Chapter 17, Problem 74E

Consider the following reaction:

${\text{N}}_2\left(g\right)+3{\text{H}}_2\left(g\right)\rightleftharpoons 2{\text{NH}}_3\left(g\right)$

Calculate ∆G for this reaction under the following conditions (assume an uncertainty of ± 1 in all quantities):

a. $T=298\text{K},P_{{\text{N}}_{\text{2}}}=P_{{\text{H}}_{\text{2}}}=200\text{atm},P_{{\text{NH}}_{\text{3}}}=50\text{atm}$

b. $T=298\text{K},P_{{\text{N}}_{\text{2}}}=200\text{atm},P_{{\text{H}}_{\text{2}}}=600\text{atm, }{P}_{{\text{NH}}_{\text{3}}}=200\text{atm}$

Interpretation Introduction

Interpretation: The reaction of ${\text{N}}_{\text{2}}$ and ${\text{3H}}_{\text{2}}$ is given. The value of $\Delta G$ is to be calculated for given conditions.

Concept introduction: Reaction quotient, $Q$, is the measure of concentration or partial pressure of reactants and products of a system before the equilibrium point is reached.

The expression for free energy change is,

$\Delta G=\Delta G°+RT\ln \left(Q\right)$

To determine: The value of $\Delta G$ for the given values.

Answer to Problem 74E

Answer

The value of $\Delta G$ for the given reaction is $\underset{\_}{-67kJ/mol}$.

Explanation of Solution

Explanation

The reaction that takes place is,

${\text{N}}_{\text{2}}\left(g\right)+{\text{3H}}_{\text{2}}\left(g\right)\stackrel{}{\to }{\text{2NH}}_{\text{3}}\left(g\right)$

Refer to Appendix $\text{4}$.

The value of $\Delta G°\left(\text{kJ/mol}\right)$ for the given reactant and product is,

Molecules	$\Delta G°\left(\text{kJ/mol}\right)$
${\text{H}}_{\text{2}}(g)$	$\text{0}$
${\text{N}}_{\text{2}}(g)$	$\text{0}$
${\text{NH}}_{\text{3}}(g)$	$-\text{17}$

The formula of $\Delta G°$ is,

$\Delta G°=\sum n_{\text{p}}\Delta G{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G{°}_{\left(\text{reactant}\right)}$

Where,

• $n_{\text{p}}$ is number of moles of each product.
• $n_{\text{r}}$ is number of moles each reactant.
• $\Delta G{°}_{\left(\text{product}\right)}$ is free energy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta G{°}_{\left(\text{reactant}\right)}$ is free energy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta G°=\sum n_{\text{p}}\Delta G{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G{°}_{\left(\text{reactant}\right)}\\ =\left[\text{2}\left(-\text{17}\right)-\text{1}\left(\text{0}\right)-\text{3}\left(\text{0}\right)\right]\text{kJ}\\ =\underset{\_}{-34kJ}\end{array}$

Given

Pressure of ${\text{H}}_{\text{2}}$, $P_{{\text{H}}_{\text{2}}}$, is $\text{200}\text{atm}$.

Pressure of ${\text{N}}_{\text{2}}$, $P_{{\text{N}}_{\text{2}}}$, is $\text{200}\text{atm}$.

Pressure of ${\text{NH}}_{\text{3}}$, $P_{{\text{NH}}_{\text{3}}}$, is $\text{50}\text{atm}$.

Formula

The reaction quotient, is expressed by the formula,

$Q=\frac{\text{Pressure}\text{of}\text{products}}{\text{Pressure}\text{of}\text{reactants}}$

Where,

• $Q$ is the reaction quotient.

The reaction quotient expression for the given reaction is,

$Q=\frac{{\left(P_{{\text{NH}}_{\text{3}}}\right)}^2}{\left(P_{{\text{N}}_{\text{2}}}\right){\left(P_{{\text{H}}_{\text{2}}}\right)}^3}$

Substitute the values of $P_{{\text{H}}_{\text{2}}},P_{{\text{N}}_{\text{2}}}$ and $P_{{\text{NH}}_{\text{3}}}$ in the above expression.

$\begin{array}{c}Q=\frac{{\left(P_{{\text{NH}}_{\text{3}}}\right)}^2}{\left(P_{{\text{N}}_{\text{2}}}\right){\left(P_{{\text{H}}_{\text{2}}}\right)}^3}\\ =\frac{{\left(50\right)}^2}{\left(200\right){\left(200\right)}^3}\\ =\underset{\_}{1.6\times 10^{-6}}\end{array}$

Reaction quotient is $\text{1}\text{.6}\times {\text{10}}^{-\text{6}}$.

The value of $\Delta G°$ is $-\text{34}\text{kJ}$.

Temperature is $\text{298}\text{K}$.

The conversion of kilo-joule $\left(\text{kJ}\right)$ into joule $\left(\text{J}\right)$ is done as,

$\text{1}\text{kJ}={\text{10}}^{\text{3}}\text{J}$

Hence,

The conversion of $-\text{34}\text{kJ}$ into joule is,

$\begin{array}{c}-\text{34}\text{kJ}=\left(-\text{34}\times {\text{10}}^{\text{3}}\right)\text{J}\\ =-\text{3}\text{.4}\times {\text{10}}^{\text{4}}\text{J}\end{array}$

Formula

The expression for free energy change is,

$\Delta G=\Delta G°+RT\ln \left(Q\right)$

Where,

• $\Delta G$ is free energy change for a reaction at specified pressure.
• $R$ is rate law constant $(\text{8}\text{.3145}\text{J/K})$.
• $T$ is absolute temperature.
• $\Delta G°$ is free energy change for a reaction at a pressure of $\text{1}\text{atm}$.

Substitute the values of $R,\Delta G°,T$ and $Q$ in the above expression.

$\begin{array}{c}\Delta G=\Delta G°+RT\ln \left(Q\right)\\ =\left(-3.4\times {10}^4\right)+(\text{8}\text{.3145}\text{J/K})\left(298\text{K}\right)\ln \left(1.6\times {10}^{-6}\right)\\ =-6.7\times {10}^4\text{J/mol}\end{array}$

The conversion of joule per mole $\left(\text{J/mol}\right)$ into kilo-joule per mole $\left(\text{kJ/mol}\right)$ is done as,

$\text{1}\text{J/mol}={\text{10}}^{-\text{3}}\text{kJ/mol}$

Hence,

The conversion of $-6.7\times {10}^4\text{J/mol}$ into kilo-joule per mole is,

$\begin{array}{c}-\text{6}\text{.7}\times {\text{10}}^{\text{4}}\text{J/mol}=\left(-\text{6}\text{.7}\times {\text{10}}^{\text{4}}\right)\text{kJ/mol}\\ =\underset{\_}{-67kJ/mol}\end{array}$

Interpretation Introduction

Interpretation: The reaction of ${\text{N}}_{\text{2}}$ and ${\text{3H}}_{\text{2}}$ is given. The value of $\Delta G$ is to be calculated for given conditions.

Concept introduction: Reaction quotient, $Q$, is the measure of concentration or partial pressure of reactants and products of a system before the equilibrium point is reached.

The expression for free energy change is,

$\Delta G=\Delta G°+RT\ln \left(Q\right)$

To determine: The value of $\Delta G$ for the given values.

Answer to Problem 74E

Answer

The value of $\Delta G$ for the given reaction is $\underset{\_}{-68.4kJ/mol}$.

Explanation of Solution

Explanation

The reaction that takes place is,

${\text{N}}_{\text{2}}\left(g\right)+{\text{3H}}_{\text{2}}\left(g\right)\stackrel{}{\to }{\text{2NH}}_{\text{3}}\left(g\right)$

Refer to Appendix $\text{4}$.

The value of $\Delta G°\left(\text{kJ/mol}\right)$ for the given reactant and product is,

Molecules	$\Delta G°\left(\text{kJ/mol}\right)$
${\text{H}}_{\text{2}}(g)$	$\text{0}$
${\text{N}}_{\text{2}}(g)$	$\text{0}$
${\text{NH}}_{\text{3}}(g)$	$-\text{17}$

The formula of $\Delta G°$ is,

$\Delta G°=\sum n_{\text{p}}\Delta G{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G{°}_{\left(\text{reactant}\right)}$

Where,

• $n_{\text{p}}$ is number of moles of each product.
• $n_{\text{r}}$ is number of moles each reactant.
• $\Delta G{°}_{\left(\text{product}\right)}$ is free energy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta G{°}_{\left(\text{reactant}\right)}$ is free energy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta G°=\sum n_{\text{p}}\Delta G{°}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G{°}_{\left(\text{reactant}\right)}\\ =\left[\text{2}\left(-\text{17}\right)-\text{1}\left(\text{0}\right)-\text{3}\left(\text{0}\right)\right]\text{kJ}\\ =\underset{\_}{-34kJ}\end{array}$

Given

Pressure of ${\text{H}}_{\text{2}}$, $P_{{\text{H}}_{\text{2}}}$, is $\text{600}\text{atm}$.

Pressure of ${\text{N}}_{\text{2}}$, $P_{{\text{N}}_{\text{2}}}$, is $\text{200}\text{atm}$.

Pressure of ${\text{NH}}_{\text{3}}$, $P_{{\text{NH}}_{\text{3}}}$, is $\text{200}\text{atm}$.

Formula

The reaction quotient is expressed by the formula,

$Q=\frac{\text{Pressure}\text{of}\text{products}}{\text{Pressure}\text{of}\text{reactants}}$

Where,

• $Q$ is the reaction quotient.

The reaction quotient expression for the given reaction is,

$Q=\frac{{\left(P_{{\text{NH}}_{\text{3}}}\right)}^2}{\left(P_{{\text{N}}_{\text{2}}}\right){\left(P_{{\text{H}}_{\text{2}}}\right)}^3}$

Substitute the values of $P_{{\text{H}}_{\text{2}}},P_{{\text{N}}_{\text{2}}}$ and $P_{{\text{NH}}_{\text{3}}}$ in the above expression.

$\begin{array}{c}Q=\frac{{\left(P_{{\text{NH}}_{\text{3}}}\right)}^2}{\left(P_{{\text{N}}_{\text{2}}}\right){\left(P_{{\text{H}}_{\text{2}}}\right)}^3}\\ =\frac{{\left(200\right)}^2}{\left(200\right){\left(600\right)}^3}\\ =\underset{\_}{9.26\times 10^{-7}}\end{array}$

Reaction quotient is $\text{9}\text{.26}\times {\text{10}}^{-\text{7}}$.

The value of $\Delta G°$ is $-\text{34}\text{kJ}$.

Temperature is $\text{298}\text{K}$.

The conversion of kilo-joule $\left(\text{kJ}\right)$ into joule $\left(\text{J}\right)$ is done as,

$\text{1}\text{kJ}={\text{10}}^{\text{3}}\text{J}$

Hence,

The conversion of $-\text{34}\text{kJ}$ into joule is,

$\begin{array}{c}-\text{34}\text{kJ}=\left(-\text{34}\times {\text{10}}^{\text{3}}\right)\text{J}\\ =-\text{3}\text{.4}\times {\text{10}}^{\text{4}}\text{J}\end{array}$

Formula

The expression for free energy change is,

$\Delta G=\Delta G°+RT\ln \left(Q\right)$

Where,

• $\Delta G$ is free energy change for a reaction at specified pressure.
• $R$ is rate law constant $(\text{8}\text{.3145}\text{J/K})$.
• $T$ is absolute temperature.
• $\Delta G°$ is free energy change for a reaction at a pressure of $\text{1}\text{atm}$.

Substitute the values of $R,\Delta G°,T$ and $Q$ in the above expression.

$\begin{array}{c}\Delta G=\Delta G°+RT\ln \left(Q\right)\\ =\left(-3.4\times {10}^4\right)+(\text{8}\text{.3145}\text{J/K})\left(298\text{K}\right)\ln \left(\text{9}\text{.26}\times {\text{10}}^{-\text{7}}\right)\\ =-6.84\times {10}^4\text{J/mol}\end{array}$

The conversion of joule per mole $\left(\text{J/mol}\right)$ into kilo-joule per mole $\left(\text{kJ/mol}\right)$ is done as,

$\text{1}\text{J/mol}={\text{10}}^{-\text{3}}\text{kJ/mol}$

Hence,

The conversion of $-6.84\times {10}^4\text{J/mol}$ into kilo-joule per mole is,

$\begin{array}{c}-\text{6}\text{.84}\times {\text{10}}^{\text{4}}\text{J/mol}=\left(-\text{6}\text{.84}\times {\text{10}}^{\text{4}}\right)\text{kJ/mol}\\ =\underset{\_}{-68.4kJ/mol}\end{array}$