Chapter 19, Problem 19.68P

(a)

Interpretation Introduction

Interpretation:

The value of $P_{C{H}_{3}OH}$ at equilibrium when $P_{H_2}$ is $0.020bar$ and $P_{CO}$ is $0.010bar$ has to be calculated.

Answer to Problem 19.68P

The value of $P_{C{H}_{3}OH}$ at equilibrium is $0.088bar.$

Explanation of Solution

The methanol synthesis reaction is given below.

  $2H_2\left(g\right)+CO\left(g\right)\rightleftharpoons C{H}_{3}OH\left(g\right)$

The expression for equilibrium constant for the above reaction can be written as given below.

  $K_P=\frac{P_{C{H}_{3}OH}}{{\left(P_{H_2}\right)}^2\left(P_{CO}\right)}=2.19\times {10}^{4}ba{r}^{-2}$

Given that the equilibrium partial pressure of $H_2,andCO$ are $0.020bar,and010bar$ respectively.

By substituting these values in the equilibrium constant expression, the value of $P_{C{H}_{3}OH}$ can be calculated.

  $\begin{array}{l}{K}_P=\frac{P_{C{H}_{3}OH}}{{\left(P_{H_2}\right)}^2\left(P_{CO}\right)}=2.19\times {10}^{4}ba{r}^{-2}\\ \\ P_{C{H}_{3}OH}=\left(2.19\times {10}^{4}ba{r}^{-2}\right){\left(0.020bar\right)}^2\left(0.010bar\right)\\ \\ P_{C{H}_{3}OH}=0.088bar\end{array}$

Therefore, the value of $P_{C{H}_{3}OH}$ at equilibrium is $0.088bar.$

(b)

Interpretation Introduction

Interpretation:

The value of $P_{CO}and{P}_{C{H}_{3}OH}$ at equilibrium when $P_{H_2}$ is $0.020bar$ and $P_{total}$ is $10.0bar$ has to be calculated.

Answer to Problem 19.68P

The value of $P_{CO}and{P}_{C{H}_{3}OH}$ at equilibrium is $1.03barand8.95bar$ respectively.

Explanation of Solution

The methanol synthesis reaction is given below.

  $2H_2\left(g\right)+CO\left(g\right)\rightleftharpoons C{H}_{3}OH\left(g\right)$

The expression for equilibrium constant for the above reaction can be written as given below.

  $K_P=\frac{P_{C{H}_{3}OH}}{{\left(P_{H_2}\right)}^2\left(P_{CO}\right)}=2.19\times {10}^{4}ba{r}^{-2}$

Given that the equilibrium partial pressure of $H_2$ are $0.020bar$ and total pressure is $10.0bar.$  By using this value two equations can be derived.

  $\begin{array}{l}{K}_P=\frac{P_{C{H}_{3}OH}}{{\left(P_{H_2}\right)}^2\left(P_{CO}\right)}=2.19\times {10}^{4}ba{r}^{-2}\\ \\ \frac{P_{C{H}_{3}OH}}{P_{CO}}={\left(P_{H_2}\right)}^2\left(2.19\times {10}^{4}ba{r}^{-2}\right)\\ \\ \frac{P_{C{H}_{3}OH}}{P_{CO}}={\left(0.020bar\right)}^2\left(2.19\times {10}^{4}ba{r}^{-2}\right)=8.76\\ \\ P_{C{H}_{3}OH}=8.76P_{CO}\mathrm{.....}\left(1\right)\\ \\ P_{total}=P_{H_2}+P_{CO}+P_{C{H}_{3}OH}\\ \\ P_{CO}+8.76P_{CO}=P_{total}-P_{H_2}\left[Fromequation(1)\right]\\ \\ 9.76P_{CO}=10.0bar-0.020bar=9.98bar\\ \\ P_{CO}=\frac{9.98bar}{9.76}=1.03bar\\ \\ P_{C{H}_{3}OH}=8.76P_{CO}=8.76\times 1.03bar=8.95bar\end{array}$

Therefore, the value of $P_{CO}and{P}_{C{H}_{3}OH}$ at equilibrium is $1.03barand8.95bar$ respectively.