Question

In a 2 L flask, the reaction takes place as:
$COC{l}_2\left(g\right)\rightleftharpoons CO\left(g\right)+C{l}_2\left(g\right)$
The equilibrium concentration of $\left[COC{l}_2\right]$ was found to be 0.4 M. If an excess of $COC{l}_2$ is added to the system, the equilibrium reestablishes and $\left[COC{l}_2\right]$ becomes 1.6 M. What is the equilibrium concentration of $\left[CO\right]$?

• A. Half of the former value
• B. Thrice of the former value
• C. Remains unaltered
• D. Twice of the former value

Answer 1

The correct option is D Twice of the former value

The given reaction is,
$COC{l}_2\left(g\right)\rightleftharpoons CO\left(g\right)+C{l}_2\left(g\right)$
$\therefore K_c=\frac{\left[CO\right]\left[C{l}_2\right]}{\left[CoC{l}_2\right]}$

Let, $\left[CO\right]=x$, then also $\left[C{l}_2\right]=x$
$\therefore K_c=\frac{x^2}{\left[CoC{l}_2\right]}=\frac{x^2}{4}$

Again, $K_c=\frac{x^2}{0.4}=\frac{\left[CO\right]\left[C{l}_2\right]}{\left[CoC{l}_2\right]}$
$\Rightarrow \frac{x^2}{0.4}=\frac{[CO{]}^2}{1.6}$
$\Rightarrow x^2\times 4=[CO{]}^2$
$\Rightarrow 2x=\left[CO\right]$