Chapter 24, Problem 97AP

Interpretation Introduction

Interpretation:

The value of equilibrium constant at given temperatures is to be determined with given $\Delta \text{H}$ and $\Delta {\text{S}}^{\circ }$

values.

Concept introduction:

The enthalpy of reaction is calculated by the expression as follows:

$\Delta {\text{H}}^{\circ }=\sum \Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{products}\right)-\sum \Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{reactants}\right)$

Here, $\Delta {\text{H}}^{\circ }$ is the standard change in enthalpy of reaction, $\sum \Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{reactants}\right)$ is the sum of standard change in enthalpy of reactants, $\sum \Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{products}\right)$ is the standard change in enthalpy of products.

The value of $\Delta {\text{S}}^{\circ }$

for the given reaction is calculated by using the relation given below:

$\Delta {\text{S}}^{\circ }=\sum {\text{S}}^{\circ }\left(\text{product}\right)-\sum {\text{S}}^{\circ }\left(\text{reactant}\right)$

Here, $\Delta {\text{S}}^{\circ }$

is the standard change in entropy,

The value of K is calculated by using the relation given below:

${\text{K=e}}^{\frac{-\Delta {\text{G}}^{\circ }}{\text{RT}}}$

Here K is the equilibrium constant, R is the gas constant, T is the temperature, and $\Delta {\text{G}}^{\circ }$

is the standard change in energy.

The value of $\Delta {\text{G}}^{\circ }$ is calculated by using the relation given below:

$\Delta {\text{G}}^{\circ }=\Delta {\text{H}}^{\circ }-\text{T}\Delta \text{S}$

Answer to Problem 97AP

Solution: K at $298\text{ K}$

is $9.61\times {10}^{-22}$

and at $373\text{ K}$

is $1.2\times {10}^{-15}$.

Explanation of Solution

Given information: The given reaction is $\text{C}\left(s\right){\text{+CO}}_2\left(g\right)\to 2\text{CO}\left(g\right)$.

The value of $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{CO}\right)$ from appendix $2$

is $\text{-110}\text{.5 kJ/mol}$.

The value of $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)$ from appendix $2$

is $-393.5\text{ kJ/mol}$.

The value of ${\text{S}}^{\circ }\left(\text{C}\right)$

from appendix 2 is $5.69\text{ J/Kmol}$.

The value of ${\text{S}}^{\circ }\left(\text{CO}\right)$

from appendix 2 is $\text{197}\text{.9 J/Kmol}$.

The value of ${\text{S}}^{\circ }\left({\text{CO}}_2\right)$

from appendix 2 is $\text{213}\text{.6 J/Kmol}$.

The value of $\Delta {\text{H}}^{\circ }$

for the given reaction is calculated by using the relation given below:

$\Delta {\text{H}}^{\circ }=2\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{CO}\right)-\left[\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{C}\right)+\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)\right]$

Here, $\Delta {\text{H}}^{\circ }$

is the standard change in enthalpy, $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{CO}\right)$

is the standard change in enthalpy of CO, $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{C}\right)$

is the standard change in enthalpy of $\text{C}$

and $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)$

is the standard change in enthalpy of ${\text{CO}}_2$.

Substitute $0$

for $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{C}\right)$, $-110.5\text{kJ/mol}$ for $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left(\text{CO}\right)$, and $-393.5\text{kJ/mol}$

for $\Delta {\text{H}}_{\text{f}}{}^{\circ }\left({\text{CO}}_2\right)$ in the equation as follows:

$\begin{array}{c}\Delta {\text{H}}^{\circ }=\left(2\right)\left(-110.5\text{kJ/mol}\right)-\left[0+\left(1\right)\left(-393.5\text{kJ/mol}\right)\right]\\ =172.5\text{ kJ/mol}\\ \text{=172}\text{.5}\times {\text{10}}^3\text{ J/mol}\end{array}$

Therefore, the value of $\Delta {\text{H}}^{\circ }$

for the given reaction is $\text{172}\text{.5}\times {\text{10}}^3\text{ J/mol}$.

The value of $\Delta {\text{S}}^{\circ }$

for the given reaction is calculated by using the relation given below:

$\Delta {\text{S}}^{\circ }=2{\text{S}}^{\circ }\left(\text{CO}\right)-\left[{\text{S}}^{\circ }\left(\text{C}\right)+{\text{S}}^{\circ }\left({\text{CO}}_2\right)\right]$

Here, $\Delta {\text{S}}^{\circ }$

is the standard change in entropy, ${\text{S}}^{\circ }\left(\text{CO}\right)$

is the standard entropy of CO, ${\text{S}}^{\circ }\left(\text{C}\right)$

is the standard entropy of $\text{C}$, and ${\text{S}}^{\circ }\left({\text{CO}}_2\right)$

is the standard entropy of ${\text{CO}}_2$.

Substitute $5.69\text{ J/Kmol}$

for ${\text{S}}^{\circ }\left(\text{C}\right)$, $\text{197}\text{.9 J/Kmol}$ for ${\text{S}}^{\circ }\left(\text{CO}\right)$, and $\text{213}\text{.6 J/Kmol}$

for ${\text{S}}^{\circ }\left({\text{CO}}_2\right)$ in the equation as follows:

$\begin{array}{c}\Delta {\text{S}}^{\circ }=\left(2\right)\left(\text{197}\text{.9 J/Kmol}\right)-\left[5.69\text{ J/Kmol}+\text{213}\text{.6 J/Kmol}\right]\\ =176.5\text{ J/Kmol}\end{array}$

Therefore, the value of $\Delta {\text{S}}^{\circ }$

for the given reaction is $176.5\text{ J/mol}$.

The value of $\Delta {\text{G}}^{\circ }$ is calculated by using the relation given below:

$\Delta {\text{G}}^{\circ }=\Delta {\text{H}}^{\circ }-\text{T}\Delta \text{S}$

The value of $\Delta {\text{G}}^{\circ }$

at $298\text{ K}\left(25{}^{\circ }\text{C}\right)$

is as follows:

Substitute $176.5\text{ J/mol}$

for $\Delta {\text{S}}^{\circ }$, $172.5\times {\text{10}}^3\text{J/mol}$ for $\Delta {\text{H}}^{\circ }$, and $298\text{ K}$

for $\text{T}$ in the above equation as follows:

$\begin{array}{c}\Delta {\text{G}}^{\circ }=172.5\times {\text{10}}^3\text{J/mol}-\left(298\text{ K}\right)\left(176.5\text{ J/mol}\right)\\ =1.99\times {10}^5\text{ J/mol}\end{array}$

The value of K is calculated by using the relation given below:

${\text{K=e}}^{\frac{-\Delta {\text{G}}^{\circ }}{\text{RT}}}$

Here K is the equilibrium constant, R is the gas constant, T is the temperature, and $\Delta {\text{G}}^{\circ }$

is the standard change in energy.

Substitute $1.99\times {10}^5\text{ J/mol}$

for $\Delta {\text{G}}^{\circ }$, $8.314\text{ J/mol}\text{.K}$ for R, and $298\text{ K}$

for $\text{T}$ in the above equation as follows:

$\begin{array}{c}{\text{K=e}}^{\frac{-1.99\times {10}^5\text{ J/mol}}{\left(8.314\text{ J/mol}\text{.K}\right)\left(298\text{ K}\right)}}\\ ={\text{e}}^{\frac{-1.99\times {10}^5\text{ J/mol}}{2477.57\text{ J/mol}}}\\ =9.61\times {10}^{-22}\end{array}$

The value of $\Delta {\text{G}}^{\circ }$ at $\text{373 K}\left({\text{100 }}^{\circ }\text{C}\right)$ is as follows:

Substitute $176.5\text{ J/mol}$

for $\Delta {\text{S}}^{\circ }$, $172.5\times {\text{10}}^3\text{J/mol}$ for $\Delta {\text{H}}^{\circ }$, and $\text{373 K}$

for $\text{T}$ in the above equation as follows:

$\begin{array}{c}\Delta {\text{G}}^{\circ }=172.5\times {\text{10}}^3\text{J/mol}-\left(\text{373 K}\right)\left(176.5\text{ J/mol}\right)\\ =1.07\times {10}^5\text{ J/mol}\end{array}$

The value of K is calculated by using the relation given below:

${\text{K=e}}^{\frac{-\Delta {\text{G}}^{\circ }}{\text{RT}}}$

Here, K is the equilibrium constant, R is the gas constant, T is the temperature, and $\Delta {\text{G}}^{\circ }$

is the standard change in energy.

Substitute $1.07\times {10}^5\text{ J/mol}$

for $\Delta {\text{G}}^{\circ }$, $8.314\text{ J/mol}\text{.K}$ for R, and $\text{373 K}$

for $\text{T}$ in the above equation as follows:

$\begin{array}{c}{\text{K=e}}^{\frac{-\left(1.07\times {10}^5\text{ J/mol}\right)}{\left(8.314\text{ J/mol}\text{.K}\right)\left(\text{373 K}\right)}}\\ ={\text{e}}^{\frac{-1.07\times {10}^5\text{ J/mol}}{\text{3101}\text{.122 J/mol}}}\\ =1.2\times {10}^{-15}\end{array}$

Conclusion

Therefore, the value of equilibrium constant K at $298\text{ K}$ is $9.61\times {10}^{-22}$ and at $373\text{ K}$ is $1.2\times {10}^{-15}$.