Chapter 9, Problem 9.116QA

Interpretation Introduction

To calculate:

The value of the $∆H_{rxn}^o$ for the overall reaction.

Answer to Problem 9.116QA

Solution:

The value of the overall reaction,$∆H_{rxn}^o=-1388 kJ$.

Explanation of Solution

1) Concept:

We can manipulate the three equations algebraically for which $∆H_{rxn}^o$ values are known so that they sum to give the equation for which $∆H_{rxn}^o$ is unknown. Then by applying Hess’s law, we can calculate the unknown value.

2) Given:

We are given three reactions with their $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ values:

Equation (1) $\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{ NiSO}_3\left(s\right)\to NiO\left(s\right)+ {SO}_2\left(g\right) ∆H_{rxn}^o=156 kJ \mathrm{ }\mathrm{ }$

Equation (2) $\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\frac{1}{8}{S}_8\left(s\right)+{ O}_2\left(g\right)\to {SO}_2\left(g\right) ∆H_{rxn}^o=-297 kJ$

Equation (3) $\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{N}\mathrm{i}\mathrm{ }\left(\mathrm{s}\right)+\frac{1}{2}{ O}_2\left(g\right)\to NiO\left(s\right) ∆H_{rxn}^o=-241 kJ$

One reaction with unknown $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$ value:

Equation (4) $\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }2 Ni \left(s\right)+ \frac{1}{4}{ S}_8\left(s\right)+3{ O}_2\left(g\right)\to 2 { NiSO}_3\left(s\right)$

3)  Calculations:

We have to manipulate equations (1), (2), and (3) to get equation (4). In equation (4), ${\mathrm{ }\mathrm{N}\mathrm{i}\mathrm{S}\mathrm{O}}_{3\mathrm{ }}$ is on the products’ side, and in equation (1), it is on the reactants’ side, so we have to reverse equation (1). Also we have to multiply (1) by $2$ because in equation (4), there are $2$ ${\mathrm{ }\mathrm{N}\mathrm{i}\mathrm{S}\mathrm{O}}_{3,\mathrm{ }}$ and in equation (1), there is only one ${\mathrm{ }\mathrm{N}\mathrm{i}\mathrm{S}\mathrm{O}}_{3\mathrm{ }}$.

$2 NiO \left(s\right)+2 {SO}_2\left(g\right)\to {2 NiSO}_3\left(s\right) 2\times \left(-∆H_{rxn}^o\right)=-312 kJ$

Now, in equation (4), there are $\mathrm{ }\frac{1}{4}{\mathrm{S}}_{8\mathrm{ }}$, and in equation (2), there are $\mathrm{ }\frac{1}{8}{\mathrm{S}}_8$, so we have to multiply equation (2) by $2$ to get $\mathrm{ }\frac{1}{4}{\mathrm{S}}_{8\mathrm{ }}$ on the reactants’ side.

$\frac{1}{4}{S}_8\left(s\right)+{ 2 O}_2\left(g\right)\to 2 {SO}_2\left(g\right) 2\times ∆H_{rxn}^o=-594 kJ$

In equation (4), there are $2\mathrm{ }{\mathrm{N}\mathrm{i},}_{\mathrm{ }}$ and in equation (3), there is only one ${\mathrm{N}\mathrm{i},}_{\mathrm{ }}$ so we have to multiply equation (3) by $2$ to get $2\mathrm{ }{\mathrm{N}\mathrm{i}}_{\mathrm{ }}$ on the reactants’ side.

$2 Ni \left(s\right)+{ O}_2\left(g\right)\to 2NiO\left(s\right) 2\times ∆H_{rxn}^o=-482 kJ$

Adding all three equations, we get

$2 {NiO}_{2 \left(s\right)}+2 {SO}_{2 \left(g\right)}\to {2 NiSO}_{3 \left(s\right)} 2\times \left(-∆H_{rxn}^o\right)=-312 kJ$

$\frac{1}{4}{S}_{8 \left(s\right)}+{ 2 O}_{2 \left(g\right)}\to 2 {SO}_{2 \left(g\right) } 2\times ∆H_{rxn}^o=-594 Kj$

$2 {Ni}_{\left(s\right)}+{ O}_{2 \left(g\right)}\to {2 NiO}_{2 \left(s\right) } 2\times ∆H_{rxn}^o=-482 kJ$

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$2 {Ni}_{\left(s\right)}+ \frac{1}{4}{ S}_{8 \left(s\right)}+3 { O}_{2 \left(g\right)}\to 2 { NiSO}_{3 \left(s\right)} ∆H_{rxn}^o=-1388 kJ$

The value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^{\mathrm{o}}$is $-1388\mathrm{ }\mathrm{k}\mathrm{J}$.

Conclusion:

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