Chapter 15.4, Problem 42SSC

Interpretation Introduction

Interpretation:

The value of $\Delta {\rm H}$ is to be calculated for the given equation.

Concept introduction:

The change in enthalpy of a reaction can be calculated using Hess Law.

According to it if one equation is the resultant of two or more different equations then change in enthalpy of the equation is sum of enthalpy change of those reactions.

Answer to Problem 42SSC

$\Delta {\rm H}$ = $-233kJ$

Explanation of Solution

Given data:

$\begin{array}{l}{O}_2\to 2O\Delta {\rm H}=+495kJ\\ 2O_3\to 3O_2\Delta {\rm H}=-427kJ\\ NO+O_3\to N{O}_2+O_2\Delta {\rm H}=-199kJ\end{array}$

Resultant equation:

$NO+O\to N{O}_2$

This equation can be formed by taking equation three and adding in it half of reverse of first equation and half of reverse of second equation.

$NO+O_3\to N{O}_2+O_2\Delta {\rm H}=-199kJ$

$O\to \frac{1}{2}{O}_2\Delta {\rm H}=\frac{1}{2}\times -495=-247.5kJ$

$\frac{3}{2}{O}_2\to O_3\Delta {\rm H}=\frac{1}{2}\times +427=+213.5kJ$

Therefore, according to Hess Law

$\begin{array}{c}\Delta {\rm H}=-199+(-247.5)+(+213.5)\\ =-233kJ\end{array}$

Conclusion

Thus, $\Delta {\rm H}$ = $-233kJ$