Thermodynamics of Reduction

Abstract

Success and efficiency of a process are depending on the thermodynamic calculation of the reactions. How much reducing gas (CO) required to reduce iron oxides at different stages and temperatures that should be known before the actual process, i.e. how much reductant is required for the process.

Appendices

Probable Questions

• 1. Discuss the thermodynamics of metal oxide reduction. What is the condition of the reduction reaction of the oxide with respect to CO/CO₂ ratio?

• 2. State and discuss the probable mechanism by which an iron ore is reduced in a CO–CO₂ mixture.

• 3. Draw the diagram showing equilibrium CO/CO₂ ratio in contact with carbon and iron oxides at various temperatures. What is the diagram known as? Below 570 °C, FeO is not formed during reduction of Fe₃O₄. Why? Discuss the significant of the diagram for reduction of iron ore.

• 4. With the help of Fe–H–O system, find out the conditions of reduction of oxide and oxidation of metal? Discuss the importance of this system.

• 5. Discuss the changes in the composition of gas in stack region of the BF.

• 6. Discuss the silicon equilibrium in BF. Explain why at high blast temperature silicon content of metal increases.

• 7. What are the thermodynamics conditions required for de-sulphurization of iron in BF? Discuss.

• 8. What are the conditions required for maximum recovery of manganese in metal of a BF?

Examples

Example 10.1

Find out the heat of reaction ($\Delta H_{r,298}^{0}$) of the following reactions:

• (i)
  $$ 3{\text{Fe}}_{2} {\text{O}}_{3} + {\text{CO}} = 2{\text{Fe}}_{3} {\text{O}}_{4} + {\text{CO}}_{2} $$
• (ii)
  $$ {\text{Fe}}_{3} {\text{O}}_{4} + {\text{CO}} = 3{\text{FeO}} + {\text{CO}}_{2} $$
• (iii)
  $$ {\text{FeO}} + {\text{CO}} = {\text{Fe}} + {\text{CO}}_{2} $$

Given: $\Delta H_{f,298}^{0}$ for Fe₂O₃, Fe₃O₄, FeO, CO and CO₂ are—821.32, −1116.71, −264.43, −110.54 and −393.51 kJ/mol, respectively.

Solution

• (i)
  $$ 3{\text{Fe}}_{2} {\text{O}}_{3} + {\text{CO}} = 2{\text{Fe}}_{3} {\text{O}}_{4} + {\text{CO}}_{2} $$

Therefore,

$$ \begin{aligned} \Delta H_{r,298}^{0} & = \left( {2\Delta H_{{{\text{Fe}}_{3} {\text{O}}_{4} }}^{0} + \Delta H^{0}_{{{\text{CO}}_{2} }} } \right) - \left( {3\Delta H_{{{\text{Fe}}_{2} {\text{O}}_{3} }}^{0} + \Delta H_{\text{CO}}^{0} } \right) \\ & = [2( - 1116.71) + ( - 393.51)] - [3( - 821.32) + ( - 110.54)] \\ & = [ - 2233.42 - 393.51] - [ - 2463.96 - 110.54] \\ & = [ - 2626.93 - ( - 2574.5)] = - {\mathbf{52}}{\mathbf{.43}}\;{\mathbf{kJ}} /{\mathbf{mol}}\,{\mathbf{of}}\,{\mathbf{CO}}/{\mathbf{CO}}_{2} \\ & = - {\mathbf{17}}{\mathbf{.48}}\,{\mathbf{kJ}} /{\mathbf{mol}}\,{\mathbf{of}}\,{\mathbf{Fe}}_{{\mathbf{2}}} {\mathbf{O}}_{{\mathbf{3}}} \\ & = - {\mathbf{26}}{\mathbf{.22}}\,{\mathbf{kJ}} /{\mathbf{mol}}\,{\mathbf{of}}\,{\mathbf{Fe}}_{{\mathbf{3}}} {\mathbf{O}}_{{\mathbf{4}}} \\ \end{aligned} $$

• (ii)
  $$ {\text{Fe}}_{3} {\text{O}}_{4} + {\text{CO}} = 3{\text{FeO}} + {\text{CO}}_{2} $$

Therefore,

$$ \begin{aligned} \Delta H_{r,298}^{0} & = \left( {3\Delta H^{0}_{\text{FeO}} } \right.\left. { + \Delta H^{0}_{{{\text{CO}}2}} } \right) - \left( {\Delta H^{0}_{{{\text{Fe}}3{\text{O}}_{4} }} + \Delta H^{0}_{\text{CO}} } \right) \\ & = [3( - 264.43) + ( - 393.51)] - [ - 1116.71 + ( - 110.54)] \\ & = [ - 7932.9 - 393.51] - [ - 1116.71 - 110.54)] \\ & = [ - 1186.8 - ( - 1227.25)] = {\mathbf{40}}{\mathbf{.45}}\;{\mathbf{kJ}} /{\mathbf{mol}}\,{\mathbf{of}}\,{\mathbf{Fe}}_{{\mathbf{3}}} {\mathbf{O}}_{{\mathbf{4}}} /{\mathbf{CO}} /{\mathbf{CO}}_{{\mathbf{2}}} \\ & = {\mathbf{13}}{\mathbf{.48}}\;{\mathbf{kJ}} /{\mathbf{mol}}\,{\mathbf{of}}\,{\mathbf{FeO}} \\ \end{aligned} $$

• (iii)
  $$ {\text{FeO}} + {\text{CO}} = {\text{Fe}} + {\text{CO}}_{2} $$

Therefore,

$$ \begin{aligned} \Delta H_{r,298}^{0} & = \left( {\Delta H_{\text{Fe}}^{0} + } \right.\left. {\Delta H_{{{\text{CO}}_{2} }}^{0} } \right) - \left( {\Delta H_{\text{FeO}}^{0} + \Delta H^{0}_{\text{CO}} } \right) \\ & = [0 + ( - 393.51)] - [ - 264.43 + ( - 110.54)] \\ & = - 393.51 - [ - 264.43 - 110.54] = [ - 393.51 - ( - 374.97)) \\ & = - {\mathbf{18}}{\mathbf{.54}}\;{\mathbf{kJ}} /{\mathbf{mol}}\,{\mathbf{of}}\,{\mathbf{FeO}} /{\mathbf{Fe}} /{\mathbf{CO}} /{\mathbf{CO}}_{{\mathbf{2}}} \\ \end{aligned} $$

Example 10.2

Reduction of iron oxide in blast furnace takes place in three stages:

$$ {\text{Fe}}_{2} {\text{O}}_{3} \to {\text{Fe}}_{3} {\text{O}}_{4} \to {\text{FeO}} \to {\text{Fe}}. $$

The reactions are as follows:

• (i) $3{\text{Fe}}_{2} {\text{O}}_{3} + {\text{CO}} = 2{\text{Fe}}_{3} {\text{O}}_{4} + {\text{CO}}_{2} ,\Delta G_{1}^{0} = - 32969.92 - 53.85T\,{\text{J}}$

• (ii) ${\text{Fe}}_{3} {\text{O}}_{4} + {\text{CO}} = 3{\text{FeO}} + {\text{CO}}_{2} ,\Delta G_{2}^{0} = 29790.08 - 38.28T\,{\text{J}}$

• (iii) ${\text{FeO}} + {\text{CO}} = {\text{Fe}} + {\text{CO}}_{2} ,\Delta G_{3}^{0} = - 22802.8 + 24.27T\,{\text{J}}$

• (iv) ${\text{Fe}}_{3} {\text{O}}_{4} + 4{\text{CO}} = 3{\text{Fe}} + 4{\text{CO}}_{2} ,\Delta G_{4}^{0} = - 38618.32 + 34.52T\,{\text{J}}$

Find out the percentage of CO in CO–CO₂ mixture in equilibrium with

• (a) ${\text{Fe}}_{2} {\text{O}}_{3}{-}{\text{Fe}}_{3} {\text{O}}_{4}$, (b) ${\text{Fe}}_{3} {\text{O}}_{4} {-} {\text{FeO}}$, (c) ${\text{FeO}} {-} {\text{Fe }}\,{\text{at}}\,1000\;{\text{K}}$.

• (d) ${\text{Fe}}_{3} {\text{O}}_{4} {-} {\text{Fe}}\,{\text{at}}\,800\;{\text{K}}$.

• (e) Find out temperature at which Fe, FeO and Fe₃O₄ exist in equilibrium.

Solution

• (a) Equilibrium constant, $k_{1} = \left[ {\frac{{a_{{{\text{Fe}}_{3} {\text{O}}_{4} }}^{2} a_{{{\text{CO}}_{2} }} }}{{a_{{{\text{Fe}}_{2} {\text{O}}_{3} }}^{3} a_{\text{CO}} }}} \right] = \left( {\frac{{p_{{{\text{CO}}_{2} }} }}{{p_{\text{CO}} }}} \right)$

At Std state, $a_{{{\text{Fe}}_{3} {\text{O}}_{4} }} = 1\quad {\text{and}}\quad a_{{{\text{Fe}}_{2} {\text{O}}_{3} }} = 1$

Since $\Delta G_{1}^{0} = - 32969.92 - 53.85T\,{\text{J}} = - 32969.92 - 53.85 \times 1000 = - 86819.92\,{\text{J}}$

Again $\Delta G_{1}^{0} = - {\text{RT}}\,\ln \,k_{1}$

Therefore, $- 86819.92 = - 8.314 \times 1000 \times \ln \,k_{1}$

$\begin{aligned} \ln \,k_{1} & = 10.4426 \\ k_{1} & = 34290.28 \\ \end{aligned}$

Let $p_{\text{CO}} = x$, since $p_{\text{CO}} + p_{{{\text{CO}}_{2} }} = 1$

Therefore, $p_{{{\text{CO}}_{2} }} = 1 - x$

Since $k_{1} = \left( {\frac{{p_{{{\text{CO}}_{2} }} }}{{p_{\text{CO}} }}} \right) = 34290.28 = \left( {\frac{1 - x}{x}} \right)$

Therefore, $x = 2.916 \times 10^{ - 5}$

Hence, the percentage of CO in CO–CO₂ mixture in equilibrium with ${\text{Fe}}_{2} {\text{O}}_{3} - {\text{ Fe}}_{3} {\text{O}}_{4} = {\mathbf{2}}{\mathbf{.916}} \times {\mathbf{10}}^{{ - {\mathbf{3}}}} {\mathbf{\% }}$

Similarly, we can calculate the following:

• (b) Equilibrium constant, $k_{2} = \left( {\frac{{p_{{{\text{CO}}_{2} }} }}{{p_{\text{CO}} }}} \right)$

Since $\Delta G_{2}^{0} = 29790.08 - 38.28\;{{T}} = 29790.08 - 38.28 \times 1000 = - 8490\;{\text{J}}$

Again $\Delta G_{2}^{0} = - {\text{RT}}\,\ln \,k_{2}$

Therefore, $- 8490 = - 8.314 \times 1000 \times \ln \,k_{2}$

$\begin{array}{*{20}l} {\ln k_{2} = 1.021} \hfill \\ {k_{2} = 2.776} \hfill \\ \end{array}$

Let $p_{\text{CO}} = x$, since $p_{\text{CO}} + p_{{{\text{CO}}_{2} }} = 1$

Therefore, $p_{{{\text{CO}}_{2} }} = 1 - x$

Since $k_{2} = \left( {\frac{{p_{{{\text{CO}}_{ 2} }} }}{{p_{\text{CO}} }}} \right) = 2.776 = \left( {\frac{1 - x}{x}} \right)$

Therefore, $x = 0.2648$

Hence, the percentage of CO in CO–CO₂ mixture in equilibrium with Fe₃O₄–FeO = 26.48%

• (c) Equilibrium constant, $k_{3} = \left( {\frac{{p_{{{\text{CO}}_{ 2} }} }}{{p_{\text{CO}} }}} \right)$

Since $\Delta G_{3}^{0} = - 22802.8 + 24.27T = - 22802.8 + 24.27 \times 1000 = 1468\,{\text{J}}$

Again $\Delta G_{3}^{0} = - {\text{RT}}\ln \,k_{3}$

Therefore, $1468 = - 8.314 \times 1000 \times \ln \,k_{3}$

$\begin{array}{*{20}l} {\ln \,k_{3} = - 0.1766} \hfill \\ {k_{3} = 0.838} \hfill \\ \end{array}$

Let $p_{\text{CO}} = x$, since $p_{\text{CO}} + p_{{{\text{CO}}_{2} }} = 1$

Therefore, $p_{{{\text{CO}}_{2} }} = 1 - x$

Since $k_{3} = \left( {\frac{{p_{{{\text{CO}}_{2} }} }}{{p_{\text{CO}} }}} \right) = 0.838 = \left( {\frac{1 - x}{x}} \right)$

Therefore, $x = 0.544$

Hence, the percentage of CO in CO–CO₂ mixture in equilibrium with FeO–Fe = 54.4%

• (d) Equilibrium constant, $k_{4} = \left( {\frac{{p_{{{\text{CO}}_{2} }} }}{{p_{\text{CO}} }}} \right)^{4}$

Since $\Delta G_{4}^{0} = - 38618.32 + 34.52T = - 38618.32 + 34.52 \times 800 = - 11002\,{\text{J}}$

Again $\Delta G_{4}^{0} = - {\text{RT}}\ln \,k_{4}$

Therefore, $- 11002 = - 8.314 \times 800 \times \ln \,k_{3}$

$\ln \,{\text{k}}_{4} = 1.654$

$k_{4} = 5.228 = \left( {\frac{{p_{{{\text{CO}}_{2} }} }}{{p_{\text{CO}} }}} \right)^{4}$, so $\left( {\frac{{p_{{{\text{CO}}_{2} }} }}{{p_{\text{CO}} }}} \right) = (5.228)^{1/4} = 1.512$

Let $p_{\text{CO}} = x$, since $p_{\text{CO}} + p_{{{\text{CO}}_{2} }} = 1$

Therefore, $p_{{{\text{CO}}_{2} }} = 1 - x$

Since $\left( {\frac{{p_{{{\text{CO}}_{2} }} }}{{p_{\text{CO}} }}} \right) = 1.512 = \left( {\frac{1 - x}{x}} \right)$

Therefore, $x = 0.3981$

Hence, the percentage of CO in CO–CO₂ mixture in equilibrium with Fe₃O₄–Fe = 39.81%

• (e) $\begin{aligned} & {\text{Fe}}_{3} {\text{O}}_{4} + {\text{CO}} = 3{\text{FeO}} + {\text{CO}}_{2} ,\Delta G_{2}^{0} = 29790.08 - 38.28T\,{\text{J}} \\ & {\text{Fe}} + {\text{CO}}_{2} = {\text{FeO}} + {\text{CO,}} - \Delta G_{3}^{0} = - 22802.8 + 24.27T\,{\text{J}} \\ \hline & {\text{Fe}}_{3} {\text{O}}_{4} + {\text{Fe}} = 4{\text{FeO}},\Delta G^{0} = \left( {\Delta G_{2}^{0} - \Delta G_{3}^{0} } \right) = 52592.88 - 62.55T \\ \end{aligned}$

At equilibrium $\Delta G^{0} = 0$

Therefore, $52592.88 - 62.55T = 0$

Hence, ${\mathbf{T}} = {\mathbf{840}}{\mathbf{.81}}\;{\mathbf{K}} \approx {\mathbf{567}}{\mathbf{.81}}\;^{\circ}{\mathbf{C}}$

Therefore, at 567.81 °C Fe, FeO and Fe₃O₄ will exist in equilibrium.

Example 10.3

Find the oxygen pressure of 75% CO and 25% CO₂ mixture at 1 atmospheric pressure and 1500 K. Will this gas be able to reduce Fe₂O₃ and FeO at 1500 K?

Given: $2{\text{CO}}_{2} = 2{\text{CO}} + {\text{O}}_{2} \quad \Delta G^{0} = 565,258.4 - 173.64T\,{\text{J}}$

Oxygen pressure for Fe₂O₃ at 1500 K = 10⁻² atm,

Oxygen pressure for FeO at 1500 K = 10⁻¹² atm.

Solution

$$ \begin{array}{*{20}l} {\Delta G^{0} = 565,258.4 - 173.64T = 565,258 - 173.64 \times 1500 = 304,798\;{\text{J}}} \hfill \\ {\Delta G^{0} = - {\text{RT}}\,\ln \,k} \hfill \\ \end{array} $$

Therefore, $304,798 = - 8.314 \times 1500 \times \ln \,k$

$\begin{array}{*{20}l} {\ln k = - 24.44} \hfill \\ {k = 2.43 \times 10^{ - 11} } \hfill \\ {2{\text{CO}}_{2} = 2{\text{CO}} + {\text{O}}_{2} } \hfill \\ \end{array}$

Equilibrium constant, $k = \left[ {\frac{{p_{\text{CO}}^{2} p_{{{\text{O}}_{2} }} }}{{p_{{{\text{CO}}_{2} }}^{2} }}} \right] = \left[ {\frac{{\left( {0.75} \right)^{2} p_{{{\text{O}}_{2} }} }}{{\left( {0.25} \right)^{2} }}} \right]$

Therefore $p_{{{\text{O}}_{2} }} = k.\left( {\frac{{p_{{{\text{CO}}_{2} }}^{2} }}{{p_{\text{CO}}^{2} }}} \right) = 2.43 \times 10^{ - 11} \times \left[ {\frac{{\left( {0.25} \right)^{2} }}{{\left( {0.75} \right)^{2} }}} \right] = 2.7 \times 10^{ - 12} \,{\text{atm}}.$

Since oxygen pressure for Fe₂O₃ at 1500 K is 10⁻² atm which is greater than equilibrium oxygen pressure, 2.7 × 10⁻¹² atm, mixture of gases will reduce the Fe₂O₃. But oxygen pressure for FeO at 1500 K is 10⁻¹² atm which is lower than equilibrium oxygen pressure, 2.7 × 10⁻¹² atm, hence mixture of gases will not reduce FeO; instate of that FeO will be oxidized.

Example 10.4

For the reaction: ${\text{C}} + {\text{CO}}_{2} = 2{\text{CO}},\Delta G^{0} = 170.71 - 0.17T\,{\text{kJ}}$

Find the temperature of which carbon would be in equilibrium with a mixture of CO and CO₂ (at one atm. pressure) containing 70% CO.

Solution

The equilibrium constant of the reaction: $k = \left[ {\frac{{p_{\text{CO}}^{2} }}{{a_{{{\text{C}} }} p_{{{\text{CO}}_{2} }} }}} \right] = \left[ {\frac{{p_{\text{CO}}^{2} }}{{p_{{{\text{CO}}_{2} }} }}} \right]$

Since a_C = 1, 70% CO, i.e. $p_{\text{CO}} = 0.7$ and $p_{{{\text{CO}}_{2} }} = 0.3$; Since total pressure = 1 atm.

Therefore, $k = \left( {\frac{{\left( {0.7} \right)^{2} }}{0.3}} \right) = 1.633$

Since

$$ \begin{array}{*{20}l} {\Delta G^{0} = - {\text{RT}}\,\ln \,k = - 8.314T \times \ln (1.633) = - (8.314 \times 0.49)T = - 4.08T} \hfill \\ {(170.71 - 0.17) \times 10^{3} T = - 4.08T} \hfill \\ {(170 - 4.08)T = 170{,}710} \hfill \\ \end{array} $$

Therefore, $T = 1028.9\;K = 755.9 = {\mathbf{756}}\;^{{\mathbf{^\circ }}} {\mathbf{C}}$

Example 10.5

• (i) ${\text{Fe}}(1) + 1 /2\;{\text{O}}_{2} ({\text{g}}) = {\text{FeO}}(1)\quad {\mathbf{\Delta G}}^{0}_{{\mathbf{i}}} = - 232714.08 + 45.31T\;{\text{J}}$

• ii) ${\text{C}}({\text{s}}) + 1 /2\;{\text{O}}_{2} ({\text{g}}) = {\text{CO}}({\text{g}})\quad {\mathbf{\Delta G}}^{{\mathbf{o}}}_{{{\mathbf{ii}}}} = - 111712.8 - 87.65T\;{\text{J}}$

• iii) ${\text{C}}({\text{s}}) + {\text{O}}_{2} ({\text{g}}) = {\text{CO}}_{2} ({\text{g}})\quad {\mathbf{\Delta G}}^{{\mathbf{^\circ }}}_{{{\mathbf{iii}}}} = - 394132.8 - 0.84T\;{\text{J}}$

From the above reactions, find out the $\Delta G^{\text{0}}_{T}$ of the reaction: ${\text{FeO}}({\text{s}}) + {\text{CO}}({\text{g}}) = {\text{Fe}}({\text{s}}) + {\text{CO}}_{2} ({\text{g}})$ at 1000 °C.

Solution

Taking reverse equation (i): ${\text{FeO}}(1) = {\text{Fe}}(1) + 1/2{\text{O}}_{2} ({\text{g}}) - {\mathbf{\Delta G}}^{{\mathbf{0}}}_{{\mathbf{i}}}$

Taking reverse equation (ii): ${\text{CO}}({\text{g}}) = {\text{C}}({\text{s}}) + 1/2{\text{O}}_{2} ({\text{g}}) - {\mathbf{\Delta G}}^{{\mathbf{o}}}_{{{\mathbf{ii}}}}$

Taking equation (iii): $\begin{aligned} & {\text{C}}({\text{s}}) + {\text{O}}_{2} ({\text{g}}) = {\text{CO}}_{2} ({\text{g}})\quad {\mathbf{\Delta G}}^{{\mathbf{0}}}_{{{\mathbf{ii}}}}\\ & \hfill \\ \hline \end{aligned}$

By adding: ${\text{FeO}}({\text{l}}) + {\text{CO}}({\text{g}}) = {\text{Fe}}(1) + {\text{CO}}_{2} ({\text{g}})$

$$ \begin{aligned} {\mathbf{\Delta G}}^{{\mathbf{o}}}_{{\mathbf{T}}} & = - {\mathbf{\Delta G}}^{{\mathbf{o}}}_{{\mathbf{i}}} - {\mathbf{\Delta G}}^{{\mathbf{o}}}_{{{\mathbf{ii}}}} + {\mathbf{\Delta G}}_{{{\mathbf{iii}}}}^{{\mathbf{o}}} = - ( - 232714.08 + 45.31T) - ( - 111712.8 - 87.65T) \\ & \quad + ( - 394132.8 - 0.84T) \\ & = \{ (232714.08 + 111712.8) - 394132.8\} + \{ - 45.31 - 0.84 + 87.65\} T \\ & = - 49705.92 + 41.5T = - 49705.92 + 41.5 \times 1273 \\ & = 3123.58\,{\text{J}} = 3.12\,{\text{kJ}} \\ \end{aligned} $$

Since value of $\Delta {\mathbf{G}}^{{\mathbf{o}}}_{{\mathbf{T}}}$ is (+) ve, reaction will not be possible in forward direction, reaction will take place in backward direction.

Problems

Problem 10.1

Find out the heat of reaction ($\Delta H_{r,298}^{0}$) of the following reactions:

• (i) $3{\text{Fe}}_{2} {\text{O}}_{3} + {\text{H}}_{2} = 2{\text{Fe}}_{3} {\text{O}}_{4} + {\text{H}}_{2} {\text{O}}$

• (ii) ${\text{Fe}}_{3} {\text{O}}_{4} + {\text{H}}_{2} = 3{\text{FeO}} + {\text{H}}_{2} {\text{O}}$

• (iii) ${\text{FeO}} + {\text{H}}_{2} = {\text{Fe}} + {\text{H}}_{2} {\text{O}}$

Given: $\Delta H_{f \, ,298}^{0}$ for Fe₂O₃, Fe₃O₄, FeO, and H₂O are −821.32, −1116.71, −264.43, and −241.81 kJ/mol respectively.

[Ans: (i) −3.48 kJ/mol of Fe₂O₃, (ii) 81.61 kJ/mol of Fe₃O₄, (iii) 22.61 kJ/mol of FeO.]

Problem 10.2

Find out the heat of reaction $(\Delta H_{r,298}^{0})$ of the following reactions:

• (a)
  $$ {\text{C}} + {\text{CO}}_{2} = 2{\text{CO}} $$
• (b)
  $$ {\text{C}} + {\text{H}}_{2} {\text{O}} = {\text{H}}_{2} + {\text{CO}} $$

Given: $\Delta H_{f \, ,298}^{0}$ for CO₂, CO, and H₂O are −393.51, −110.54 and −241.81 kJ/mol respectively.

[Ans: (a) 172.42 kJ/mol of C, (b) 131.27 kJ/mol of C.]

Problem 10.3

For the reaction: ${\text{CO}}_{2} + {\text{C}} = 2{\text{CO}}$, $\Delta G^{0} = 170707.2 - 174.47T\;{\text{J}}$

• (a) Find the temperature at which graphite will be in equilibrium with CO and CO₂ at 1 atm pressure containing (i) 30% CO, (ii) 60% CO.

• (b) Find similar values for a total pressure of 0.1 atm.

• (c) Find composition of CO and CO₂ in equilibrium with graphite at 1000 K.

[Ans: (a) (i) 618.31 °C, (ii) 700.55 °C; (b) (i) 537.31 °C, (ii) 656.95 °C; (c) 69.38% CO and 30.62% CO₂].

Problem 10.4

For the reaction: C (s) + CO₂ (g) = 2CO (g) at 1100 K and 1 atm, the equilibrium mixture contains 91.6% CO and 8.4% CO₂ by volume. (i) Calculate equilibrium constant for the reaction and (ii) the partial pressure of CO₂ gas if partial pressure of CO is changed to 10⁻⁴ atm.

[Ans: (i) 9.989, (ii) 1.0 × 10⁻⁹].

Problem 10.5

The reactions are as follows:

• (i) $3{\text{Fe}}_{2} {\text{O}}_{3} + {\text{CO}} = 2{\text{Fe}}_{3} {\text{O}}_{4} + {\text{CO}}_{2} ,\Delta G_{1}^{0} = - 32969.92 - 53.85T\;{\text{J}}$

• (ii) ${\text{Fe}}_{3} {\text{O}}_{4} + {\text{CO}} = 3{\text{FeO}} + {\text{CO}}_{2} ,\Delta {\text{G}}_{2}^{0} = 29790.08 - 38.28T\;{\text{J}}$

• (iii) ${\text{FeO}} + {\text{CO}} = {\text{Fe}} + {\text{CO}}_{2} ,\Delta G_{3}^{0} = - 22802.8 + 24.27T\;{\text{J}}$

Find out the percentage of CO in CO–CO₂ mixture in equilibrium with

• (a) Fe₂O₃–Fe₃O₄, (b) Fe₃O₄–FeO, (c) FeO–Fe at 1200 K.

[Ans: (a) 5.647 × 10^−3%, (b) 16.55% and (c) 65.33%].

Problem 10.6

Calculate $\Delta G_{T}^{0}$ and the equilibrium CO/CO₂ ratio for the reduction of wustite at 900 °C according to the reaction: FeO(s) + CO (g) = Fe (s) + CO₂ (g).

Given:

• (i) ${\text{Fe}}({\text{I}}) + 1/2{\text{O}}_{2} ({\text{g}}) = {\text{FeO}}({\text{l}})\quad {\mathbf{\Delta G}}_{{\mathbf{i}}}^{{\mathbf{o}}} = - 232714.08 + 45.31T\;{\text{J}}$

• (ii) ${\text{C}}({\text{s}}) + 1/2{\text{O}}_{2} ({\text{g}}) = {\text{CO}}({\text{g}})\quad {\mathbf{\Delta G}}^{{\mathbf{o}}}_{{{\mathbf{ii}}}} = - 111712.8 - 87.65T\;{\text{J}}$

• (iii)
  $$ {\text{C}}({\text{s}}) + {\text{O}}_{2} ({\text{g}}) = {\text{CO}}_{2} ({\text{g}})\quad {\mathbf{\Delta G}}_{{{\mathbf{iii}}}}^{{\mathbf{o}}} = - 394132.8 - 0.84T\;{\text{J}} $$

[Ans: −1026.42 J and 0.9].