Chapter 14, Problem 14.93AP

Interpretation Introduction

Interpretation: The percentage decomposition of ${\text{CO}}_{\text{2}}$ increases with increase in temperature. The given reaction is endothermic is to be predicted. The value of $K_{\text{p}}$ at each temperature is to be calculated.

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

$K_{\text{c}}=\frac{{\left[\text{Product}\right]}^y}{{\left[\text{Reactant}\right]}^x}$

To determine: The given reaction is endothermic and the value of $K_{\text{p}}$ at each temperature.

Answer to Problem 14.93AP

Solution

The decomposition reaction of ${\text{CO}}_{\text{2}}$ is endothermic. The value of $K_{\text{p}}$ at $1500\text{K}$ is $\underset{\_}{5.53\times 10^{-11}}$, the value of $K_{\text{p}}$ at $2500\text{K}$ is $\underset{\_}{4.014\times 10^{-3}}$ and the value of $K_{\text{p}}$ at $3000\text{K}$ is $\underset{\_}{0.403}$. The decomposition of ${\text{CO}}_{\text{2}}$ is not an antidote for global warming as decomposition of ${\text{CO}}_{\text{2}}$ leads to the production of $\text{CO}$ which is still toxic.

Explanation of Solution

Explanation

Given

The balanced chemical equation is,

${\text{2CO}}_2\left(g\right)\rightleftharpoons 2\text{CO}\left(g\right)+{\text{O}}_2\left(g\right)$

Initial partial pressure of ${\text{CO}}_2$ is $1\text{atm}$.

The decomposition percentage of ${\text{CO}}_2$ at $1500\text{K}$ is $0.048\%$.

The decomposition percentage of ${\text{CO}}_2$ at $2500\text{K}$ is $17.6\%$.

The decomposition percentage of ${\text{CO}}_2$ at $3000\text{K}$ is $54.8\%$.

The equilibrium constant is calculated by the formula,

$K_{\text{p}}=\frac{{\left(P_{\text{CO}\left(g\right)}\right)}^2\left(P_{{\text{O}}_2\left(g\right)}\right)}{{\left(P_{{\text{CO}}_2\left(g\right)}\right)}^2}$ (1)

Where,

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

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

$\begin{array}{cccccc}& 2{\text{CO}}_2\left(g\right)& \rightleftharpoons & 2\text{CO}\left(g\right)& +& {\text{O}}_2\left(g\right)\\ \text{Initial}& 1\text{atm}& & 0& & 0\\ \text{Equilibrium}& 1-2x& & +x& & +x\end{array}$

Where,

• $x$ is the percentage decomposition of the molecules.

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

$\begin{array}{c}{K}_{\text{p}}=\frac{{\left(2x\right)}^2\left(x\right)}{{\left(1-2x\right)}^2}\\ =\frac{4x^3}{{\left(1-2x\right)}^2}\end{array}$ (2)

The value of percentage decomposition of ${\text{CO}}_{\text{2}}$ at $1500\text{K}$ is $0.048\%$.

$\begin{array}{c}2x=\frac{0.048}{100}\\ x=0.024\times {10}^{-2}\end{array}$

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

$\begin{array}{c}{K}_{\text{p}}=\frac{4{\left(0.024\times {10}^{-2}\right)}^3}{{\left(1-2\times 0.024\times {10}^{-2}\right)}^2}\\ =\underset{\_}{5.53\times 10^{-11}}\end{array}$

Therefore, the value of $K_{\text{p}}$ at $1500\text{K}$ is $\underset{\_}{5.53\times 10^{-11}}$.

The value of percentage decomposition of ${\text{CO}}_{\text{2}}$ at $2500\text{K}$ is $17.6\%$.

$\begin{array}{c}2x=\frac{17.6}{100}\\ x=8.8\times {10}^{-2}\end{array}$

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

$\begin{array}{c}{K}_{\text{p}}=\frac{4{\left(8.8\times {10}^{-2}\right)}^3}{{\left(1-2\times 8.8\times {10}^{-2}\right)}^2}\\ =\underset{\_}{4.014\times 10^{-3}}\end{array}$

Therefore, the value of $K_{\text{p}}$ at $2500\text{K}$ is $\underset{\_}{4.014\times 10^{-3}}$.

The value of percentage decomposition of ${\text{CO}}_{\text{2}}$ at $3000\text{K}$ is $54.8\%$.

$\begin{array}{c}2x=\frac{54.8}{100}\\ x=27.4\times {10}^{-2}\end{array}$

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

$\begin{array}{c}{K}_{\text{p}}=\frac{4{\left(27.4\times {10}^{-2}\right)}^3}{{\left(1-2\times 27.4\times {10}^{-2}\right)}^2}\\ =\underset{\_}{0.403}\end{array}$

Therefore, the value of $K_{\text{p}}$ at $3000\text{K}$ is $\underset{\_}{0.403}$.

The value of $K_{\text{p}}$ increases with increase in temperature, this indicates that the decomposition reaction of ${\text{CO}}_{\text{2}}$ is endothermic. The decomposition of ${\text{CO}}_{\text{2}}$ is not an antidote for global warming as decomposition of ${\text{CO}}_{\text{2}}$ leads to the production of $\text{CO}$ which is still toxic.

Conclusion

The decomposition reaction of ${\text{CO}}_{\text{2}}$ is endothermic. The value of $K_{\text{p}}$ at $1500\text{K}$ is $\underset{\_}{5.53\times 10^{-11}}$, the value of $K_{\text{p}}$ at $2500\text{K}$ is $\underset{\_}{4.014\times 10^{-3}}$ and the value of $K_{\text{p}}$ at $3000\text{K}$ is $\underset{\_}{0.403}$. The decomposition of ${\text{CO}}_{\text{2}}$ is not an antidote for global warming as decomposition of ${\text{CO}}_{\text{2}}$ leads to the production of $\text{CO}$ which is still toxic.