Chapter 17, Problem 57E

From data in Appendix 4, calculate ∆H°, ∆S°, and ∆G° for each of the following reactions at 25°C.

a. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

b. $6{\text{CO}}_2\left(g\right)+6{\text{H}}_2\text{O}\left(l\right)\to \underset{\text{Glucose}}{C_6{\text{H}}_{12}{\text{O}}_6\left(s\right)}+6{\text{O}}_2\left(g\right)$

c. P₄O₁₀(s) + 6H₂O(l) → 4H₃PO₄(s)

d. HCl(g) + NH₃(g) → NH₄Cl(s)

Interpretation Introduction

Interpretation: The values of $\Delta S^{\circ },\Delta H^{\circ }$ and $\Delta G^{\circ }$ is to be calculated at $25°\text{C}$ in each case.

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

Explanation

The stated reaction is,

$C{H}_4\left(g\right)+2O_2\left(g\right)\stackrel{}{\to }C{O}_2\left(g\right)+2H_{2}O\left(g\right)$

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	$\Delta S^{\circ }\left(\text{J/K}\cdot \text{mol}\right)$
${\text{O}}_{\text{2}}(g)$	$\text{205}$
$C{H}_4\left(g\right)$	$\text{186}$
${\text{CO}}_{\text{2}}(g)$	$\text{214}$
${\text{H}}_{\text{2}}\text{O}(g)$	$\text{189}$

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.

$\begin{array}{c}\Delta S^{\circ }=\sum n_{\text{p}}\Delta S^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S^{\circ }{}_{\left(\text{reactant}\right)}\\ =\left[\left(\text{214}\right)+\text{2}\left(\text{189}\right)-\left\{\text{2}\left(\text{205}\right)+\left(\text{186}\right)\right\}\right]\text{J/K}\\ =\underset{\_}{-4J/K}\end{array}$

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

Molecules	$\Delta H^{\circ }\left(\text{kJ/mol}\right)$
${\text{O}}_{\text{2}}(g)$	$\text{0}$
$C{H}_4\left(g\right)$	$-\text{75}$
${\text{CO}}_{\text{2}}(g)$	$-\text{393}\text{.5}$
${\text{H}}_{\text{2}}\text{O}(g)$	$-\text{242}$

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.

$\begin{array}{c}\Delta H^{\circ }=\sum n_{\text{p}}\Delta H^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H^{\circ }{}_{\left(\text{reactant}\right)}\\ =\left[\left(-\text{393}\text{.5}\right)+\text{2}\left(-\text{242}\right)-\left\{\left(-\text{75}\right)+\text{2}\left(\text{0}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{-802.5kJ}\end{array}$

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

Molecules	$\Delta G^{\circ }\left(\text{kJ/mol}\right)$
${\text{O}}_{\text{2}}(g)$	$\text{0}$
$C{H}_4\left(g\right)$	$-\text{51}$
${\text{CO}}_{\text{2}}(g)$	$-\text{394}$
${\text{H}}_{\text{2}}\text{O}(g)$	$-\text{229}$

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.

$\begin{array}{c}\Delta G^{\circ }=\sum n_{\text{p}}\Delta G^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G^{\circ }{}_{\left(\text{reactant}\right)}\\ =\left[\left(-\text{394}\right)+\text{2}\left(-\text{229}\right)-\left\{\left(-\text{51}\right)+\text{2}\left(\text{0}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{-801kJ}\end{array}$

Interpretation Introduction

Interpretation: The values of $\Delta S^{\circ },\Delta H^{\circ }$ and $\Delta G^{\circ }$ is to be calculated at $25°\text{C}$ in each case.

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

Explanation

The stated reaction is,

$6C{O}_2\left(g\right)+6H_{2}O\left(g\right)\stackrel{}{\to }{C}_6{H}_{12}{O}_6\left(s\right)+6O_2\left(g\right)$

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	$\Delta S^{\circ }\left(\text{J/K}\cdot \text{mol}\right)$
${\text{O}}_{\text{2}}(g)$	$\text{205}$
$C_6{H}_{12}{O}_6\left(s\right)$	$\text{212}$
${\text{CO}}_{\text{2}}(g)$	$\text{214}$
${\text{H}}_{\text{2}}\text{O}(l)$	$\text{70}$

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.

$\begin{array}{c}\Delta S^{\circ }=\sum n_{\text{p}}\Delta S^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S^{\circ }{}_{\left(\text{reactant}\right)}\\ =\left[\left(\text{212}\right)+\text{6}\left(\text{205}\right)-\left\{\text{6}\left(\text{214}\right)+\text{6}\left(\text{70}\right)\right\}\right]\text{J/K}\\ =\underset{\_}{-262J/K}\end{array}$

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

Molecules	$\Delta H^{\circ }\left(\text{kJ/mol}\right)$
${\text{O}}_{\text{2}}(g)$	$\text{0}$
$C_6{H}_{12}{O}_6\left(s\right)$	$-\text{1275}$
${\text{CO}}_{\text{2}}(g)$	$-\text{393}\text{.5}$
${\text{H}}_{\text{2}}\text{O}(l)$	$-\text{286}$

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.

$\begin{array}{c}\Delta H^{\circ }=\sum n_{\text{p}}\Delta H^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H^{\circ }{}_{\left(\text{reactant}\right)}\\ =\left[\left(-\text{1275}\right)+\text{6}\left(\text{0}\right)-\left\{\text{6}\left(-\text{393}\text{.5}\right)+\text{6}\left(-\text{286}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{2802kJ}\end{array}$

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

Molecules	$\Delta G^{\circ }\left(\text{kJ/mol}\right)$
${\text{O}}_{\text{2}}(g)$	$\text{0}$
$C_6{H}_{12}{O}_6\left(s\right)$	$-\text{911}$
${\text{CO}}_{\text{2}}(g)$	$-\text{394}$
${\text{H}}_{\text{2}}\text{O}(l)$	$-\text{237}$

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.

$\begin{array}{c}\Delta G^{\circ }=\sum n_{\text{p}}\Delta G^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G^{\circ }{}_{\left(\text{reactant}\right)}\in \\ =\left[\left(-\text{911}\right)+\text{6}\left(\text{0}\right)-\left\{\text{6}\left(-\text{394}\right)+\text{6}\left(-\text{237}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{2875kJ}\end{array}$

Interpretation Introduction

Interpretation: The values of $\Delta S^{\circ },\Delta H^{\circ }$ and $\Delta G^{\circ }$ is to be calculated at $25°\text{C}$ in each case.

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

Explanation

The stated reaction is,

$P_4{O}_{10}\left(s\right)+6H_{2}O\left(g\right)\stackrel{}{\to }4H_{3}P{O}_4\left(s\right)$

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	$\Delta S^{\circ }\left(\text{J/K}\cdot \text{mol}\right)$
$H_{3}P{O}_4\left(s\right)$	$\text{110}$
$P_4{O}_{10}\left(s\right)$	$\text{229}$
${\text{H}}_{\text{2}}\text{O }(l)$	$\text{70}$

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.

$\begin{array}{c}\Delta S^{\circ }=\sum n_{\text{p}}\Delta S^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S^{\circ }{}_{\left(\text{reactant}\right)}\\ =\left[\text{4}\left(\text{110}\right)-\left\{\left(\text{229}\right)+\text{6}\left(\text{70}\right)\right\}\right]\text{J/K}\\ =\underset{\_}{-209J/K}\end{array}$

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

Molecules	$\Delta H^{\circ }\left(\text{kJ/mol}\right)$
$H_{3}P{O}_4\left(s\right)$	$-\text{1279}$
$P_4{O}_{10}\left(s\right)$	$-\text{2984}$
${\text{H}}_{\text{2}}\text{O }(l)$	$-\text{286}$

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.

$\begin{array}{c}\Delta H^{\circ }=\sum n_{\text{p}}\Delta H^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H^{\circ }{}_{\left(\text{reactant}\right)}\\ =\left[\text{4}\left(-\text{1279}\right)-\left\{\left(-\text{2984}\right)+\text{6}\left(-\text{286}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{-416kJ}\end{array}$

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

Molecules	$\Delta G^{\circ }\left(\text{kJ/mol}\right)$
$H_{3}P{O}_4\left(s\right)$	$-\text{1119}$
$P_4{O}_{10}\left(s\right)$	$-\text{2698}$
${\text{H}}_{\text{2}}\text{O }(l)$	$-\text{237}$

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.

$\begin{array}{c}\Delta G^{\circ }=\sum n_{\text{p}}\Delta G^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G^{\circ }{}_{\left(\text{reactant}\right)}\in \\ =\left[\text{4}\left(-\text{1119}\right)-\left\{\left(-\text{2698}\right)+\text{6}\left(-\text{237}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{-356kJ}\end{array}$

Interpretation Introduction

Interpretation: The values of $\Delta S^{\circ },\Delta H^{\circ }$ and $\Delta G^{\circ }$ is to be calculated at $25°\text{C}$ in each case.

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

Explanation

The stated reaction is,

$P_4{O}_{10}\left(s\right)+6H_{2}O\left(g\right)\stackrel{}{\to }4H_{3}P{O}_4\left(s\right)$

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	$\Delta S^{\circ }\left(\text{J/K}\cdot \text{mol}\right)$
$N{H}_{4}Cl$	$\text{96}$
$HCl\left(g\right)$	$\text{187}$
${\text{NH}}_{\text{3}}(g)$	$\text{193}$

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.

$\begin{array}{c}\Delta S^{\circ }=\sum n_{\text{p}}\Delta S^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta S^{\circ }{}_{\left(\text{reactant}\right)}\\ =\left[\left(\text{96}\right)-\left\{\left(\text{193}\right)+\left(\text{187}\right)\right\}\right]\text{J/K}\\ =\underset{\_}{-284J/K}\end{array}$

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

Molecules	$\Delta H^{\circ }\left(\text{kJ/mol}\right)$
$N{H}_{4}Cl$	$-\text{314}$
$HCl\left(g\right)$	$-\text{92}$
${\text{NH}}_{\text{3}}(g)$	$-\text{46}$

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.

$\begin{array}{c}\Delta H^{\circ }=\sum n_{\text{p}}\Delta H^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta H^{\circ }{}_{\left(\text{reactant}\right)}\\ =\left[\left(-\text{314}\right)-\left\{\left(-\text{92}\right)+\left(-\text{46}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{-176kJ}\end{array}$

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

Molecules	$\Delta G^{\circ }\left(\text{kJ/mol}\right)$
$N{H}_{4}Cl$	$-\text{203}$
$HCl\left(g\right)$	$-\text{95}$
${\text{NH}}_{\text{3}}(g)$	$-\text{17}$

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.

$\begin{array}{c}\Delta G^{\circ }=\sum n_{\text{p}}\Delta G^{\circ }{}_{\left(\text{product}\right)}-\sum n_{\text{f}}\Delta G^{\circ }{}_{\left(\text{reactant}\right)}\in \\ =\left[\left(-\text{203}\right)-\left\{\left(-\text{95}\right)+\left(-\text{17}\right)\right\}\right]\text{kJ}\\ =\underset{\_}{-91kJ}\end{array}$