Question 15.11: Finding Equilibrium Partial Pressures When You Are Given the...
Finding Equilibrium Partial Pressures When You Are Given the Equilibrium Constant and Initial Partial Pressure
Consider the reaction:
I_2(g) + Cl_2(g) \rightleftharpoons 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.
Follow the procedure used in Examples 15.5 and 15.6 (using partial pressures in place of concentrations) to solve the problem.
1. Using the balanced equation as a guide, prepare a table showing the known initial partial pressures of the reactants and products.
I_2(g) + Cl_2(g) \rightleftharpoons 2 \ ICl(g)
| P_{I_2} (atm) | P_{Cl_2} (atm) | P_{ICl}(atm) | |
| Initial | 0.1 | 0.1 | 0.1 |
| Change | |||
| Equil |
2. Use the initial partial pressures to calculate the reaction quotient (\mathcal{Q} ). Compare \mathcal{Q} to K to predict the direction in which the reaction will proceed.
\mathcal{Q_p} =\frac{(P_{ICI})^2}{P_{I_2}P_{CI_2}}=\frac{(0.100)^2}{(0.100)(0.100)} = 1
K_p = 81.9 (given)
\mathcal{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) \rightleftharpoons 2 \ ICl(g)
| P_{I_2} (atm) | P_{Cl_2} (atm) | P_{ICl}(atm) | |
| Initial | 0.1 | 0.1 | 0.1 |
| Change | -x | -x | +2x |
| Equil |
4. Sum each column for each reactant and product to determine the equilibrium partial pressures in terms of the initial partial pressures and the variable x.
I_2(g) + Cl_2(g) \rightleftharpoons 2 \ ICl(g)
| P_{I_2} (atm) | P_{Cl_2} (atm) | P_{ICl}(atm) | |
| Initial | 0.1 | 0.1 | 0.1 |
| 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 equilibrium constant. Use the given value of the equilibrium constant to solve the expression for the variable x.
K_p =\frac{(P_{ICI})^2}{P_{I_2}P_{CI_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 partial 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 calculated equilibrium partial pressures into the equilibrium expression. The calculated value of K should match the given value of K.
K_p =\frac{(P_{ICI})^2}{P_{I_2}P_{CI_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.