Chapter 15.4, Problem 33PP

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 different equations then change in enthalpy of the equation is sum of enthalpy change of both reactions.

Answer to Problem 33PP

$\Delta {\rm H}(b)=-521kJ$

Explanation of Solution

Given data:

$\begin{array}{l}4Al+3Mn{O}_2\to 2A{l}_2{O}_3+3Mn\Delta {\rm H}=-1789kJ\\ (a)4Al+3O_2\to 2A{l}_2{O}_3\Delta {\rm H}=-3352kJ\\ (b)Mn+O_2\to Mn{O}_2\Delta {\rm H}=?\end{array}$

According to Hess Law,

Adding first equation to reverse of three times second equation, resultant equation is obtained.

Therefore

$\begin{array}{c}\Delta {\rm H}=\Delta {\rm H}(a)-3\Delta {\rm H}(b)\\ -1789=-3352-3\Delta {\rm H}(b)\\ 3\Delta {\rm H}(b)=-3352+1789\\ 3\Delta {\rm H}(b)=-1563\\ \Delta {\rm H}(b)=-521kJ\end{array}$

Conclusion

Thus, $\Delta {\rm H}(b)=-521kJ$