Chapter 15.4, Problem 35PP

Interpretation Introduction

Interpretation:

The $\Delta {{\rm H}}_{rxn}^0$ for the given reaction is to be calculated.

Concept introduction:

$\Delta {{\rm H}}_{rxn}^0$ is the standard enthalpy of reaction.

Standard enthalpy of a reaction is calculated by subtracting sum of heat of formation of reactants from sum of heat of formation of products.

$\Delta {{\rm H}}_{rxn}^0={\displaystyle \sum \Delta {{\rm H}}_f^0(products)-}{\displaystyle \sum \Delta {{\rm H}}_f^0(reac\tan ts)}$

Answer to Problem 35PP

$\Delta {{\rm H}}_{rxn}^0=-1398.4kJ/mol$

Explanation of Solution

The given reaction is:

$4N{H}_3(g)+7O_2(g)\to 4N{O}_2(g)+6H_{2}O(l)$

$\Delta {{\rm H}}_{rxn}^0=\left\{4{{\rm H}}_f^0(N{O}_2)+6{{\rm H}}_f^0(H_{2}O)\right\}-\left\{4{{\rm H}}_f^0(N{H}_3)+7{{\rm H}}_f^0(O_2)\right\}$

From table:

$\begin{array}{l}{{\rm H}}_f^0(N{O}_2)=33.2kJ/mol\\ {{\rm H}}_f^0(H_{2}O)=-285.8kJ/mol\\ {{\rm H}}_f^0(N{H}_3)=-45.9kJ/mol\\ {{\rm H}}_f^0(O_2)=0\end{array}$

Therefore,

$\begin{array}{c}\Delta {{\rm H}}_{rxn}^0=\left\{4{{\rm H}}_f^0(N{O}_2)+6{{\rm H}}_f^0(H_{2}O)\right\}-\left\{4{{\rm H}}_f^0(N{H}_3)+7{{\rm H}}_f^0(O_2)\right\}\\ =\left\{4\times 33.2+6\times -285.8\right\}-\left\{4\times -45.9+7\times 0\right\}\\ =-1582+183.6\\ =-1398.4kJ/mol\end{array}$

Conclusion

$\Delta {{\rm H}}_{rxn}^0=-1398.4kJ/mol$