Chapter 10, Problem 32A

Interpretation Introduction

Interpretation:For the given reaction $\Delta \text{H}$ is to becalculated. 

  $\text{S}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{SO}}_{\text{2}}\left(g\right)$

Concept introduction:According to Hess' law for any process, the change in enthalpy will not depend on the initial and final states and it is same whether a reaction takes place in one or more than one step.

Answer to Problem 32A

  $\Delta \text{H}$ for the reaction is -296.1kJ

Explanation of Solution

The given equations are as follows:

  $\text{S}\left(s\right)+\frac{3}{2}{\text{O}}_{\text{2}}\left(g\right)\to {\text{SO}}_3\left(g\right)\Delta \text{H=}-395.2\text{kJ}$ …… (1)

  ${\text{2SO}}_{\text{2}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{SO}}_3\left(g\right)\Delta \text{H=}-198.2\text{kJ}$ …… (2)

Divide equation (2) with 2 is as follows:

  ${\text{SO}}_{\text{2}}\left(g\right)+\frac{1}{2}{\text{O}}_{\text{2}}\left(g\right)\to {\text{SO}}_3\left(g\right)\Delta \text{H=}-99.1\text{kJ}$ …… (3)

Subtract (1)-(3) is as follows:

  $\begin{array}{l}\text{S}+\frac{3}{2}{\text{O}}_{\text{2}}-{\text{SO}}_{\text{2}}-\frac{1}{2}{\text{O}}_{\text{2}}{\text{=SO}}_3-{\text{SO}}_{\text{3}}\Delta \text{H=}-395.2\text{kJ}-\left(-99.1\text{kJ}\right)\\ {\text{S+O}}_{\text{2}}={\text{SO}}_{\text{2}}\Delta \text{H=}-296.1\text{kJ}\end{array}$

Thus $\Delta \text{H}$ for the reaction is -296.1kJ

Conclusion

Given overall equation is as follows:

  $\text{S}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{SO}}_{\text{2}}\left(g\right)$

For this reaction $\Delta \text{H}$ is -296.1kJ