Chapter 16, Problem 54E

The major industrial use of hydrogen is in the production of ammonia by the Haber process:

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

a. Using data from Appendix 4, calculate ∆H°, ∆S°, and ∆G° for the Haber process reaction.

b. Is the reaction spontaneous at standard conditions?

c. At what temperatures is the reaction spontaneous at standard conditions? Assume ∆H° and ∆S° do not depend on temperature.

Interpretation Introduction

Interpretation: The values of $\Delta S^{\circ },\Delta H^{\circ }$ and $\Delta G^{\circ }$ is to be calculated at $25°\text{C}$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $\Delta S^{\circ }$ is,

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

The expression for $\Delta H^{\circ }$ is,

$\Delta H^{\circ }=\sum n_{\text{p}}\Delta H^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H^{\circ }{}_{\left(\text{reactant}\right)}$

The expression for $\Delta G^{\circ }$ is,

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

Explanation of Solution

The stated reaction is,

$3H_2\left(g\right)+N_2\left(g\right)\stackrel{}{\to }2N{H}_3\left(g\right)$

Refer to Appendix $4$.

The value of $\Delta S^{\circ }\left(\text{J/K}\cdot \text{mol}\right)$ for the given reactant and product is,

Molecules	$\Delta S^{\circ }\left(\text{J/K}\cdot \text{mol}\right)$
${\text{NH}}_{\text{3}}(g)$	$\text{193}$
$N_2\left(g\right)$	$\text{192}$
${\text{H}}_{\text{2}}(g)$	$\text{131}$

The formula of $\Delta S^{\circ }$ is,

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

Where,

• $\Delta S^{\circ }$ is the standard entropy of reaction.
• $n_{\text{p}}$ is the number of moles of each product.
• $n_{\text{r}}$ is the number of moles each reactant.
• $\Delta S^{\circ }{}_{\left(\text{product}\right)}$ is the standard entropy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta S^{\circ }{}_{\left(\text{reactant}\right)}$ is the standard entropy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta S^{\circ }=\sum n_{\text{p}}\Delta S^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S^{\circ }{}_{\left(\text{reactant}\right)}\\ =\left[\text{2}\left(\text{193}\right)-\left\{\text{3}\left(\text{131}\right)+\left(\text{192}\right)\right\}\right]\text{J/K}\\ =\underset{\_}{-199J/K}\end{array}$

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

Molecules	$\Delta H^{\circ }\left(\text{kJ/mol}\right)$
${\text{NH}}_{\text{3}}(g)$	$-\text{46}$
$N_2\left(g\right)$	$\text{0}$
${\text{H}}_{\text{2}}(g)$	$\text{0}$

The formula of $\Delta H^{\circ }$ is,

$\Delta H^{\circ }=\sum n_{\text{p}}\Delta H^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H^{\circ }{}_{\left(\text{reactant}\right)}$

Where,

• $\Delta H^{\circ }$ is the standard enthalpy of reaction.
• $n_{\text{p}}$ is the number of moles of each product.
• $n_{\text{r}}$ is the number of moles each reactant.
• $\Delta H^{\circ }{}_{\left(\text{product}\right)}$ is the standard enthalpy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta H^{\circ }{}_{\left(\text{reactant}\right)}$ is the standard enthalpy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

$\begin{array}{c}\Delta H^{\circ }=\sum n_{\text{p}}\Delta H^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H^{\circ }{}_{\left(\text{reactant}\right)}\\ =\left[\text{2}\left(-\text{46}\right)-\left\{\text{3}\left(\text{0}\right)+\left(\text{0}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{-92kJ}\end{array}$

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

Molecules	$\Delta G^{\circ }\left(\text{kJ/mol}\right)$
${\text{NH}}_{\text{3}}(g)$	$-\text{17}$
$N_2\left(g\right)$	$\text{0}$
${\text{H}}_{\text{2}}(g)$	$\text{0}$

The formula of $\Delta G^{\circ }$ is,

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

Where,

• $\Delta G^{\circ }$ is the standard Gibb’s free energy of reaction.
• $n_{\text{p}}$ is the number of moles of each product.
• $n_{\text{r}}$ is the number of moles each reactant.
• $\Delta G^{\circ }{}_{\left(\text{product}\right)}$ is the standard Gibb’s free energy of product at a pressure of $\text{1}\text{atm}$.
• $\Delta G^{\circ }{}_{\left(\text{reactant}\right)}$ is the standard Gibb’s free energy of reactant at a pressure of $\text{1}\text{atm}$.

Substitute all values from the table in the above equation.

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

Interpretation Introduction

Interpretation: The values of $\Delta S^{\circ },\Delta H^{\circ }$ and $\Delta G^{\circ }$ is to be calculated at $25°\text{C}$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $\Delta S^{\circ }$ is,

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

The expression for $\Delta H^{\circ }$ is,

$\Delta H^{\circ }=\sum n_{\text{p}}\Delta H^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H^{\circ }{}_{\left(\text{reactant}\right)}$

The expression for $\Delta G^{\circ }$ is,

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

Explanation of Solution

A reaction is said to be spontaneous if the value of $\Delta G$ is negative.

Since, the value of $\Delta G$ for the given reaction is $-\text{34}\text{kJ}$, therefore, the given reaction is spontaneous.

The formula of $\Delta G$ is,

$\Delta G=\Delta H-T\Delta S$

Where,

• $\Delta H$ is the enthalpy of reaction.
• $\Delta G$ is the free energy change.
• $T$ is the temperature.
• $\Delta S$ is the entropy of reaction.

Since, the value of $\Delta H$ is negative, therefore, the reaction is spontaneous at lower temperature.

Interpretation Introduction

Interpretation: The values of $\Delta S^{\circ },\Delta H^{\circ }$ and $\Delta G^{\circ }$ is to be calculated at $25°\text{C}$ for the given reaction. If the reaction is spontaneous is to be stated. The temperature at which the reaction is spontaneous is to be calculated.

Concept introduction:

The expression for $\Delta S^{\circ }$ is,

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

The expression for $\Delta H^{\circ }$ is,

$\Delta H^{\circ }=\sum n_{\text{p}}\Delta H^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H^{\circ }{}_{\left(\text{reactant}\right)}$

The expression for $\Delta G^{\circ }$ is,

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

Answer to Problem 54E

Solutions are as follows.

Explanation of Solution

The value of $\Delta S^{\circ }$ for the given reaction is $-\text{199}\text{J/K}$.

The value of $\Delta H^{\circ }$ for the given reaction is $-\text{92}\text{kJ}$.

The negative value of $\Delta H^{\circ }$ suggests that the give reaction is exothermic. But $\Delta S^{\circ }$ favors opposite process. These opposite tendencies cancel out that means that the reaction has achieved equilibrium. The value of $\Delta G^{\circ }$ at equilibrium is zero.

The formula of $\Delta G$ is,

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

Where,

• $\Delta H^{\circ }$ is the enthalpy of reaction.
• $\Delta G^{\circ }$ is the free energy change.
• $T$ is the temperature.
• $\Delta S^{\circ }$ is the entropy of reaction.

Substitute the values of $\Delta S^{\circ },\Delta G^{\circ }$ and $\Delta H^{\circ }$ in the above equation.

$\begin{array}{c}\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }\\ T=\frac{\Delta H^{\circ }}{\Delta S^{\circ }}-\Delta G^{\circ }\\ =\frac{-\text{92}\times {\text{10}}^{\text{3}}\text{J}}{-\text{199}\text{J/K}}-\text{0}\\ =\underset{\_}{462.3K}\end{array}$

The given reaction will be spontaneous below $\underset{\_}{462.3K}$.