Chapter 13, Problem 13.76QP

Interpretation Introduction

Interpretation: The standard enthalpy of the given reactions from the appropriate standard enthalpies of formation is to be calculated.

Concept introduction: The energy which is required in a molecule to break one mole of a covalent bond in its gaseous phase is known as bond energy of that bond. It is expressed in terms of enthalpy change denoted by $\left(\Delta H\right)$.

To determine: The standard enthalpy of the given reaction from the appropriate standard enthalpies of formation.

Answer to Problem 13.76QP

The standard enthalpy of the given reaction is $\underset{\_}{-3767.28kJ/mol}$.

The standard enthalpy of the given reaction is $\underset{\_}{-3980.44kJ/mol}$.

Explanation of Solution

The standard enthalpy of formation $\left(\Delta H_{{\text{f,CH}}_{\text{3}}{\text{NH}}_{\text{2}}}^{\text{ο}}\right)$ of ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ is $-23.0\text{kJ/mol}$.

The given reaction is,

${\text{4CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+13{\text{O}}_2\left(g\right)\stackrel{}{\to }4{\text{CO}}_{\text{2}}\left(g\right)+4{\text{NO}}_2\left(g\right)+10{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O}$ is $-241.8\text{kJ/mol}$.

Standard enthalpy of formation of ${\text{O}}_{\text{2}}$ is $0\text{kJ/mol}$.

Standard enthalpy of formation of ${\text{CO}}_{\text{2}}$ is $-393.5\text{kJ/mol}$.

Standard enthalpy of formation of ${\text{NO}}_2$ is $33.18\text{kJ/mol}$.

The standard enthalpy for the given reaction is calculated by using the formula,

$\Delta H_{\text{rxn}}^{\text{ο}}={\displaystyle \sum \Delta H_{\text{bond}\text{breaking}}^{\text{ο}}-{\displaystyle \sum \Delta H_{\text{bond}\text{forming}}^{\text{ο}}}}$

Substitute the values of standard enthalpies of formation of reactants and products in the above formula to calculate the standard enthalpy of the reaction.

$\begin{array}{c}\Delta H_{\text{rxn}}^{\text{ο}}=\left(4\times {\text{CO}}_{\text{2}}+4\times {\text{NO}}_2+10\times {\text{H}}_{\text{2}}\text{O}\right)-\left(4\times {\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}+13\times {\text{O}}_2\right)\\ =\left(\begin{array}{c}\left(4\times \left(-393.5\right)\text{kJ/mol}+4\times 33.18\text{kJ/mol}+10\times \left(-241.8\right)\text{kJ/mol}\right)-\\ \left(4\times -23\text{kJ/mol}+13\times 0\text{kJ/mol}\right)\end{array}\right)\\ =\left(-1574+132.72-2418\right)\text{kJ/mol}-\left(-92-0\right)\text{kJ/mol}\\ =-3859.28-\left(-92\right)\text{kJ/mol}\end{array}$

Simplify the above equation,

$\begin{array}{c}\Delta H_{\text{rxn}}^{\text{ο}}=-3859.28+92\text{kJ/mol}\\ =\underset{\_}{-3767.28kJ/mol}\end{array}$

Hence, the standard enthalpy of the given reaction is $\underset{\_}{-3767.28kJ/mol}$.

The given reaction is,

${\text{4CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+6{\text{O}}_2\left(g\right)\stackrel{}{\to }4{\text{CO}}_{\text{2}}\left(g\right)+4{\text{NH}}_3\left(g\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O}$ is $-241.8\text{kJ/mol}$.

Standard enthalpy of formation of ${\text{O}}_{\text{2}}$ is $0\text{kJ/mol}$.

Standard enthalpy of formation of ${\text{CO}}_{\text{2}}$ is $-393.5\text{kJ/mol}$.

Standard enthalpy of formation of ${\text{NH}}_{\text{3}}$ is $-382.81\text{kJ/mol}$.

The standard enthalpy for the given reaction is calculated by using the formula,

$\Delta H_{\text{rxn}}^{\text{ο}}={\displaystyle \sum \Delta H_{\text{bond}\text{breaking}}^{\text{ο}}-{\displaystyle \sum \Delta H_{\text{bond}\text{forming}}^{\text{ο}}}}$

Substitute the values of standard enthalpies of formation of reactants and products in the above formula to calculate the standard enthalpy of the reaction.

$\begin{array}{c}\Delta H_{\text{rxn}}^{\text{ο}}=\left(4\times {\text{CO}}_{\text{2}}+4\times {\text{NH}}_3+4\times {\text{H}}_{\text{2}}\text{O}\right)-\left(4\times {\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}+6\times {\text{O}}_2\right)\\ =\left(\begin{array}{c}\left(4\times \left(-393.5\right)\text{kJ/mol}+4\times \left(-382.81\right)\text{kJ/mol}+4\times \left(-241.8\right)\text{kJ/mol}\right)-\\ \left(4\times -23\text{kJ/mol}+6\times 0\text{kJ/mol}\right)\end{array}\right)\\ =\left(-1574-1531.24-967.2\right)\text{kJ/mol}-\left(-92-0\right)\text{kJ/mol}\\ =-4072.44-\left(-92\right)\text{kJ/mol}\end{array}$

Simplify the above equation,

$\begin{array}{c}\Delta H_{\text{rxn}}^{\text{ο}}=-4072.44+92\text{kJ/mol}\\ =\underset{\_}{-3980.44kJ/mol}\end{array}$

Hence, the standard enthalpy of the given reaction is $\underset{\_}{-3980.44kJ/mol}$.

Conclusion

The standard enthalpy of the given reaction is $\underset{\_}{-3767.28kJ/mol}$.

The standard enthalpy of the given reaction is $\underset{\_}{-3980.44kJ/mol}$.