Questions: Assume that the change in concentration of COCl2 is small enough to be neglected in the following problem.
(a) Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of COCl2 with an initial concentration of 0.3166 M.
COCl2(g) ⇌ CO(g) + Cl2(g)
Kc=2.2 × 10^-10
(b) Confirm that the change is small enough to be neglected.
Transcript text: 3)
Assume that the change in concentration of $\mathrm{COCl}_{2}$ is small enough to be neglected in the following problem.
(a) Calculate the equilibrium concentration of all species in an equilibrium mixture that results from the decomposition of $\mathrm{COCl}_{2}$ with an initial concentration of 0.3166 M .
\[
\begin{array}{l}
\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \\
K_{c}=2.2 \times 10^{-10}
\end{array}
\]
(b) Confirm that the change is small enough to be neglected.
(from OpenStax)
Solution
Solution Steps
Step 1: Write the Equilibrium Expression
Given the reaction:
\[
\mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g) + \mathrm{Cl}_{2}(g)
\]
The equilibrium constant expression \( K_c \) is:
\[
K_c = \frac{[\mathrm{CO}][\mathrm{Cl}_2]}{[\mathrm{COCl}_2]}
\]
Step 2: Define Initial and Equilibrium Concentrations
Let the initial concentration of \(\mathrm{COCl}_2\) be \( [\mathrm{COCl}_2]_0 = 0.3166 \) M. At equilibrium, let the change in concentration of \(\mathrm{COCl}_2\) be \( x \). Therefore, the equilibrium concentrations are:
\[
[\mathrm{COCl}_2] = 0.3166 - x
\]
\[
[\mathrm{CO}] = x
\]
\[
[\mathrm{Cl}_2] = x
\]
Step 3: Substitute into the Equilibrium Expression
Substitute the equilibrium concentrations into the equilibrium expression:
\[
K_c = \frac{[\mathrm{CO}][\mathrm{Cl}_2]}{[\mathrm{COCl}_2]} = \frac{x \cdot x}{0.3166 - x} = \frac{x^2}{0.3166 - x}
\]
Given \( K_c = 2.2 \times 10^{-10} \):
\[
2.2 \times 10^{-10} = \frac{x^2}{0.3166 - x}
\]
Step 4: Approximate and Solve for \( x \)
Since \( K_c \) is very small, we assume \( x \) is very small compared to 0.3166, so \( 0.3166 - x \approx 0.3166 \):
\[
2.2 \times 10^{-10} = \frac{x^2}{0.3166}
\]
\[
x^2 = 2.2 \times 10^{-10} \times 0.3166
\]
\[
x^2 = 6.9652 \times 10^{-11}
\]
\[
x = \sqrt{6.9652 \times 10^{-11}} = 8.347 \times 10^{-6}
\]
To confirm the change is negligible, compare \( x \) to the initial concentration:
\[
\frac{x}{0.3166} = \frac{8.347 \times 10^{-6}}{0.3166} \approx 2.637 \times 10^{-5}
\]
Since \( 2.637 \times 10^{-5} \) is much less than 5%, the change is indeed negligible.