Q17.90CP - Chemistry: Molecular Nature Of Matter And Change

Question

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

Step-by-Step Solution

Answer

The equilibrium concentrations of the gases are as follows –

$\begin{array}{c}\text{[CO]=0.287 mol/L}\\ \left[{\text{Cl}}_{\text{2}}\right]\text{=0.287 mol/L}\\ {\text{[COCl}}_2\text{]=0.413mol/L}\end{array}$

Step 1: Concept Introduction

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.

Step 2: Equilibrium Constant

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  $\frac{\text{0.350 mol}}{\text{0.500 L}}\text{=0.70 mol/L}$. Write the reaction table.

	$\text{CO(g)}$	${\mathrm{Cl}}_2\text{(g)}$	${\mathrm{COCl}}_2\text{(g)}$
Initial	$0.70$	$0.70$	$0$
Change	$-\mathrm{x}$	$-\mathrm{x}$	$+\mathrm{x}$
Equilibrium	$0.70-\mathrm{x}$	$0.70-\mathrm{x}$	$\mathrm{x}$

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 –

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

Step 3: Equilibrium Concentration Values

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

$\begin{array}{c}\text{x=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}\\ \text{=}\frac{\text{-(-7.93)±}\sqrt{{\text{(-7.93)}}^{\text{2}}\text{-4(4.95)(2.43)}}}{\text{2(4.95)}}\\ \text{x=1.189 and 0.413}\end{array}$

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  $\mathrm{CO}$–

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

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

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

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 $\text{0.287 mol/L, 0.287 mol/L}$  and $\text{0.413mol/L}$ respectively.