Chapter 14, Problem 14.109QA

Interpretation Introduction

To calculate:

a) The value of $∆H_{rxn}^0$ for this reaction using the thermodynamic data in Appendix 4.

b) The value of $K_P$ for this reaction at ${298}^0\mathrm{K}$.

c) $∆G_{rxn}^0$ at ${298}^0\mathrm{K}$ using the value of $K_P$  from part (b) and compare it to the value obtained using the $∆G_f^0$ values in Appendix 4.

Answer to Problem 14.109QA

Solution:

a) The value of $∆H_{rxn}^0$ for this reaction is $-197.8\mathrm{ }\mathrm{k}\mathrm{J}$.

b) The value of $K_P$ for this reaction at $298\mathrm{ }\mathrm{K}$ is $7.4\times {10}^{24}$.

c) The value of $∆G_{rxn}^0$ at ${298}^0\mathrm{K}$ using the value of $K_P$  from part (b) is $-142 kJ/mol$ and the value obtained using the $∆G_f^0$ values in Appendix 4 for $∆G_{rxn}^0=-142.0 kJ/mol$.

Explanation of Solution

a) To calculate the value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$:

1) Formula and concept:

We know the formula to calculate $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$

$∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0=\sum {\mathrm{n}}_{\left(\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{s}\right)}∆{\mathrm{H}}_{\mathrm{f}\mathrm{ }\left(\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{s}\right)}^0-\sum {\mathrm{n}}_{\left(\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{s}\right)}∆{\mathrm{H}}_{\mathrm{f}\mathrm{ }\left(\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{s}\right)}^0$

Using the values from Appendix 4 we can calculate $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$

2) Given:

We are given the reaction of ${SO}_{2 \left(g\right)}$ combining with oxygen to make ${SO}_{3 \left(g\right)}$ at $1000 K$.

$2 {SO}_{2 \left(g\right)}+O_{2 \left(g\right)}\rightleftharpoons 2 {SO}_{3 \left(g\right)}$

From appendix 4, we got the values of $∆H_{f }^0$ values for reactants and products.

	$∆H_{f }^{0}kJ/mol$
${SO}_{2 \left(g\right)}$	$-296.8$
$O_{2 \left(g\right)}$	$0.0$
${SO}_{3 \left(g\right)}$	$-395.7$

3) Calculations:

$∆H_{rxn}^0=\sum n_{\left(products\right)}∆H_{f \left(products\right)}^0-\sum n_{\left(reactants\right)}∆H_{f \left(reactants\right)}^0$

$∆H_{rxn}^0=\left[2\times \left(-395.7 kJ/mol\right)\right]-[0+2\times \left(-296.8 kJ/mol\right)]$

$∆H_{rxn}^0=-791.4 kJ/mol+593.6kJ/mol$

$∆H_{rxn}^0=-197.8 kJ/mol$

b) The value of $K_P$ for this reaction at $298 K$:

1) Formula and concept:

To calculate $K_P$ or $K_2$ we use the van’t Hoff equation.

$ln\left(\frac{K_2}{K_1}\right)=\frac{-∆H^0}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

2) Given:

To find out $K_2$ we have to plug ${ K}_1=3.4$,$T_1=1000 K$,$T_2=298 K$ and $∆H_{rxn}^0=-197.8 kJ/mol$ in the above equation.

3) Calculations

$ln\left(\frac{K_2}{K_1}\right)=\frac{-∆H^0}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

$ln\left(\frac{K_2}{3.4}\right)=\frac{-(-197.8\times {10}^{3}J/mol)}{(8.314 J/K\times mol)}\left(\frac{1}{298}-\frac{1}{1000}\right)$

$ln\left(\frac{K_2}{3.4}\right)=56.045$

$\frac{K_2}{3.4}={2.188\times 10}^{24}$

$K_2={7.439\times 10}^{24}$

$K_2={7.4\times 10}^{24}$

c) To calculate $∆G_{rxn}^0$ at $298 K$:

1) Formula and concept:

To calculate $∆G_{rxn}^0$ using the value of $K_2={7.4\times 10}^{24}$ we use the formula

$∆G^0=-RTln K$

Then, to find out $∆{\mathrm{G}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$ using the values from Appendix 4, the formula is

$∆G_{rxn}^0=\sum n_{\left(products\right)}∆G_{f \left(products\right)}^0-\sum n_{\left(reactants\right)}∆G_{f \left(reactants\right)}^0$

2) Given:

From appendix 4, we got the values of $∆G_{f }^0$ values for reactants and products.

	$∆G_{f }^{0}kJ/mol$
${SO}_{2 \left(g\right)}$	$-300.1$
$O_{2 \left(g\right)}$	$0$
${SO}_{3 \left(g\right)}$	$-371.1$

3) Calculations:

$∆G_{rxn}^0$ using the value of $K_2={7.4\times 10}^{24}$

$∆G^0=-RTln K$

$∆G^0=-(8.314 J/K.mol\left)\right(298 K)\times ln({7.4\times 10}^{24}$ )

$∆G^0=-141887 J/mol$

$∆G^0=-142 kJ/mol$

$∆G_{rxn}^0$ Using the values from Appendix 4

$∆G_{rxn}^0=\sum n_{\left(products\right)}∆G_{f \left(products\right)}^0-\sum n_{\left(reactants\right)}∆G_{f \left(reactants\right)}^0$

$∆G_{rxn}^0=\left[2\times \left(-371.1 kJ/mol\right)\right]-[0+2\times \left(-300.1 kJ/mol\right)]$

$∆G_{rxn}^0=-142.0 kJ/mol$

Conclusion:

Using the thermodynamic data in Appendix 4 the value of $∆{\mathrm{H}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$ and $∆{\mathrm{G}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$ is calculated. The value of ${\mathrm{K}}_{\mathrm{P}}$ is calculated using special case of Clausius-Clapeyron equation. The value of $∆{\mathrm{G}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$ is calculated using equation which relates $∆{\mathrm{G}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$ and ${\mathrm{K}}_{\mathrm{P}}$. Compared this value of $∆{\mathrm{G}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$ with the calculated value of $∆{\mathrm{G}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$ from thermodynamic data in Appendix 4.