Chapter 14, Problem 14.80QP

(a)

Interpretation Introduction

Interpretation: The equilibrium partial pressure of the products and reactants at $1000°\text{C}$ is to be calculated and to predict the equilibrium partial pressure of the reactants and products after sufficient addition of $\text{CO}$ and ${\text{H}}_2$ to the equilibrium mixture.

Concept introduction: The equilibrium constant $\left(K_{\text{p}}\right)$ is expressed as,

$K_{\text{p}}=\frac{{\left(P_{\text{C}}\right)}^c{\left(P_{\text{D}}\right)}^d}{{\left(P_{\text{A}}\right)}^a{\left(P_{\text{B}}\right)}^b}$

To determine: The equilibrium partial pressure of the products and reactants at $1000°\text{C}$.

Answer to Problem 14.80QP

Solution

The equilibrium partial pressure of ${\text{H}}_2\text{O}$ is $\underset{\_}{0.292atm}$, the equilibrium partial pressure of $\text{CO}$ is $\underset{\_}{5.15atm}$ and the equilibrium partial pressure of ${\text{H}}_2$ is $\underset{\_}{0.15atm}$.

Explanation of Solution

Explanation

Given

The balanced chemical equation is,

${\text{H}}_2\text{O}\left(g\right)+\text{C}\left(s\right)\rightleftharpoons \text{CO}\left(g\right)+{\text{H}}_2\left(g\right)$

The equilibrium constant at $1000°\text{C}$ is $3.0\times {10}^{-2}$.

The initial partial pressure of ${\text{H}}_2\text{O}$ is $0.442\text{atm}$.

The initial partial pressure of $\text{CO}$ is $5.0\text{atm}$.

The relationship between $K_{\text{p}}$ and $K_{\text{c}}$ is given by,

$K_{\text{p}}=K_{\text{c}}{\left(\text{R}T\right)}^{\Delta n}$

Where,

• $\text{R}$ is the gas constant.
• $T$ is the temperature.
• $\Delta n$ is the change in number of moles during the reaction.

The change in number of moles is $\begin{array}{c}\Delta n=\text{moles}\text{of}\text{product}-\text{moles}\text{of}\text{reactant}\\ \text{=}2-1\\ =1\end{array}$

Convert $1000°\text{C}$ to $\text{K}$.

$\begin{array}{c}1000°\text{C}=1000+273\text{K}\\ \text{=1273}\text{K}\end{array}$

Substitute the value of equilibrium constant, number of moles, temperature in above formula.

$\begin{array}{c}{K}_{\text{p}}=3.0\times {10}^{-2}\times 0.08206\times 1273\text{K}\\ =3.1\end{array}$

The value of $K_{\text{p}}$ is $3.1$.

The equilibrium constant for given reaction is calculated by the formula,

$K_{\text{p}}=\frac{\left(P_{\text{CO}\left(g\right)}\right)\left(P_{{\text{H}}_2\left(g\right)}\right)}{\left(P_{{\text{H}}_2\text{O}\left(g\right)}\right)\left(P_{\text{C}\left(s\right)}\right)}$

The partial pressure of pure solids and liquids do not change during the reaction, so they are excluded from the equilibrium expression.

$K_{\text{p}}=\frac{\left(P_{\text{CO}\left(g\right)}\right)\left(P_{{\text{H}}_2\left(g\right)}\right)}{\left(P_{{\text{H}}_2\text{O}\left(g\right)}\right)}$ (1)

Where,

• $\left(P_{\text{CO}\left(g\right)}\right)$ is the partial pressure of $\text{CO}$ at equilibrium.
• $\left(P_{{\text{H}}_2\left(g\right)}\right)$ is the partial pressure of ${\text{H}}_2$ at equilibrium.
• $\left(P_{{\text{H}}_2\text{O}\left(g\right)}\right)$ is the partial pressure of ${\text{H}}_2\text{O}$ at equilibrium.

Consider the following table showing the values of partial pressure at initial and at equilibrium stage,

$\begin{array}{cccccccc}& {\text{H}}_2\text{O}\left(g\right)& +& \text{C}\left(s\right)& \rightleftharpoons & \text{CO}\left(g\right)& +& {\text{H}}_2\left(g\right)\\ \text{Initial}& 0.442& & & & 5.0& & 0\\ \text{Equilibrium}& 0.442-x& & & & 5.0+x& & x\end{array}$ (2)

Where,

• $x$ is the change in partial pressure occurred during the reaction.

Substitute the values of partial pressure of ${\text{H}}_2\text{O}$, $\text{CO}$, ${\text{H}}_2$ at equilibrium and equilibrium constant  in equation (1).

$\begin{array}{c}3.1=\frac{\left(5.0+x\right)\left(x\right)}{\left(0.442-x\right)}\\ 3.1\left(0.442-x\right)=x^2+5.0x\\ x^2+8.1x-1.4=0\end{array}$

Solve this quadratic equation by the formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where,

• $a$ is the coefficient of $x^2$.
• $b$ is the coefficient of $x$.
• $c$ is the coefficient of $1$.

Substitute the values of $a$, $b$ and $c$ in the above formula.

$\begin{array}{c}x=\frac{-8.1\pm \sqrt{{\left(8.1\right)}^2-4\times 1\times \left(-1.4\right)}}{2\times 1}\\ =\frac{-8.1\pm 8.4}{2}\\ =\frac{0.3}{2}\\ =0.15\end{array}$

The value if $x$ is $0.15$.

Substitute the value of $x$  in equation (2).

The equilibrium Partial pressure of ${\text{H}}_2\text{O}$ is $\begin{array}{l}=0.442-x\\ =0.442-0.15\\ =\underset{\_}{0.292atm}\end{array}$

The equilibrium Partial pressure of $\text{CO}$ is $\begin{array}{l}=5.0+x\\ =5.0+0.15\\ =\underset{\_}{5.15atm}\end{array}$

The equilibrium Partial pressure of ${\text{H}}_2$ is $\begin{array}{l}=x\\ =\underset{\_}{0.15atm}\end{array}$

Therefore, the equilibrium partial pressure of ${\text{H}}_2\text{O}$ is $\underset{\_}{0.292atm}$, the equilibrium partial pressure of $\text{CO}$ is $\underset{\_}{5.15atm}$ and the equilibrium partial pressure of ${\text{H}}_2$ is $\underset{\_}{0.15atm}$.

(b)

Interpretation Introduction

To determine: The equilibrium partial pressure of the reactants and products after sufficient addition of $\text{CO}$ and ${\text{H}}_2$ to the equilibrium mixture.

Answer to Problem 14.80QP

Solution

The equilibrium partial pressure of ${\text{H}}_2\text{O}$ is $\underset{\_}{0.411atm}$, the equilibrium partial pressure of $\text{CO}$ is $\underset{\_}{5.106atm}$ and the equilibrium partial pressure of ${\text{H}}_2$ is $\underset{\_}{0.106atm}$.

Explanation of Solution

Explanation

The addition $\text{CO}$ and ${\text{H}}_2$ to the equilibrium mixture increases the concentration of products. The equilibrium of the reaction shifts toward the left of the reaction and reaction will proceed in the reverse direction.

$\text{CO}\left(g\right)+{\text{H}}_2\left(g\right)\rightleftharpoons {\text{H}}_2\text{O}\left(g\right)+\text{C}\left(s\right)$

The partial pressure of $\text{CO}$ and ${\text{H}}_2$ is increased by $0.075\text{atm}$.

The new partial pressure of $\text{CO}$ is $\begin{array}{l}=5.15\text{atm}+0.075\text{atm}\\ =5.225\text{atm}\end{array}$

The new partial pressure of ${\text{H}}_2$ is $\begin{array}{l}=0.15\text{atm}+0.075\text{atm}\\ \text{=0}\text{.225}\text{atm}\end{array}$

The equilibrium constant in reverse direction is calculated by formula,

$K_{\text{r}}=\frac{1}{K_{\text{p}}}$

Where,

• $K_{\text{p}}$ is the equilibrium constant in forward direction.

Substitute the value of $K_{\text{p}}$ in the above formula.

$\begin{array}{c}{K}_{\text{r}}=\frac{1}{3.1}\\ =0.32\end{array}$

The value of equilibrium constant in reverse direction is $0.32$.

The equilibrium constant in reverse direction is calculated by the formula,

$K_{\text{r}}=\frac{\left(P_{{\text{H}}_2\text{O}\left(g\right)}\right)\left(P_{\text{C}\left(s\right)}\right)}{\left(P_{\text{CO}\left(g\right)}\right)\left(P_{{\text{H}}_2\left(g\right)}\right)}$

The partial pressure of pure solids and liquids do not change during the reaction, so they are excluded from the equilibrium expression.

$K_{\text{r}}=\frac{\left(P_{{\text{H}}_2\text{O}\left(g\right)}\right)}{\left(P_{\text{CO}\left(g\right)}\right)\left(P_{{\text{H}}_2\left(g\right)}\right)}$ (3)

Where,

• $\left(P_{\text{CO}\left(g\right)}\right)$ is the partial pressure of $\text{CO}$ at equilibrium.
• $\left(P_{{\text{H}}_2\left(g\right)}\right)$ is the partial pressure of ${\text{H}}_2$ at equilibrium.
• $\left(P_{{\text{H}}_2\text{O}\left(g\right)}\right)$ is the partial pressure of ${\text{H}}_2\text{O}$ at equilibrium.

Consider the following table showing the values of partial pressure at initial and at equilibrium stage,

$\begin{array}{cccccccc}& \text{CO}\left(g\right)& +& {\text{H}}_2\left(g\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(g\right)& +& \text{C}\left(s\right)\\ \text{Initial}& 5.225& & 0.225& & 0.292& & \\ \text{Equilibrium}& 5.225-x& & 0.225-x& & 0.292+x& & \end{array}$ (4)

Where,

• $x$ is the change in partial pressure occurred during the reaction.

Substitute the values of partial pressure of ${\text{H}}_2\text{O}$, $\text{CO}$, ${\text{H}}_2$ at equilibrium and equilibrium constant in reverse direction in equation (3).

$\begin{array}{c}0.32=\frac{\left(0.292+x\right)}{\left(5.225-x\right)\left(0.225-x\right)}\\ =\frac{\left(0.292+x\right)}{x^2-5.45x+1.175}\\ 0.32\left(x^2-5.45x+1.175\right)=\left(0.292+x\right)\\ 0.32x^2-0.74x+0.084=0\end{array}$

Solve this quadratic equation by the formula,

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where,

• $a$ is the coefficient of $x^2$.
• $b$ is the coefficient of $x$.
• $c$ is the coefficient of $1$.

Substitute the values of $a$, $b$ and $c$ in the above formula.

$\begin{array}{c}x=\frac{-\left(-0.74\right)\pm \sqrt{{\left(-0.74\right)}^2-4\times 1\times \left(0.084\right)}}{2\times 0.084}\\ =0.119\end{array}$

The value if $x$ is $0.119$.

Substitute the value of $x$  in equation (4).

The equilibrium Partial pressure of ${\text{H}}_2\text{O}$ is $\begin{array}{l}=0.292+x\\ =0.292+0.119\\ =\underset{\_}{0.411atm}\end{array}$

The equilibrium Partial pressure of $\text{CO}$ is $\begin{array}{l}=5.225-x\\ =5.225-0.119\\ =\underset{\_}{5.106atm}\end{array}$

The equilibrium Partial pressure of ${\text{H}}_2$ is $\begin{array}{l}=0.225-x\\ =0.225-0.119\\ =\underset{\_}{0.106atm}\end{array}$

Therefore, the equilibrium partial pressure of ${\text{H}}_2\text{O}$ is $\underset{\_}{0.411atm}$, the equilibrium partial pressure of $\text{CO}$ is $\underset{\_}{5.106atm}$ and the equilibrium partial pressure of ${\text{H}}_2$ is $\underset{\_}{0.106atm}$.

Conclusion

1. a. The equilibrium partial pressure of ${\text{H}}_2\text{O}$ is $\underset{\_}{0.292atm}$, the equilibrium partial pressure of $\text{CO}$ is $\underset{\_}{5.15atm}$ and the equilibrium partial pressure of ${\text{H}}_2$ is $\underset{\_}{0.15atm}$.
2. b. The equilibrium partial pressure of ${\text{H}}_2\text{O}$ is $\underset{\_}{0.411atm}$, the equilibrium partial pressure of $\text{CO}$ is $\underset{\_}{5.106atm}$ and the equilibrium partial pressure of ${\text{H}}_2$ is $\underset{\_}{0.106atm}$.