Chapter 16, Problem 47QAP

Interpretation Introduction

Interpretation:

The reaction that proceeds spontaneously at a lower temperature needs to be identified from the given options:

$\begin{array}{l}{\text{a) Fe}}_{\text{2}}{\text{O}}_3\text{(s) + }\frac{3}{2}\text{C(s) }\to \text{2Fe(s) + }\frac{3}{2}{\text{CO}}_2\text{(g)}\\ {\text{b) Fe}}_{\text{2}}{\text{O}}_3{\text{(s) + 3H}}_{\text{2}}\text{(g) }\to \text{2Fe(s) + }3{\text{H}}_2\text{O(g)}\end{array}$

Concept introduction:

The change in the Gibbs free energy, ΔG is a thermodynamic function which governs the spontaneity of a chemical reaction. The negative value of ΔG represents that the reaction is spontaneous, whereas, the positive value represents that the reaction is non-spontaneous, and if ΔG = 0, then it represents that the reaction is at equilibrium. The standard Gibbs free energy, ${\text{ΔG}}^{\text{0}}$ is the value measured under standard conditions i.e. Pressure = 1 atm and Temperature The standard Gibbs free energy ${\text{ΔG}}^{\text{0}}$ for a given chemical reaction can be expressed as a function of temperature, T via the Gibbs-Helmholtz equation:

${\text{ΔG}}^{\text{0}}{\text{ = ΔH}}^{\text{0}}{\text{ - TΔS}}^{\text{0}}\text{ -------(1)}$

Where, ${\text{ΔH}}^{\text{0}}$ is the standard enthalpy change, and ${\text{ΔS}}^{\text{0}}$ is the standard entropy change

Answer to Problem 47QAP

Reaction (a)

Explanation of Solution

The chemical equation corresponding to the reaction between iron ore and hydrogen is given as:

${\text{Fe}}_{\text{2}}{\text{O}}_3\text{(s) + }\frac{3}{2}\text{C(s) }\to \text{2Fe(s) + }\frac{3}{2}{\text{CO}}_2\text{(g)}$

For the above reaction, the ${\text{ΔG}}^{\text{0}}$ value can be deduced by calculating the ${\text{ΔH}}^{\text{0}}$ and ${\text{ΔS}}^{\text{0}}$ values from the standard enthalpy and entropy of formation of the reactants and products.

Step 1: Calculate ${\text{ΔH}}^{\text{0}}$

The standard enthalpy change for a reaction ${\text{ΔH}}^{\text{0}}$ is given in terms of the difference in the standard enthalpy of formation of the products and that of reactants.

i.e. $\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) }$

Where n_p and n_r are the number of moles of the products and reactants

For the given reaction:

$\begin{array}{l}\Delta {\text{H}}^0{\text{ = [2ΔH}}_{\text{f}}^{\text{0}}{\text{(Fe(s)) + 3/2ΔH}}_{\text{f}}^{\text{0}}{\text{(CO}}_{\text{2}}{\text{(g))] - [1ΔH}}_{\text{f}}^{\text{0}}{\text{(Fe}}_{\text{2}}{\text{O}}_3{\text{(s)) + 3/2ΔH}}_{\text{f}}^{\text{0}}\text{(C(s))]}\\ {\text{Based on the ΔH}}_{\text{f}}^{\text{0}}\text{:}\\ \Delta {\text{H}}^0\text{ = [2 mole }\times (0\text{ kJ/mol) + 3/2 moles }\times \text{(-393}\text{.5 kJ/mol)]-[1 moles }\times \text{(}-824.3\text{ kJ/mol) }\\ \text{ + 3/2moles }\times \text{(}0\text{ kJ/mol)]}\\ \text{ = 0 kJ -590}\text{.25 kJ + 824}\text{.3 kJ + 0 kJ = 234}\text{.1 kJ }\end{array}$

Step 2: Calculate ${\text{ΔS}}^{\text{0}}$

The standard entropy change for a reaction ${\text{ΔS}}^{\text{0}}$ is given in terms of the difference in the standard entropy of formation of the products and that of reactants.

i.e. $\Delta {\text{S}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{S}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{S}}_{\text{f}}^{\text{0}}\text{(reactants) }$

Where n_p and n_r are the number of moles of the products and reactants

For the given reaction:

$\begin{array}{l}\Delta {\text{S}}^0{\text{ = [2S}}_{\text{f}}^{\text{0}}{\text{(Fe(s)) + 3/2S}}_{\text{f}}^{\text{0}}{\text{(CO}}_{\text{2}}{\text{(g))] - [1S}}_{\text{f}}^{\text{0}}{\text{(Fe}}_{\text{2}}{\text{O}}_3{\text{(s)) + 3/2S}}_{\text{f}}^{\text{0}}\text{(C(s))]}\\ {\text{Based on the S}}_{\text{f}}^{\text{0}}\text{ values:}\\ \Delta {\text{S}}^0\text{ = [2 mole }\times (27.28\text{ J/mol-K) + 3/2 moles }\times \text{(213}\text{.68 J/mol-K)]-[1 moles }\times \text{(}87.4\text{ J/mol-K) }\\ \text{ + 3/2moles }\times \text{(}5.69\text{ J/mol-K)]}\\ \text{ = 54}\text{.56 + 320}\text{.52 - 87}\text{.4 -8}\text{.54 = 471}\text{.0 J/K }\end{array}$

Step 3: Estimate the temperature when ${\text{ΔG}}^{\text{0}}$ = 0

${\text{ΔH}}^{\text{0}}$ = 234.1 kJ and ${\text{ΔS}}^{\text{0}}$ = 0.471 kJ/K

When ${\text{ΔG}}^{\text{0}}$ = 0, equation (1) becomes:

$\begin{array}{l}{\text{ΔH}}^{\text{0}}{\text{ = TΔS}}^{\text{0}}\\ \text{T = }\frac{{\text{ ΔH}}^{\text{0}}}{{\text{ΔS}}^{\text{0}}}=\frac{\text{234}\text{.1 kJ}}{\text{0}\text{.471 kJ/K}}=497\text{ K}\end{array}$

Thus, ${\text{ΔG}}^{\text{0}}$ = 0 at T = 497 K. The reaction will be spontaneous at temperatures greater than

497 K.

The chemical equation corresponding to the reaction between iron ore and carbon is given as:

${\text{ Fe}}_{\text{2}}{\text{O}}_3{\text{(s) + 3H}}_{\text{2}}\text{(g) }\to \text{2Fe(s) + }3{\text{H}}_2\text{O(g)}$

For the above reaction, the ${\text{ΔG}}^{\text{0}}$ value can be deduced by calculating the ${\text{ΔH}}^{\text{0}}$ and ${\text{ΔS}}^{\text{0}}$ values from the standard enthalpy and entropy of formation of the reactants and products.

Step 1: Calculate ${\text{ΔH}}^{\text{0}}$

The standard enthalpy change for a reaction ${\text{ΔH}}^{\text{0}}$ is given in terms of the difference in the standard enthalpy of formation of the products and that of reactants.

i.e. $\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) }$

Where n_p and n_r are the number of moles of the products and reactants

For the given reaction:

$\begin{array}{l}\Delta {\text{H}}^0{\text{ = [2ΔH}}_{\text{f}}^{\text{0}}{\text{(Fe(s)) + 3ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}{\text{O(g))] - [1ΔH}}_{\text{f}}^{\text{0}}{\text{(Fe}}_{\text{2}}{\text{O}}_3{\text{(s)) + 3ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}_2\text{(g))]}\\ {\text{Based on the ΔH}}_{\text{f}}^{\text{0}}\text{ we have:}\\ \Delta {\text{H}}^0\text{ = [2 mole }\times (0\text{ kJ/mol) + 3 moles }\times \text{(-241}\text{.82 kJ/mol)]-[1 moles }\times \text{(}-824.3\text{ kJ/mol) }\\ \text{ + 3 moles }\times \text{(}0\text{ kJ/mol)]}\\ \text{ = 0 kJ -725}\text{.46 kJ + 824}\text{.3 kJ + 0 kJ = 98}\text{.8 kJ }\end{array}$

Step 2: Calculate ${\text{ΔS}}^{\text{0}}$

The standard entropy change for a reaction ${\text{ΔS}}^{\text{0}}$ is given in terms of the difference in the standard entropy of formation of the products and that of reactants.

i.e. $\Delta {\text{S}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{S}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{S}}_{\text{f}}^{\text{0}}\text{(reactants) }$

Where n_p and n_r are the number of moles of the products and reactants

For the given reaction:

$\begin{array}{l}\Delta {\text{S}}^0{\text{ = [2S}}_{\text{f}}^{\text{0}}{\text{(Fe(s)) + 3S}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}{\text{O(g))] - [1S}}_{\text{f}}^{\text{0}}{\text{(Fe}}_{\text{2}}{\text{O}}_3{\text{(s)) + 3S}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}\text{(g))]}\\ {\text{Based on the S}}_{\text{f}}^{\text{0}}\text{ values:}\\ \Delta {\text{S}}^0\text{ = [2 mole }\times (27.28\text{ J/mol-K) + 3 moles }\times \text{(188}\text{.72 J/mol-K)]-[1 moles }\times \text{(}87.4\text{ J/mol-K) }\\ \text{ + 3 moles }\times \text{(}130.59\text{ J/mol-K)]}\\ \text{ = 54}\text{.56 + 566}\text{.16 - 87}\text{.4 -391}\text{.77= 141}\text{.6 J/K }\end{array}$

Step 3: Estimate the temperature when ${\text{ΔG}}^{\text{0}}$ = 0

${\text{ΔH}}^{\text{0}}$ = 98.8 kJ and ${\text{ΔS}}^{\text{0}}$ = 0.1416 kJ/K

When ${\text{ΔG}}^{\text{0}}$ = 0, equation (1) becomes:

$\begin{array}{l}{\text{ΔH}}^{\text{0}}{\text{ = TΔS}}^{\text{0}}\\ \text{T = }\frac{{\text{ ΔH}}^{\text{0}}}{{\text{ΔS}}^{\text{0}}}=\frac{\text{98}\text{.8 kJ}}{\text{0}\text{.1416 kJ/K}}=698\text{ K}\end{array}$

Thus, ${\text{ΔG}}^{\text{0}}$ = 0 at T = 698 K. The reaction will be spontaneous at temperatures greater than

698K.

Therefore, reaction (a) would proceed spontaneously at a lower temperature

Conclusion

Reaction (a) proceeds spontaneously at 497 K while reaction (b) requires a temperature of 698 K Thus, the reaction of iron ore with carbon i.e. reaction (a) will proceed spontaneously at a lower temperature relative to (b).