If you cannot make a simplifying assumption, many times you will end up with a quadratic equation for an equilibrium constant expression.
You can end up with a 3rd, 4th, 5th, etc. order polynomial, but I will not hold you responsible for being able to solve those as there is no simple formula for the solution.
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A sample problem.A mixture of 0.00250 mol H2 (g) and 0.00500
mol of I2 (g) was placed in a 1.00 L stainless steel flask at 430 °C. The equilibrium constant, based on concentration, for the creation of HI from hydrogen and iodine is 54.3 at this temperature.
What are the equilibrium concentrations of all 3 species?
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Determining the concentrationsICE - ICE - BABY - ICE – ICE
The easiest way to solve this problem is by using an ICE chart.
We just need a BALANCED EQUATION
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An ICE Chart
H2 (g) + I2 (g) 2 HI (g)
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An ICE Chart
H2 (g) + I2 (g) 2 HI (g)
I
C
E
0.00250 M 0.00500 M 0 M
-x -x +2x
0.00250 – x 0.00500 – x 2x
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Plug these numbers into the equilibrium constant expression.
I could start by assuming x<<0.00250, it is always worth taking a look at the “easy” solution.
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Are we good to the K-equationA. YesB. No, please talk moreC. I really can’t get past my test grade, so I
can’t be bothered with your stupid problem.
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Assuming x is small…
IF x<<0.00250
6.788x10-4 = 4x2
1.697x10-4 = x2
0.0130 = x
THE ASSUMPTION DOES NOT WORK!
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X=0.0130 (from the solution)
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We’re going to have to use the quadratic formula
54.3(x2-0.00750 x+1.25x10-5) = 4x2
54.3x2 -0.407 x + 6.788x10-4 = 4x2
50.3x2 -0.407 x + 6.788x10-4 = 0
On to the Quadratic Formula
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Using the quadratic formula50.3x2 -0.407 x + 6.788x10-4 = 0
X = 0.005736 OR 0.002356
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There are 2 roots…All 2nd order polynomials have 2 roots, BUT only one will make
sense in the equilibrium problem
x = 0.005736 OR 0.002356
Which is correct?
Look at the ICE chart and it will be clear.
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x = 0.005736 OR 0.002356
H2 (g) + I2 (g) 2 HI (g)
I
C
E
If x = 0.005736, then the equilibrium concentrations of the reactants would be NEGATIVE! This is a physical impossibility.
0.00250 M 0.00500 M 0 M
-x -x +2x
0.00250 – x 0.00500 – x 2x
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SO x = 0.002356
H2 (g) + I2 (g) 2 HI (g)
I
C
E
And you are done!
0.00250 M 0.00500 M 0 M
-0.002356 -0.002356 +2(0.002356)
0.000144 M 0.002644 M 0.00471 M
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`
X IS NOT THE ANSWER
X IS A WAY TO GET TO THE ANSWER
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Another Itty Bitty ProblemCaCO3 (s) will decompose to give CaO (s) and
CO2 (g) at 350°C. A sample of calcium carbonate is sealed in an evacuated 1 L flask and heated to 350 °C. When equilibrium is established, the total pressure in the flask is 0.105 atm. What is Kc and Kp?
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Another Itty Bitty ProblemCaCO3 (s) will decompose to give CaO (s) and
CO2 (g) at 350°C. A sample of calcium carbonate is sealed in an evacuated 1 L flask and heated to 350 °C. When equilibrium is established, the total pressure in the flask is 0.105 atm. What is Kc and Kp?
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As always, we 1st need a balanced equation:
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As always, we 1st need a balanced equation:
CaCO3 (s) CaO (s) + CO2 (g)
Then we can immediately write the equilibrium constant expressions:
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As always, we 1st need a balanced equation:
CaCO3 (s) CaO (s) + CO2 (g)
Then we can immediately write the equilibrium constant expressions:
Kc = [CO2]
Kp = PCO2
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Another Itty Bitty ProblemCaCO3 (s) will decompose to give CaO (s) and
CO2 (g) at 350°C. A sample of calcium carbonate is sealed in an evacuated 1 L flask and heated to 350 °C. When equilibrium is established, the total pressure in the flask is 0.105 atm. What is Kc and Kp?
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Kc = [CO2]
[CO2] = moles CO2/L
How do we determine the # of moles?
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All of the pressure must be due to the carbon dioxide.
As a gas, carbon dioxide should obey the ideal gas law.
P V = n R T
And we know P, V, R, and T!!
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P V = n R T
And we know P, V, R, and T!!
In fact, we could calculate moles/L directly:
PV = n R T
Either way will work.
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2.05x10-3 M =We can use this to directly calculate Kc
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Kc = [CO2]
Kc = 2.05x10-3
And we’re done!!! (Boring when there’s no exponents, isn’t it? )
What about Kp?
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Kp = PCO2
Kp = 0.105
And we’re essentially done!
Now, that may have seemed simple, but it does point out something interesting about the relationship between Kc
and Kp
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Kp is NOT A PRESSURE
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Kc vs Kp
Kc depends on concentration (moles/L)
Kp depends on pressure (atm)
For gases, pressure and concentration are directly related.
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Kc vs Kp
P V = nRT
=M
The only difference between P and M is (RT)
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Kc vs Kp
Kc = [CO2] =
Or, alternatively
Kp = Kc (RT)
This can be generalized.
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Kc vs Kp - in generalConsider a general reaction:
x A (g) + y B (g) z C (g)
And I can quickly write Kc and Kp:
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Using the Ideal Gas Law…Kc =
If I collect all the (1/RT) terms separately
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If I collect all the (1/RT) terms separately
KC = KP
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Simplifying…KC = KP
Z-Y-X is just the change in the stoichiometry of the reaction:
Z moles of products – Total moles of reactants (Y+X)
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Simplifying…KC = Kp
Δn = total moles of product gas – total moles of reactant gas
This is the general relationship between Kp and Kc for all gas phase reactions.
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Another Kp vs Kc problem
2 SO3 (g) 2 SO2 (g) + O2 (g)
The above reaction has a Kp value of 1.8x10-5 at 360°C. What is Kc for the reaction?
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If we recall that:
KC = Kp
The solution is simple. Δn = 3 moles product gas – 2 moles reactant gas
So:KC = 1.8x10-5
Kc = 3.46x10-7