Chapter 14, Problem 14.110QA

Interpretation Introduction

To calculate:

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

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

Answer to Problem 14.110QA

Solution:

a) The value of $K_P$ for this reaction at $298\mathrm{ }\mathrm{K}$ is $5.22\times {10}^{-6}$.

b) The value of $∆{\mathrm{G}}_{\mathrm{r}\mathrm{x}\mathrm{n}}^0$ at $298\mathrm{ }\mathrm{K}$ using the value of $K_P$  from part (b) is $30 kJ/mol$ and the value obtained using the $∆G_f^0$ values in Appendix 4 for $∆G_{rxn}^0=28.6 kJ/mol$

Explanation of Solution

a) To calculate the value of $K_P$ for this reaction at $298 K$:

1. Formula and concept:

We know the formula to calculate $∆H_{rxn}^0$

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

Using the values from Appendix 4 we can calculate $∆H_{rxn}^0$

To calculate $K_P$ or $K_2$, we use the van’t Hoff equation and $∆H_{rxn}^0$.

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

2. Given:

We are given the reaction and equilibrium constant $K_1=0.534$ at $700^0 C=\left(700+273.15\right)= 973.15 K$

$H_{2 \left(g\right)}+{CO}_{2 \left(g\right)}\rightleftharpoons H_2{O}_{\left(g\right)}+{CO}_{\left(g\right)}$

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

	$∆H_{f }^{0}kJ/mol$
$H_{2 \left(g\right)}$	$0.0$
${CO}_{2 \left(g\right)}$	$-393.5$
$H_2{O}_{\left(g\right)}$	$-241.8$
${CO}_{\left(g\right)}$	$-110.5$

To find out $K_2$ we have to plug ${ K}_1=0.534$,$T_1=973 K,T_2=298 K$ and $∆H_{rxn}^0$ obtained from Appendix 4 in the van’t Hoff equation.

3. Calculations:

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

$∆H_{rxn}^0=\left[\left(-241.8 kJ/mol\right)+(-110.5 kJ/mol)\right]-[0+\left(-393.5 kJ/mol\right)]$

$∆H_{rxn}^0=-352.3 kJ/mol+393.5kJ/mol$

$∆H_{rxn}^0=41.2 kJ/mol$

Now equilibrium constant $K_P$ for this reaction at $298 K$

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

$ln\left(\frac{K_2}{0.534}\right)=\frac{-(41.2\times {10}^{3}J/mol)}{(8.314 J/K.mol)}\left(\frac{1}{298}-\frac{1}{973}\right)$

$ln\left(\frac{K_2}{0.534}\right)=-11.5362$

$\frac{K_2}{0.534}={9.77\times 10}^{-6}$

$K_2={5.22\times 10}^{-6}$

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

1. Formula and concept:

To calculate $∆G_{rxn}^0$ using the value of $K_2={5.22\times 10}^{-6}$ we use the formula

$∆G^0=-RTln K$

Then, to find out $∆G_{rxn}^0$, using the values from Appendix 4, we get the formula

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

2. Given:

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

	$∆G_{f }^{0}kJ/mol$
$H_{2 \left(g\right)}$	$0$
${CO}_{2 \left(g\right)}$	$-394.4$
$H_2{O}_{\left(g\right)}$	$-228.6$
${CO}_{\left(g\right)}$	$-137.2$

3. Calculations:

$∆G_{rxn}^0$ Using the value of $K_2={5.22\times 10}^{-6}$

$∆G^0=-RTln K$

$∆G^0=-(8.314 J/K.mol\left)\right(298 K)\times ln({5.22\times 10}^{-6})$ )

$∆G^0=30136J/mol$

$∆G^0=30 kJ/mol$

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

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

$∆H_{rxn}^0=\left[\left(-228.6 kJ/mol\right)+(-137.2 kJ/mol)\right]-[0+\left(-394.4 kJ/mol\right)]$

$∆H_{rxn}^0=(-365.8kJ/mol)+394.4 kJ/mol$

$∆G_{rxn}^0=28.6 kJ/mol$

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.