Chapter 18, Problem 19PE

Interpretation Introduction

Interpretation:

The value of equilibrium constant $\text{K}$ for the reaction $\text{CO}\left(\text{g}\right)+{\text{Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{COCl}}_{\text{2}}\left(\text{g}\right)$ is to be calculated.

Concept introduction:

The acids that do not undergo complete dissociation when dissolved in water are known as weak acids. The equilibrium constant in the case of dissociation of a weak acid is known as the acid constant and is denoted by ${\text{K}}_{\text{a}}$.

Answer to Problem 19PE

The value of equilibrium constant $\text{K}$ for the reaction $\text{CO}\left(\text{g}\right)+{\text{Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{COCl}}_{\text{2}}\left(\text{g}\right)$ is $14$.

Explanation of Solution

The equilibrium reaction is shown below.

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

The volume of the reaction chamber is $\text{3}\text{.00}\text{L}$.

The formula to calculate the molar concentration of the compounds is given below.

$\text{Molar}\text{concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{(L)}}$…(1)

Substitute the value of initial number of moles of $\text{CO}$ and the volume of the reaction chamber in equation (1) to calculate the initial molar concentration of $\text{CO}$.

$\begin{array}{rll}\text{Molar}\text{concentration}\text{of}\text{CO}& =& \frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{(L)}}\\ & =& \frac{\text{9}\text{mol}}{\text{3}\text{L}}\\ & =& 3\text{mol}/\text{L}\end{array}$

Substitute the value of initial number of moles of ${\text{Cl}}_{\text{2}}$ and the volume of the reaction chamber in equation (1) to calculate the initial molar concentration of ${\text{Cl}}_{\text{2}}$.

$\begin{array}{rll}\text{Molar}\text{concentration}\text{of}{\text{Cl}}_{\text{2}}& =& \frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{(L)}}\\ & =& \frac{\text{15}\text{mol}}{\text{3}\text{L}}\\ & =& 5\text{mol}/\text{L}\end{array}$

Substitute the value of final number of moles of ${\text{Cl}}_{\text{2}}$ and the volume of the reaction chamber in equation (1) to calculate the final molar concentration of ${\text{Cl}}_{\text{2}}$.

$\begin{array}{rll}\text{Molar}\text{concentration}\text{of}{\text{Cl}}_{\text{2}}& =& \frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{(L)}}\\ & =& \frac{\text{6}\text{.3}\text{mol}}{\text{3}\text{L}}\\ & =& 2.1\text{mol}/\text{L}\end{array}$

The change in concentration of ${\text{Cl}}_{\text{2}}$ is calculated by the formula as shown below.

$\text{Change}\text{in}\text{concentration}=\text{Final}\text{concentration}-\text{Initial}\text{concentration}$

Substitute the values of final and initial concentrations of ${\text{Cl}}_{\text{2}}$ in the above equation.

$\begin{array}{rll}\text{Change}\text{in}\text{concentration}& =& \text{Final}\text{concentration}-\text{Initial}\text{concentration}\\ & =& \text{2}\text{.1}-\text{5}\\ & =& -2.9\text{mol}/\text{L}\end{array}$

The negative sign indicates that the concentration of ${\text{Cl}}_{\text{2}}$ is decreasing.

According to the equilibrium reaction, $\text{1}\text{mol}$ of ${\text{Cl}}_{\text{2}}$ produces $\text{1}\text{mol}$ of ${\text{COCl}}_{\text{2}}$. Therefore, dissociation of $2.9\text{mol}$ of ${\text{Cl}}_{\text{2}}$ produces $2.9\text{mol}$ of ${\text{COCl}}_{\text{2}}$.

The ICE table for the given reaction is shown below.

$\begin{array}{rlll}& \text{CO}\left(\text{g}\right)+& {\text{Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons & {\text{COCl}}_{\text{2}}\left(\text{g}\right)\\ \text{Initial}& \text{3}& \text{5}& \text{0}\\ \text{Reacting}& -\text{2}\text{.9}& -\text{2}\text{.9}& +\text{2}\text{.9}\\ \text{Equilibrium}& \text{0}\text{.1}& \text{2}\text{.1}& \text{2}\text{.9}\end{array}$

The value of $\text{K}$ can be calculated by the formula given below.

$\text{K}=\frac{\left[{\text{COCl}}_{\text{2}}\right]}{\left[\text{CO}\right]\left[{\text{Cl}}_{\text{2}}\right]}$

Substitute the values of concentrations of $\text{CO}$, ${\text{Cl}}_{\text{2}}$ and ${\text{COCl}}_{\text{2}}$ in the above equation.

$\begin{array}{rll}\text{K}& =& \frac{\left[{\text{COCl}}_{\text{2}}\right]}{\left[\text{CO}\right]\left[{\text{Cl}}_{\text{2}}\right]}\\ & =& \frac{2.9}{0.1\times 2.1}\\ & =& 13.809\\ & \approx & 14\end{array}$

Therefore, the value of $\text{K}$ is $14$.

Conclusion

The value of equilibrium constant $\text{K}$ for the reaction $\text{CO}\left(\text{g}\right)+{\text{Cl}}_{\text{2}}\left(\text{g}\right)\rightleftharpoons {\text{COCl}}_{\text{2}}\left(\text{g}\right)$ is $14$.