91Ó°ÊÓ

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.

(a)

The reaction given is –

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

The$∆H$ of the given reaction is –

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

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

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

The$\mathrm{Δ}{G}^0$ for the reaction is –

$\mathrm{Δ}{G}^0=\mathrm{Δ}{H}^0-T\mathrm{Δ}{S}^0$

For spontaneous process$\mathrm{Δ}{G}^0<0$ , hence $\mathrm{Δ}{H}^0<T\mathrm{Δ}{S}^0$.

Assume that $\mathrm{Δ}{H}^0=T\mathrm{Δ}{S}^0$-

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

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

(b)

The reaction given is –

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

The $∆H$of the given reaction is –

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

Similarly, the$\mathrm{Δ}S$of the given reaction is –

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

The$\mathrm{Δ}{G}^0$for the reaction is –

$\mathrm{Δ}{G}^0=\mathrm{Δ}{H}^0-T\mathrm{Δ}{S}^0$

For spontaneous process $\mathrm{Δ}{G}^0<0$, hence $\mathrm{Δ}{H}^0<T\mathrm{Δ}{S}^0$.

Assume that $\mathrm{Δ}{H}^0=T\mathrm{Δ}{S}^0$-

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

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

(c)

The reaction given is –

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

The$∆H$of the given reaction is –

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

Similarly, the$\mathrm{Δ}S$of the given reaction is –

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

The $\mathrm{Δ}{G}^0$for the reaction is –

$\mathrm{Δ}{G}^0=\mathrm{Δ}{H}^0-T\mathrm{Δ}{S}^0$

For spontaneous process $\mathrm{Δ}{G}^0<0$, hence $\mathrm{Δ}{H}^0<T\mathrm{Δ}{S}^0$.

Assume that localid="1664191262067" $\mathrm{Δ}{H}^0=T\mathrm{Δ}{S}^0$-

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

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

(d)

The reaction given is –

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

The $\mathrm{Δ}H$of the given reaction is –

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

Similarly, the$\mathrm{Δ}S$ of the given reaction is –

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

The$\mathrm{Δ}{G}^0$ for the reaction is –

$\mathrm{Δ}{G}^0=\mathrm{Δ}{H}^0-T\mathrm{Δ}{S}^0$

For spontaneous process $\mathrm{Δ}{G}^0<0$, hence $\mathrm{Δ}{H}^0<T\mathrm{Δ}{S}^0$.

Assume that $\mathrm{Δ}{H}^0=T\mathrm{Δ}{S}^0$-

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

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