Chapter 9, Problem 9.69QA

Interpretation Introduction

To find:

The value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ of the overall reaction by using Hess’s law

Answer to Problem 9.69QA

Solution:

The value of the overall reaction is $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}=-297\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$.

Explanation of Solution

1) Concept:

To calculate the value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ of the overall reaction, we can use Hess’s law. We can algebraically manipulate the two equations for which $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ values are known so that we can calculate the $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ of the overall reaction. Then by applying Hess’s law, we can calculate the unknown value.

2) Given:

$Equation 1 {2 SO}_{2 \left(g\right)}+O_{2 \left(g\right)}\to 2 {SO}_{3 \left(g\right)} ∆H_{rxn}^o=-196 kJ$

$Equation 2 \frac{1}{4}{S}_{8 \left(s\right)}+{3 O}_{2 \left(g\right)}\to 2 {SO}_{3 \left(g\right) } ∆H_{rxn}^o=-790 kJ$

To calculate $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ value of this reaction:

$Equation 3 \frac{1}{8}{S}_{8 \left(s\right)}+{ O}_{2 \left(g\right)}\to {SO}_{2 \left(g\right) }$

3)  Calculations:

To get equation $3,$ we need to add equation $1$ and equation $2$. We need to reverse the first reaction to get SO₂ on the product side. We also need to divide equation $1$ by $2$ because we need one mole of SO2 in the final reaction.

${SO}_{3 \left(g\right)}\to {SO}_{2 \left(g\right)}+\frac{1}{2}{O}_{2 \left(g\right)} ∆H_{rxn}^o=98 kJ equation 4$

We need to divide equation $2$ by $2$ because we need $\frac{1}{8} {\mathrm{S}}_8$ in the final reaction.

$\frac{1}{8}{S}_{8 \left(s\right)}+{\frac{3}{2} O}_{2 \left(g\right)}\to {SO}_{3 \left(g\right) } ∆H_{rxn}^o=-395 kJ equation 5$

Adding equation 4 and equation 5 by cancelling SO₃ in the final reaction, we get

${SO}_{3 \left(g\right)}\to {SO}_{2 \left(g\right)}+\frac{1}{2}{O}_{2 \left(g\right)} ∆H_{rxn}^o=98 kJ equation 4$

$\frac{1}{8}{S}_{8 \left(s\right)}+{\frac{3}{2} O}_{2 \left(g\right)}\to {SO}_{3 \left(g\right) } ∆H_{rxn}^o=-395 kJ equation 5$

 

$\frac{1}{8}{S}_{8 \left(s\right)}+{ O}_{2 \left(g\right)}\to {SO}_{2 \left(g\right) } ∆H_{rxn}^o=(98+\left(-395\right))= -297 kJ$

The value of the overall reaction is $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}=-297\mathrm{ }\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$.

Conclusion:

We have calculated the value of $\mathrm{ }∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ by using Hess’s law and by manipulating the reactions.