Chapter 10, Problem 10.55PAE

Using tabulated thermodynamic data, calculate $\Delta \text{G}°$ for these reactions.

(a) $\text{Fe}\left(\text{s}\right)+\text{2HCl}\left(\text{g}\right)\to {\text{FeCl}}_{\text{2}}\left(\text{s}\right)+{\text{H}}_{\text{2}}\left(\text{g}\right)$

(b) ${\text{3NO}}_{\text{2}}\left(\text{g}\right)+{\text{H}}_{\text{2}}\text{O}(\mathcal{l})\to {\text{2HNO}}_{\text{3}}(\mathcal{l})+\text{NO}\left(\text{g}\right)$

(c) $\text{2K}\left(\text{s}\right)+{\text{Cl}}_{\text{2}}\left(\text{g}\right)\to \text{2KCl}\left(\text{s}\right)$

(d) ${\text{Cl}}_{\text{2}}\left(\text{g}\right)+\text{2NO}\left(\text{g}\right)\to \text{2NOCl}\left(\text{g}\right)$

(e) ${\text{SiCl}}_{\text{4}}\left(\text{g}\right)\to \text{Si}\left(\text{s}\right)+{\text{2Cl}}_{\text{2}}\left(\text{g}\right)$

Interpretation Introduction

Interpretation:

Using tabulated thermodynamic data, value of $\text{ΔG}$ for the following reactions needs to be calculated.

$\text{Fe}\left(\text{s}\right)\text{ + 2HCl }\left(\text{g}\right)\to {\text{FeCl}}_{\text{2}}\left(\text{s}\right){\text{ + H}}_{\text{2}}\left(\text{g}\right)$

${\text{3NO}}_2{\text{(g) + H}}_2\text{O(l)}\to {\text{2HNO}}_3\text{(l) + NO(g)}$

${\text{2K(s) + Cl}}_2\text{(g)}\to \text{ 2KCl(s)}$

${\text{Cl}}_2\text{(g) + 2NO(g)}\to 2\text{NOCl(g)}$

${\text{SiCl}}_{\text{4}}\left(\text{g}\right)\to \text{Si }\left(\text{s}\right){\text{ + 2 Cl}}_{\text{2}}\left(\text{g}\right)$

Concept introduction:

The standard change of Gibbs free energy is defined as the formation of 1 mole of a substance in its standard state from its constituent elements in their standard state.

For a reaction where reactant gives product, change in standard Gibbs free energy of reaction is as calculated as follows:

${\text{ΔG}}_{\text{(reaction)}}^0{\text{ = ΔG}}_{\text{products}}^0-{\text{ΔG}}_{\text{reactants}}^0$

Here, ${\text{ΔG}}_{\text{products}}^0$ is change in standard Gibbs free energy of product and ${\text{ΔG}}_{\text{reactants}}^0$ is change in standard Gibbs free energy of reactant.

Answer to Problem 10.55PAE

Solution:

1. -111.702 kJ
2. 8.329 kJ
3. -818.28 kJ
4. -40.94 kJ
5. 616 kJ

(a)

Explanation of Solution

For a reaction where reactant gives product, change in standard Gibbs free energy of reaction is as calculated as follows:

${\text{ΔG}}_{\text{(reaction)}}^0{\text{ = ΔG}}_{\text{products}}^0-{\text{ ΔG}}_{\text{reactants}}^0$

Thu, for the following reaction:

$\text{Fe}\left(\text{s}\right)\text{ + 2HCl }\left(\text{g}\right)\to {\text{FeCl}}_{\text{2}}\left(\text{s}\right){\text{ + H}}_{\text{2}}\left(\text{g}\right)$

The change in standard Gibbs free energy of reaction is as calculated as follows:

$\begin{array}{c}{\text{ΔG}}_{\text{(reaction)}}^0{\text{ = ΔG}}_{{\text{FeCl}}_2\text{(s)}}^0{\text{+ ΔG}}_{{\text{H}}_2\text{(g)}}^0-{\text{ ΔG}}_{\text{Fe(s)}}^0-{\text{ 2ΔG}}_{\text{HCl(g)}}^0\\ \text{ = }\left[\begin{array}{c}\left(1\text{ mol}\right)\left(-302.3\frac{\text{kJ}}{\text{mol}}\right)+\left(1\text{ mol}\right)\left(0\frac{\text{kJ}}{\text{mol}}\right)\\ -\left(1\text{ mol}\right)\left(0\frac{\text{kJ}}{\text{mol}}\right)-\left(\text{2 mol}\right)\left(95.299\frac{\text{kJ}}{\text{mol}}\right)\end{array}\right]\\ \text{ = }\left(1\text{ mol}\right)\left(-302.3\frac{\text{kJ}}{\text{mol}}\right)+190.598\text{ kJ}\\ \text{ = }-\text{111}\text{.702 kJ}\end{array}$

(b)

For a reaction where reactant gives product, change in standard Gibbs free energy of reaction is as calculated as follows:

${\text{ΔG}}_{\text{(reaction)}}^0{\text{ = ΔG}}_{\text{products}}^0-{\text{ ΔG}}_{\text{reactants}}^0$

Thu, for the following reaction:

${\text{3NO}}_2{\text{(g) + H}}_2\text{O(l)}\to {\text{2HNO}}_3\text{(l) + NO(g)}$

The change in standard Gibbs free energy of reaction is as calculated as follows:

$\begin{array}{c}{\text{ΔG}}_{\text{(reaction)}}^0{\text{ = 2ΔG}}_{{\text{HNO}}_3\text{(l)}}^0{\text{+ ΔG}}_{\text{NO(g)}}^0-{\text{ ΔG}}_{{\text{H}}_2\text{O(l)}}^0-{\text{ 3ΔG}}_{{\text{NO}}_2\text{(g)}}^0\\ \text{ = }\left[\begin{array}{l}\left(2\text{ mol}\right)\left(-80.71\frac{\text{kJ}}{\text{mol}}\right)+\left(1\text{ mol}\right)\left(-86.55\frac{\text{kJ}}{\text{mol}}\right)\\ -\left(1\text{ mol}\right)\left(-237.129\frac{\text{kJ}}{\text{mol}}\right)-\left(3\text{ mol}\right)\left(+51.31\frac{\text{kJ}}{\text{mol}}\right)\end{array}\right]\\ \text{ = }-161.42\text{ kJ}+86.55\text{ kJ}+237.129\text{ kJ}-153.93\text{ kJ}\\ \text{ = }8.329\text{ kJ}\end{array}$

(c)

For a reaction where reactant gives product, change in standard Gibbs free energy of reaction is as calculated as follows:

${\text{ΔG}}_{\text{(reaction)}}^0{\text{ = ΔG}}_{\text{products}}^0-{\text{ ΔG}}_{\text{reactants}}^0$

Thu, for the following reaction:

${\text{2K(s) + Cl}}_2\text{(g)}\to \text{ 2KCl(s)}$

The change in standard Gibbs free energy of reaction is as calculated as follows:

$\begin{array}{l}{\text{ΔG}}_{\text{(reaction)}}^0{\text{ = 2ΔG}}_{\text{KCl(s)}}^0-{\text{ 2ΔG}}_{\text{K(s)}}^0-{\text{ ΔG}}_{{\text{Cl}}_2\text{(g)}}^0\\ \text{ = }\left(2\text{ mol}\right)\left(-409.14\frac{\text{kJ}}{\text{mol}}\right)-\left(2\text{ mol}\right)\left(0\frac{\text{kJ}}{\text{mol}}\right)-\left(\text{1 mol}\right)\left(\text{0}\frac{\text{kJ}}{\text{mol}}\right)\\ \text{ = }-818.28\text{ kJ}\end{array}$

(d)

For a reaction where reactant gives product, change in standard Gibbs free energy of reaction is as calculated as follows:

${\text{ΔG}}_{\text{(reaction)}}^0{\text{ = ΔG}}_{\text{products}}^0-{\text{ ΔG}}_{\text{reactants}}^0$

Thu, for the following reaction:

${\text{Cl}}_2\text{(g) + 2NO(g)}\to 2\text{NOCl(g)}$

The change in standard Gibbs free energy of reaction is as calculated as follows:

$\begin{array}{c}{\text{ΔG}}_{\text{(reaction)}}^0{\text{ = 2ΔG}}_{\text{NOCl(g)}}^0-{\text{ 2ΔG}}_{\text{NO(g)}}^0-{\text{ ΔG}}_{{\text{Cl}}_2\text{(g)}}^0\\ \text{ =}\left(2\text{ mol}\right)\left(66.08\frac{\text{kJ}}{\text{mol}}\right)-\left(2\text{ mol}\right)\left(86.55\frac{\text{kJ}}{\text{mol}}\right)-\left(2\text{ mol}\right)\left(0\frac{\text{kJ}}{\text{mol}}\right)\\ \text{ = }132.16\text{ kJ}-173.1\text{ kJ}\\ \text{ = }-40.94\text{ kJ}\end{array}$

(e)

For a reaction where reactant gives product, change in standard Gibbs free energy of reaction is as calculated as follows:

${\text{ΔG}}_{\text{(reaction)}}^0{\text{ = ΔG}}_{\text{products}}^0-{\text{ ΔG}}_{\text{reactants}}^0$

Thu, for the following reaction:

${\text{SiCl}}_{\text{4}}\left(\text{g}\right)\to \text{Si }\left(\text{s}\right){\text{ + 2 Cl}}_{\text{2}}\left(\text{g}\right)$

The change in standard Gibbs free energy of reaction is as calculated as follows:

$\begin{array}{c}{\text{ΔG}}_{\text{(reaction)}}^0{\text{ = ΔG}}_{\text{Si(s)}}^0{\text{+2ΔG}}_{{\text{Cl}}_2\text{(g)}}^0-{\text{ ΔG}}_{{\text{SiCl}}_4\text{(g)}}^0\\ =\left(1\text{ mol}\right)\left(0\frac{\text{kJ}}{\text{mol}}\right)+\left(2\text{ mol}\right)\left(0\frac{\text{kJ}}{\text{mol}}\right)-\left(1\text{ mol}\right)\left(-616.98 \frac{\text{kJ}}{\text{mol}}\right)\\ \text{ = }616.98 \text{ kJ}\\ \end{array}$

Conclusion

The standard change of Gibbs free energy is defined as the formation of 1 mole of a substance in its standard state from its constituent elements in their standard state. Based on the thermodynamic table we have determined the below values: -

• -111.702 kJ
• 8.329 kJ
• -818.28 kJ
• -40.94 kJ
• 616kJ