Chapter 13, Problem 13.112QP

(a)

Interpretation Introduction

To determine: The enthalpy change of the given reaction.

Interpretation: The enthalpy change of the given reaction is to be calculated.

Concept Introduction: The enthalpy change of the reaction is calculated by the difference of sum of enthalpy of formation of products and reactants.

Answer to Problem 13.112QP

The enthalpy change of the given reaction is $-252.96\text{kJ/mol}$.

Explanation of Solution

The balanced chemical equation for the given reaction is,

${\text{4H}}_2\left(g\right)+{\text{CO}}_2\left(g\right)\to {\text{CH}}_4\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

The enthalpy change of the given reaction is calculated by the formula,

$\begin{array}{c}\Delta H^{\circ }={\displaystyle \sum n\Delta H^{\circ }\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^{\circ }\left(\text{reactants}\right)}\\ =\left(2\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }+\Delta H_{{\text{CH}}_4\left(g\right)}^{\circ }\right)-\left(4\Delta H_{{\text{H}}_2\left(g\right)}^{\circ }+\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }\right)\end{array}$

Substitute the values of $\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }$, $\Delta H_{{\text{CH}}_4\left(g\right)}^{\circ }$, $\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }$ and $\Delta H_{{\text{H}}_2\left(g\right)}^{\circ }$ from appendix C in equation.

$\begin{array}{c}\Delta H^{\circ }=\left(\begin{array}{c}\left(2\times -285.83\text{kJ/mol}+1\times -78.4\text{kJ/mol}\right)\\ -\left(4\times 0.0\text{kJ/mol}+1\times -393.5\text{kJ/mol}\right)\end{array}\right)\\ =-252.96\text{kJ/mol}\end{array}$

(b)

Interpretation Introduction

To determine: The enthalpy change of the given reaction.

Interpretation: The enthalpy change of the given reaction is to be calculated.

Concept Introduction: The enthalpy change of the reaction is calculated by the difference of sum of enthalpy of formation of products and reactants.

Answer to Problem 13.112QP

The enthalpy change of the given reaction is $-234.76\text{kJ/mol}$.

Explanation of Solution

The balanced chemical equation for the given reaction is,

${\text{4CH}}_3\text{OH}\left(l\right)\to 3{\text{CH}}_4\left(g\right)+{\text{CO}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

The enthalpy change of the given reaction is calculated by the formula,

$\begin{array}{c}\Delta H^{\circ }={\displaystyle \sum n\Delta H^{\circ }\left(\text{products}\right)}-{\displaystyle \sum m\Delta H^{\circ }\left(\text{reactants}\right)}\\ =\left(\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }+3\Delta H_{{\text{CH}}_4\left(g\right)}^{\circ }+2\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }\right)-4\Delta H_{{\text{CH}}_3\text{OH}\left(l\right)}^{\circ }\end{array}$

Substitute the values of $\Delta H_{{\text{CO}}_2\left(g\right)}^{\circ }$, $\Delta H_{{\text{CH}}_4\left(g\right)}^{\circ }$, $\Delta H_{{\text{H}}_2\text{O}\left(l\right)}^{\circ }$ and $\Delta H_{{\text{CH}}_3\text{OH}\left(l\right)}^{\circ }$ from appendix C in equation.

$\begin{array}{c}\Delta H^{\circ }=\left(1\times -393.5\text{kJ/mol}+3\times -74.8\text{kJ/mol}+2\times -285.83\text{kJ/mol}\right)\\ -4\times -238.7\text{kJ/mol}\\ =-234.76\text{kJ/mol}\end{array}$

Conclusion

1. a. The enthalpy change of the given reaction is $-252.96\text{kJ/mol}$.
2. b. The enthalpy change of the given reaction is $-234.76\text{kJ/mol}$.