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Question 14.11: Finding Equilibrium Partial Pressures when we are given the ...

Finding Equilibrium Partial Pressures when we are given the Equilibrium Constant and Initial Partial Pressures
Consider the reaction.
I_{2}(g) + Cl_{2}(g) \xrightleftharpoons[]{} 2 ICl(g)     K_{p} = 81.9 (at 25 °C)
A reaction mixture at 25 °C initially contains P_{I_{2}} = 0.100 atm, P_{Cl_{2}} = 0.100 atm, and P_{ICl} = 0.100 atm. Find the equilibrium partial pressures of I_{2}, Cl_{2}, and ICl at this temperature.

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Follow the procedure outlined in Examples 14.5 and 14.6 (using the pressures in place of the concentrations) to solve the problem.

1. Using the balanced equation as a guide, prepare a table with the known initial partial pressures of the reactants and products. I_{2}(g) + Cl_{2}(g) \xrightleftharpoons[]{} 2 ICl(g)

P_{i_{2}}  (atm) P_{Cl_{2}}  (atm) P_{iCl_{2}}  (atm)
Initial 0.100 0.100 0.100
Change
Equil
2. Use the initial partial pressures to calculate the reaction quotient (Q). Compare Q to K and predict the direction inwhich the reaction will proceed. Q_{p} = \frac{P^{2}_{ICl}}{P_{I_{2}}P_{Cl_{2}}}= \frac{0.100^{2}}{(0.100)(0.100)} = 1
K_{p} = 81.9 (given)
Q < K; therefore, the reaction will proceed to the right.
3. Represent the change in the partial pressure of one of the reactants or products with the variable x. Define the changes in the partial pressures of the other reactants or products in terms of x. I_{2}(g) + Cl_{2}(g) \xrightleftharpoons[]{} 2 ICl(g)

P_{i_{2}}  (atm) P_{Cl_{2}}  (atm) P_{iCl_{2}}  (atm)
Initial 0.100 0.100 0.100
Change -x -x +2x
Equil
4. Sum each column for each reactant and product to deter mine the equilibrium partial pressures in terms of the ini-tial partial pressures and the variable x. I_{2}(g) + Cl_{2}(g) \xrightleftharpoons[]{} 2 ICl(g)

P_{i_{2}}  (atm) P_{Cl_{2}}  (atm) P_{iCl_{2}}  (atm)
Initial 0.100 0.100 0.100
Change -x -x +2x
Equil 0.100-x 0.100-x 0.100+2x
5. Substitute the expressions for the equilibrium partial pressures (from step 4) into the expression for the equi-librium constant. Use the given value of the equilibrium constant to solve the expression for the variable x. K_{p} = \frac{P^{2}_{ICl}}{P_{I_{2}}P_{Cl_{2}}}= \frac{(0.100 + 2x)^{2}}{(0.100 – x)(0.100 – x)}
81.9 = \frac{(0.100 + 2x)^{2}}{(0.100 – x)^{2}} (perfect square)
\sqrt{81.9} = \frac{(0.100 + 2x)}{(0.100 – x)}
\sqrt{81.9}(0.100 – x) = 0.100 + 2x
\sqrt{81.9}(0.100) – \sqrt{81.9}x = 0.100 + 2x
\sqrt{81.9}(0.100) – 0.100 = 2x + \sqrt{81.9}x
0.805 = 11.05x
x = 0.0729
6. Substitute x into the expressions for the equilibrium par tial pressures of the reactants and products (from step 4)and calculate the partial pressures. P_{I_{2}} = 0.100 – 0.0729 = 0.027 atm
P_{Cl_{2}} = 0.100 – 0.0729 = 0.027 atm
P_{ICl} = 0.100 + 2(0.0729) = 0.246 atm
7. Check your answer by substituting the computed equilibri um partial pressures into the equilibrium expression. The calculated value of K should match the given value of K. K_{p} = \frac{P^{2}_{ICl}}{P_{I_{2}}P_{Cl_{2}}}= \frac{0.246^{2}}{(0.027)(0.027)}
= 83
Since the calculated value of K_{p} matches the given value(within the uncertainty indicated by the significant figures),the answer is valid.

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