Chapter 19, Problem 19.81P

Interpretation Introduction

Interpretation:

The total pressure after equilibrium is reestablished for the given reaction has to be calculated.

  $CO\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons COC{l}_2\left(g\right)$

Explanation of Solution

The reaction is given below.

  $\begin{array}{l}CO\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons COC{l}_2\left(g\right)\\ \\ K_P=\frac{P_{COC{l}_2}}{\left(P_{CO}\right)\left(P_{C{l}_2}\right)}\end{array}$

The first set of equilibrium partial pressures are $P_{CO}=P_{C{l}_2}=1.09barand{P}_{COC{l}_2}=0.144bar.$

By substituting these values in the above expression, the value of equilibrium constant can be calculated.

  $K_P=\frac{0.144bar}{\left(1.09bar\right)\left(1.09bar\right)}=0.1212ba{r}^{-1}$

The pressure of $CO\left(g\right)$ immediately after addition of $CO\left(g\right)$ can be calculated from the relation for the total pressure.

  $\begin{array}{l}{P}_{total}=P_{CO}+P_{C{l}_2}+P_{COC{l}_2}\\ \\ 3.31bar=P_{CO}+1.09bar+0.144bar\\ \\ P_{CO}=2.076bar\end{array}$

According to Le Chatelier’s principle, the position of the equilibrium will shift towards right in response to the addition of $CO\left(g\right).$  Thus, some of $CO\left(g\right)$ and $C{l}_2\left(g\right)$ will decompose to produce $COC{l}_2\left(g\right).$  Let it be x.

To determine the new equilibrium conditions, a pressure table can be set up as given below.

  $\begin{array}{cccccccc}\text{Pressure}\left(bar\right)& CO\left(g\right)& +& C{l}_2\left(g\right)& \rightleftharpoons & COC{l}_2\left(g\right)& & \\ Initial& 2.076& & 1.09& & 0.144& & \\ Change& -x& & +x& & +x& & \\ Equilibrium& 2.076-x& & 1.09-x& & 0.144+x& & \end{array}$

Substitute the equilibrium pressures into the equilibrium – constant expression as shown below.

  $\begin{array}{c}{K}_P=\frac{0.144bar+x}{\left(2.076bar-x\right)\left(1.09bar-x\right)}=0.1212ba{r}^{-1}\\ \\ 0.144bar+x=\left(2.076bar-x\right)\left(1.09bar-x\right)\left(0.1212ba{r}^{-1}\right)\\ \\ \left(0.1212ba{r}^{-1}\right)x^2-1.3837x+0.1299bar=0\\ \\ x=11.32bar,0.0946bar\end{array}$

The value $11.32bar$ results into negative value of concentration, thus this must be rejected.  Therefore, the value of x is $0.0946bar.$

Then,

  $\begin{array}{cccccccc}\text{Pressure}\left(bar\right)& CO\left(g\right)& +& C{l}_2\left(g\right)& \rightleftharpoons & COC{l}_2\left(g\right)& & \\ Initial& 2.076& & 1.09& & 0.144& & \\ Change& -0.0946& & -0.0946& & +0.0946& & \\ Equilibrium& 1.9814& & 0.9954& & 0.2386& & \end{array}$

The total pressure after equilibrium is re-established can be calculated as given below.

  $\begin{array}{c}{P}_{total}=P_{CO}+P_{C{l}_2}+P_{COC{l}_2}\\ \\ =1.9814bar+0.9954bar+0.2386bar\\ \\ =3.2154bar\approx 3.22bar\end{array}$

As a final check, it can be shown that,

  $\frac{P_{COC{l}_2}}{\left(P_{CO}\right)\left(P_{C{l}_2}\right)}=\frac{0.2386bar}{\left(1.9814bar\right)\left(0.9954bar\right)}=0.121ba{r}^{-1}$

which agrees with $K_P=0.1212ba{r}^{-1}.$

Therefore, the total pressure after equilibrium is re-established is $3.22bar.$