Question

: Many metal oxides are converted to the free metal by reduction with other elements, such as $\mathbf{\text{C}}$or$\mathbf{\text{Si}}$. For each of the following reactions, calculate the temperature at which the reduction occurs spontaneously:

(a)${\mathbf{\text{MnO}}}_{\mathbf{\text{2}}}\mathbf{\text{(s)}}\mathbf{\to }{\mathbf{\text{Mn(s) +O}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

(b)${\mathbf{\text{MnO}}}_{\mathbf{\text{2}}}\mathbf{\text{(s) + 2C(graphite)}}\mathbf{\to }\mathbf{\text{Mn(s) + 2CO(g)}}$

(c)${\mathbf{\text{MnO}}}_{\mathbf{\text{2}}}\mathbf{\text{(s) + C(graphite)}}\mathbf{\to }{\mathbf{\text{Mn(s) + CO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

(d)${\mathbf{\text{MnO}}}_{\mathbf{\text{2}}}\mathbf{\text{(s) + Si(s)}}\mathbf{\to }{\mathbf{\text{Mn(s) + SiO}}}_{\mathbf{\text{2}}}\mathbf{\text{(s)}}$

Short answer

(a) For the reaction ${\text{MnO}}_{\text{2}}\text{(s)}\to {\text{Mn(s) +O}}_{\text{2}}\text{(g)}$, the reduction occurs spontaneously at the temperature of $\text{2835.6 K}$.

(b) For the reaction ${\text{MnO}}_{\text{2}}\text{(s) + 2C(graphite)}\to \text{Mn(s) + 2CO(g)}$, the reduction occurs spontaneously at the temperature of $\text{827.7 K}$.

(c) For the reaction ${\text{MnO}}_{\text{2}}\text{(s) + C(graphite)}\to {\text{Mn(s) + CO}}_{\text{2}}\text{(g)}$, the reduction occurs spontaneously at the temperature of $\text{682.3 K}$.

(d) For the reaction ${\text{MnO}}_{\text{2}}\text{(s) + Si(s)}\to {\text{Mn(s) + SiO}}_{\text{2}}\text{(s)}$, the reduction occurs spontaneously at the temperature of $\text{- 199823.5 K}$.

Step-by-step solution

Concept Introduction

According to the classical or earlier notion, reduction is a process that involves the addition of hydrogen or any other electropositive element, as well as the removal of oxygen or any other electronegative element.

According to the electrical idea, reduction is the process of gaining one or more electrons by an atom or ion.

Calculation of temperature for part (a)

(a)

The reaction given is –

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

The$∆H$ of the given reaction is –

$$\begin{gathered} \Delta H^0=\sum n\Delta H_f^0\mathrm{Pr}oducts-\sum n\Delta H_f^{0}Reac\tan ts \\ =\left[\Delta H_f^{0}Mn\left(s\right)+\Delta H_f^0{O}_2\left(g\right)\right]-\left[\Delta H_f^{0}Mn{O}_2\left(s\right)\right] \\ =[0+0]-[-520.9KJ/mol] \\ =520.9KJ/mol \end{gathered}$$

Similarly, the$∆S$ of the given reaction is –

$$\begin{gathered} \Delta S^0=\sum n\Delta S^0\mathrm{Pr}oducts-\sum n\Delta S^{0}Reac\tan ts \\ =\left[\Delta S^{0}Mn\left(s\right)+\Delta S^0{O}_2\left(g\right)\right]-\left[\Delta S^{0}Mn{O}_2\left(s\right)\right] \\ =[31.8J/mol·K+205.0J/mol·K]-[53.1J/mol·K] \\ =183.7J/mol·K \end{gathered}$$

The$\Delta G^0$ for the reaction is –

$\Delta G^0=\Delta H^0-T\Delta S^0$

For spontaneous process$\Delta G^0<0$ , hence $\Delta H^0<T\Delta S^0$.

Assume that $\Delta H^0=T\Delta S^0$-

$$\begin{gathered} T=\frac{\Delta H^0}{\Delta S^0} \\ T=\frac{520.9KJ/mole}{183.7·{10}^{-3}KJ/mole} \\ T=2835.6K \end{gathered}$$

Therefore, the value for temperature is obtained as $\text{2835.6 K}$.

Calculation of temperature for part (b)

(b)

The reaction given is –

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

The $∆H$of the given reaction is –

$$\begin{gathered} \Delta H^0=\sum n\Delta H_f^0\mathrm{Pr}oducts-\sum n\Delta H_f^{0}Reac\tan ts \\ =\left[\Delta H_f^{0}Mn\left(s\right)+2\Delta H_f^{0}CO\left(g\right)\right]-\left[\Delta H_f^{0}Mn{O}_2\left(s\right)+2\Delta H_f^{0}C\right] \\ =[0+2(-110.5KJ/mol\left)\right]-[-520.9KJ/mol+2(0\left)\right] \\ =299.9KJ/mol \end{gathered}$$

Similarly, the$\Delta S$of the given reaction is –

$$\begin{gathered} \Delta S^0=\sum n\Delta S^0\mathrm{Pr}oducts-\sum n\Delta S^{0}Reac\tan ts \\ =\left[\Delta S^{0}Mn\left(s\right)+\Delta S^{0}2CO\left(g\right)\right]-\left[\Delta S^{0}Mn{O}_2\left(s\right)+\Delta S^{0}2C\right] \\ =[31.8J/mol·K+2(197.5J/mol·K\left)\right]-[53.1J/mol·K+2(5.686J/mol·K\left)\right] \\ =362.328mol·K \end{gathered}$$

The$\Delta G^0$for the reaction is –

$\Delta G^0=\Delta H^0-T\Delta S^0$

For spontaneous process $\Delta G^0<0$, hence $\Delta H^0<T\Delta S^0$.

Assume that $\Delta H^0=T\Delta S^0$-

localid="1663320359498" $$\begin{gathered} T=\frac{\Delta H^0}{\Delta S^0} \\ T=\frac{299.9KJ/mole}{362.328·{10}^{-3}KJ/mole} \\ T=827.7K \end{gathered}$$

Therefore, the value for temperature is obtained as $\text{827.7 K}$.

Calculation of temperature for part (c)

(c)

The reaction given is –

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

The$∆H$of the given reaction is –

$$\begin{gathered} \Delta H^0=\sum n\Delta H_f^0\mathrm{Pr}oducts-\sum n\Delta H_f^{0}Reac\tan ts \\ =\left[\Delta H_f^{0}Mn\left(s\right)+\Delta H_f^{0}C{O}_2\left(g\right)\right]-\left[\Delta H_f^{0}Mn{O}_2\left(s\right)+\Delta H_f^{0}C\right] \\ =[0+(-393.5KJ/mol\left)\right]-[-520.9KJ/mol+0] \\ =127.4KJ/mol \end{gathered}$$

Similarly, the$\Delta S$of the given reaction is –

$$\begin{gathered} \Delta S^0=\sum n\Delta S^0\mathrm{Pr}oducts-\sum n\Delta S^{0}Reac\tan ts \\ =\left[\Delta S^{0}Mn\left(s\right)+\Delta S^{0}C{O}_2\left(g\right)\right]-\left[\Delta S^{0}Mn{O}_2\left(s\right)+\Delta S^{0}C\right] \\ =[31.8J/mol·K+213.7J/mol·K]-[53.1J/mol·K+5.686J/mol·K] \\ =186.714J/mol·K \end{gathered}$$

The $\Delta G^0$for the reaction is –

$\Delta G^0=\Delta H^0-T\Delta S^0$

For spontaneous process $\Delta G^0<0$, hence $\Delta H^0<T\Delta S^0$.

Assume that localid="1664191262067" $\Delta H^0=T\Delta S^0$-

localid="1664191270747" $$\begin{gathered} T=\frac{\Delta H^0}{\Delta S^0} \\ T=\frac{127.4KJ/mole}{186.714·{10}^{-3}KJ/mole} \\ T=682.3K\backslash \mathrm{hfill} \end{gathered}$$

Therefore, the value for temperature is obtained as $\text{682.3 K}$.

Calculation of temperature for part (d)

(d)

The reaction given is –

${\text{MnO}}_{\text{2}}\text{(s) + Si(s)}\to {\text{Mn(s) + SiO}}_{\text{2}}\text{(s)}$

The $\Delta H$of the given reaction is –

$$\begin{gathered} \Delta H^0=\sum n\Delta H_f^0\mathrm{Pr}oducts-\sum n\Delta H_f^{0}Reac\tan ts \\ =\left[\Delta H_f^{0}Mn\left(s\right)+\Delta H_f^{0}Si{O}_2\left(s\right)\right]-\left[\Delta H_f^{0}Mn{O}_2\left(s\right)+\Delta H_f^{0}Si\left(s\right)\right] \\ =[0+(-859.4KJ/mol\left)\right]-[-520.9KJ/mol+0] \\ =-339.7KJ/mol \end{gathered}$$

Similarly, the$\Delta S$ of the given reaction is –

$$\begin{gathered} \Delta S^0=\sum n\Delta S^0\mathrm{Pr}oducts-\sum n\Delta S^{0}Reac\tan ts \\ =\left[\Delta S^{0}Mn\left(s\right)+\Delta S^{0}Si{O}_2\left(s\right)\right]-\left[\Delta S^{0}Mn{O}_2\left(s\right)+\Delta S^{0}Si\left(s\right)\right] \\ =[31.8J/mol·K+41.8J/mol·K]-[53.1J/mol·K+18.8J/mol·K] \\ =1.7J/mol·K \end{gathered}$$

The$\Delta G^0$ for the reaction is –

$\Delta G^0=\Delta H^0-T\Delta S^0$

For spontaneous process $\Delta G^0<0$, hence $\Delta H^0<T\Delta S^0$.

Assume that $\Delta H^0=T\Delta S^0$-

$$\begin{gathered} T=\frac{\Delta H^0}{\Delta S^0} \\ T=\frac{-339.7KJ/mole}{1.7\times {10}^{-3}KJ/mole} \\ T=-199823.5K \end{gathered}$$

Therefore, the value for temperature is obtained as$\text{- 199823.5 K}$.