Question

1. consider the following reaction: so₂cl₂(g) ⇌ so₂(g) + cl₂(g) a reaction vessel initially contains only 0.020 m so₂cl₂. at equilibrium, it was found that the vessel contained 1.2 x 10⁻² m cl₂. determine kc for the equilibrium at the temperature that the experiment was carried out.

Explanation:

Step1: Set up the ICE table

Let the initial concentration of $SO_2Cl_2$ be $[SO_2Cl_2]_0 = 0.020\ M$, and initial concentrations of $SO_2$ and $Cl_2$ be $0\ M$. At equilibrium, $[Cl_2]=1.2\times 10^{- 2}\ M$. Since the stoichiometry of the reaction $SO_2Cl_2(g)
ightleftharpoons SO_2(g)+Cl_2(g)$ is 1:1:1, the change in concentration of $SO_2Cl_2$ is $-\Delta x$, and the changes in concentrations of $SO_2$ and $Cl_2$ are $+\Delta x$. Since $[Cl_2]=\Delta x = 1.2\times 10^{-2}\ M$ at equilibrium.

Step2: Calculate the equilibrium concentrations

$[SO_2]=\Delta x=1.2\times 10^{-2}\ M$, and $[SO_2Cl_2]=[SO_2Cl_2]_0-\Delta x=0.020 - 1.2\times 10^{-2}=0.020 - 0.012 = 0.008\ M$

Step3: Calculate the equilibrium constant $K_c$

The expression for $K_c$ for the reaction $SO_2Cl_2(g)
ightleftharpoons SO_2(g)+Cl_2(g)$ is $K_c=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]}$. Substitute the equilibrium - concentrations: $K_c=\frac{(1.2\times 10^{-2})\times(1.2\times 10^{-2})}{0.008}$.
$K_c=\frac{1.44\times 10^{-4}}{0.008}=0.018$

Answer:

$0.018$