Chapter 16, Problem 48QAP

Interpretation Introduction

(a)

Interpretation:

Based on thermodynamic calculations an explanation needs to be provided to substantiate the low temperature requirement for the production of Mn metal by the decomposition of MnO₂

Concept introduction:

• The change in the Gibbs free energy, ΔG is a thermodynamic function which governs the spontaneity of a chemical reaction. If ΔG is negative, the reaction is spontaneous. Positive value indicates that the reaction is non-spontaneous and if ΔG = 0, then the reaction is said to be 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 48QAP

In this method, the reaction will become spontaneous at T >2827 K. This temperature is much higher than that deduced for the method described in part (b), hence it is not the recommended method for the extraction of Mn from MnO₂

Explanation of Solution

The given reaction is:

${\text{MnO}}_{\text{2}}\text{(s) }\rightleftharpoons {\text{ Mn(s) + O}}_{\text{2}}\text{(g)}$

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:

$\begin{array}{l}\Delta {\text{H}}^0{\text{ = [1ΔH}}_{\text{f}}^{\text{0}}{\text{(Mn(s)) + 1ΔH}}_{\text{f}}^{\text{0}}{\text{(O}}_{\text{2}}{\text{(g)) ] - [1ΔH}}_{\text{f}}^{\text{0}}{\text{(MnO}}_2\text{(s})\text{)]}\\ {\text{Substituting the ΔH}}_{\text{f}}^{\text{0}}\text{ values from the standard thermochemical data:}\\ \Delta {\text{H}}^0\text{ = [1 mole }\times 0\text{ kJ/mol + 1 mole }\times 0\text{ kJ/mol] - [1 mole }\times -520.03\text{ kJ/mol]}\\ \text{ = 0 + 0 + 520}\text{.03 = 520}\text{.03 kJ }\end{array}$

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:

$\begin{array}{l}\Delta {\text{S}}^0{\text{ = [1S}}_{\text{f}}^{\text{0}}{\text{(Mn(s)) + 1S}}_{\text{f}}^{\text{0}}{\text{(O}}_{\text{2}}{\text{(g)) ] - [1S}}_{\text{f}}^{\text{0}}{\text{(MnO}}_2\text{(s})\text{)]}\\ {\text{Substituting the S}}_{\text{f}}^{\text{0}}\text{ values from the standard thermochemical data:}\\ \Delta {\text{S}}^0\text{ = [1 mole }\times 32.01\text{ J/mol-K + 1 mole }\times 205.02\text{ J/mol-K] - [1 mole }\times 53.05\text{ J/mol-K]}\\ \text{ = 32}\text{.01 + 205}\text{.02 - 53}\text{.05 = 183}\text{.98 J/K }\end{array}$

Step 3: Estimate the temperature when ${\text{ΔG}}^{\text{0}}$ = 0 i.e. at equilibrium

${\text{ΔH}}^{\text{0}}$ = 520.03 kJ and ${\text{ΔS}}^{\text{0}}$ = 0.18398 kJ/K

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

$\begin{array}{l}{\text{ΔH}}^{\text{0}}{\text{ = TΔS}}^{\text{0}}\\ \text{T = }\frac{{\text{ ΔH}}^{\text{0}}}{{\text{ΔS}}^{\text{0}}}=\frac{\text{520}\text{.03 kJ}}{\text{0}\text{.18398 kJ/K}}=+2827\text{ K}\end{array}$

Thus, at equilibrium i.e. when ${\text{ΔG}}^{\text{0}}$ = 0 the temperature, T = +2827 K. The reaction will be spontaneous at temperatures greater than 2827 K

Interpretation Introduction

(b)

Interpretation:

Based on thermodynamic calculations an explanation needs to be provided to substantiate the low temperature requirement for the production of Mn metal by the reaction of MnO₂ with hydrogen gas.

Concept introduction:

• The change in the Gibbs free energy, ΔG is a thermodynamic function which governs the spontaneity of a chemical reaction. If ΔG is negative, the reaction is spontaneous. Positive value indicates that the reaction is non-spontaneous and if ΔG = 0, then the reaction is said to be 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 48QAP

The production of Mn by the reaction between MnO₂ and H₂ is the recommended method as it becomes spontaneous at comparatively lower temperatures i.e. above 382 K. This temperature requirement is much lower than that calculated for methods described in (a) and (c)

Explanation of Solution

The given reaction is:

${\text{MnO}}_{\text{2}}{\text{(s) + 2H}}_{\text{2}}\text{(g)}\rightleftharpoons {\text{ Mn(s) + 2H}}_{\text{2}}\text{O(g)}$

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:

$\begin{array}{l}\Delta {\text{H}}^0{\text{ = [1ΔH}}_{\text{f}}^{\text{0}}{\text{(Mn(s)) + 2ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}{\text{O(g)) ] - [1ΔH}}_{\text{f}}^{\text{0}}{\text{(MnO}}_2\text{(s}){\text{) + 2ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}\text{(g)) ]}\\ {\text{Substituting the ΔH}}_{\text{f}}^{\text{0}}\text{ values from the standard thermochemical data:}\\ \Delta {\text{H}}^0\text{ = [1 mole }\times 0\text{ kJ/mol + 2 moles }\times -241.82\text{ kJ/mol] - [1 mole }\times -520.03\text{ kJ/mol }\\ \text{ +2 moles }\times 0\text{ kJ/mol]}\\ \text{ = 0 - 483}\text{.64 + 520}\text{.03 + 0 = 36}\text{.39 kJ }\end{array}$

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:

$\begin{array}{l}\Delta {\text{S}}^0{\text{ = [1S}}_{\text{f}}^{\text{0}}{\text{(Mn(s)) + 2S}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}{\text{O(g)) ] - [1S}}_{\text{f}}^{\text{0}}{\text{(MnO}}_2\text{(s}){\text{) + 2S}}_{\text{f}}^{\text{0}}{\text{(H}}_{\text{2}}\text{(g))]}\\ {\text{Substituting the S}}_{\text{f}}^{\text{0}}\text{ values from the standard thermochemical data:}\\ \Delta {\text{S}}^0\text{ = [1 mole }\times 32.01\text{ J/mol-K + 2 moles }\times 188.72\text{ J/mol-K] - [1 mole }\times 53.05\text{ J/mol-K }\\ \text{ + 2 moles }\times 130.59\text{ J/mol-K]}\\ \text{ = 32}\text{.01 + 377}\text{.44 - 53}\text{.05 -261}\text{.18 = +95}\text{.22 J/K }\end{array}$

Step 3: Estimate the temperature when ${\text{ΔG}}^{\text{0}}$ = 0 i.e. at equilibrium

${\text{ΔH}}^{\text{0}}$ = +36.39 kJ and ${\text{ΔS}}^{\text{0}}$ = +0.09522 kJ/K

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

$\begin{array}{l}{\text{ΔH}}^{\text{0}}{\text{ = TΔS}}^{\text{0}}\\ \text{T = }\frac{{\text{ ΔH}}^{\text{0}}}{{\text{ΔS}}^{\text{0}}}=\frac{\text{36}\text{.39 kJ}}{\text{0}\text{.09522 kJ/K}}=+382\text{ K}\end{array}$

Thus, at equilibrium i.e. when ${\text{ΔG}}^{\text{0}}$ = 0 the temperature, T = +382 K. The reaction will be spontaneous at temperatures greater than 382 K. This is in accordance with the low temperature requirement and hence the recommended method.

Interpretation Introduction

(c)

Interpretation:

Based on thermodynamic calculations an explanation needs to be provided to substantiate the low temperature requirement for the production of Mn metal by the reaction of MnO₂ with coke i.e. carbon.

Concept introduction:

• The change in the Gibbs free energy, ΔG is a thermodynamic function which governs the spontaneity of a chemical reaction. If ΔG is negative, the reaction is spontaneous. Positive value indicates that the reaction is non-spontaneous and if ΔG = 0, then the reaction is said to be 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 48QAP

In this method, the reaction will become spontaneous at T >677 K. This temperature is much higher than that deduced for the method described in part (b), hence it is not the recommended method for the extraction of Mn from MnO₂

Explanation of Solution

The given reaction is:

${\text{MnO}}_{\text{2}}\text{(s) + C(s)}\rightleftharpoons {\text{ Mn(s) + CO}}_2\text{(g)}$

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:

$\begin{array}{l}\Delta {\text{H}}^0{\text{ = [1ΔH}}_{\text{f}}^{\text{0}}{\text{(Mn(s)) + 1ΔH}}_{\text{f}}^{\text{0}}{\text{(CO}}_{\text{2}}{\text{(g)) ] - [1ΔH}}_{\text{f}}^{\text{0}}{\text{(MnO}}_2\text{(s}){\text{) + 1ΔH}}_{\text{f}}^{\text{0}}\text{(C(s)) ]}\\ {\text{Substituting the ΔH}}_{\text{f}}^{\text{0}}\text{ values from the standard thermochemical data:}\\ \Delta {\text{H}}^0\text{ = [1 mole }\times 0\text{ kJ/mol + 1 mole }\times -393.51\text{ kJ/mol] - [1 mole }\times -520.03\text{ kJ/mol }\\ \text{ +1 mole }\times 0\text{ kJ/mol]}\\ \text{ = 0 -393}\text{.51 + 520}\text{.03 + 0 = 126}\text{.52 kJ }\end{array}$

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:

$\begin{array}{l}\Delta {\text{S}}^0{\text{ = [1S}}_{\text{f}}^{\text{0}}{\text{(Mn(s)) + 1S}}_{\text{f}}^{\text{0}}{\text{(CO}}_{\text{2}}{\text{(g)) ] - [1S}}_{\text{f}}^{\text{0}}{\text{(MnO}}_2\text{(s}){\text{) + 1S}}_{\text{f}}^{\text{0}}\text{(C(s))]}\\ {\text{Substituting the S}}_{\text{f}}^{\text{0}}\text{ values from the standard thermochemical data:}\\ \Delta {\text{S}}^0\text{ = [1 mole }\times 32.01\text{ J/mol-K + 1 mole }\times 213.68\text{ J/mol-K] - [1 mole }\times 53.05\text{ J/mol-K }\\ \text{ + 1 mole }\times 5.69\text{ J/mol-K]}\\ \text{ = 32}\text{.01 + 213}\text{.68 - 53}\text{.05 -5}\text{.69 = +186}\text{.95 J/K }\end{array}$

Step 3: Estimate the temperature when ${\text{ΔG}}^{\text{0}}$ = 0 i.e. at equilibrium

${\text{ΔH}}^{\text{0}}$ = +126.52 kJ and ${\text{ΔS}}^{\text{0}}$ = +0.18695 kJ/K

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

$\begin{array}{l}{\text{ΔH}}^{\text{0}}{\text{ = TΔS}}^{\text{0}}\\ \text{T = }\frac{{\text{ ΔH}}^{\text{0}}}{{\text{ΔS}}^{\text{0}}}=\frac{\text{126}\text{.52 kJ}}{\text{0}\text{.18695 kJ/K}}=+676.7\text{ K}\end{array}$

Thus, at equilibrium i.e. when ${\text{ΔG}}^{\text{0}}$ = 0 the temperature, T = +677 K. The reaction will be spontaneous at temperatures greater than +677 K. This is not in accordance with the low temperature requirement and hence not a recommended method.