Phosgene $\mathbf{(}{\mathbf{\text{COCl}}}_{\mathbf{\text{2}}}\mathbf{\text{)}}$is a toxic substance that forms readily from carbon monoxide and chlorine at elevated temperatures:

${\mathbf{\text{CO(g)+Cl}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\rightleftharpoons }{\mathbf{\text{COCl}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

If$\mathbf{\text{0.350mol}}$of each reactant is placed in a$\mathbf{\text{0.500-L}}$flask at$\mathbf{\text{600K}}$, what are the concentrations of all three substances at equilibrium (${\mathbf{\text{K}}}_{\mathbf{\text{c}}}\mathbf{\text{=4.95}}$at this temperature)?

Chemical equilibrium is the state of a system in which the concentration of the reactant and the concentration of the products do not change over time and the system's attributes do not change.

The reaction given is –

${\text{CO(g)+Cl}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{COCl}}_{\text{2}}{\text{(g)K}}_{\text{c}}\text{=4.95}$

The concentration of each reactant is equal to $\frac{\text{0.350mol}}{\text{0.500L}}\text{=0.70mol/L}$. Write the reaction table.

Write the expression for the equilibrium constant of the reaction in terms of concentration –

$\begin{array}{l}{\text{K}}_{\text{c}}\text{=}\frac{\text{Products}}{\text{Reactants}}\\ {\text{K}}_{\text{c}}\text{=}\frac{{\text{[COCl}}_{\text{2}}\text{]}}{{\text{[CO][Cl}}_{\text{2}}\text{]}}\end{array}$

Substitute the equilibrium equations from the reaction table to solve for –

$\begin{array}{c}{\text{K}}_{\text{c}}\text{=}\frac{{\text{[COCl}}_{\text{2}}\text{]}}{{\text{[CO][Cl}}_{\text{2}}\text{]}}\\ \text{4.95=}\frac{\text{x}}{\text{(0.70-x)(0.70-x)}}\\ \text{4.95=}\frac{\text{x}}{{\text{x}}^{\text{2}}\text{-1.40x+0.49}}\\ {\text{4.95x}}^{\text{2}}\text{-6.93x+2.43=x}\\ {\text{4.95x}}^{\text{2}}\text{-7.93x+2.43=0}\end{array}$

The ${\text{K}}_{\text{c}}$is big so we cannot neglect the $\mathrm{x}$in the denominator. Use the quadratic formula to solve for $\mathrm{x}$.

$\begin{array}{c}\text{x=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}\\ \text{=}\frac{\text{-(-7.93)±}\sqrt{{\text{(-7.93)}}^{\text{2}}\text{-4(4.95)(2.43)}}}{\text{2(4.95)}}\\ \text{x=1.189and0.413}\end{array}$

In this case, use$\text{x=0.413}$since$\text{x=1.189}$will return a negative value. Now solve for the concentration of each reaction component using the equilibrium formula from the reaction table.

Calculate the equilibrium concentration of $\mathrm{CO}$–

$\begin{array}{c}\text{[CO]=0.70-x}\\ \text{=0.70-0.413}\\ \text{[CO]=0.287mol/L}\end{array}$

Now calculate the equilibrium concentration of ${\mathrm{Cl}}_2$–

$\begin{array}{c}\left[{\text{Cl}}_{\text{2}}\right]\text{=0.70-x}\\ \text{=0.70-0.413}\\ \left[{\text{Cl}}_{\text{2}}\right]\text{=0.287mol/L}\end{array}$

Now calculate the equilibrium concentration of${\text{COCl}}_{\text{2}}$–

$\begin{array}{c}\left[{\text{COCl}}_{\text{2}}\right]\text{=x}\\ {\text{[COCl}}_2\text{]=0.413mol/L}\end{array}$

Therefore, the values for equilibrium concentrations for${\text{CO,Cl}}_2$ and ${\text{COCl}}_{\text{2}}$are$\text{0.287mol/L,0.287mol/L}$ and$\text{0.413mol/L}$ respectively.