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Data given for the following reaction is as follows : $\mathrm{FeO}_{(s)}+\mathrm{C}_{(\text {graphite })} \longrightarrow \mathrm{Fe}_{(s)}+\mathrm{CO}_{(g)}$

Substance $\mathrm{\Delta_{f} \mathrm{H}^{\degree} (kJ \mathrm{mol}^{-1} )}$ $\mathrm{ \Delta{S\degree} \left(\mathrm{Jmol}^{-1} \mathrm{~K}^{-1}\right) }$

$\mathrm{FeO}_{(s)}$ $-266.3$ $57.49$

${ C{(graphite)} }$ $0$ $5.74$

$\mathrm{Fe}_{(\mathrm{s}\rangle}$ $0$ $27.28$

$\mathrm{CO}_{(g)}$ $-110.5$ $197.6$

The minimum temperature in $\mathrm{K}$ at which the reaction becomes spontaneous is____________ (Integer answer)

Answers (1)

best_answer

For the reaction,

$\mathrm{FeO\left ( s \right )+C\left ( graphite \right )\rightarrow Fe\left ( s \right ) + CO\left ( g \right )}$

$\mathrm{\Delta H^{\circ}= -110.5+266.3= 155.8KJmol^{-1}}$

$\mathrm{\Delta S^{\circ}= 27.28+197.6-5.74-57.49=161.65Jmol^{-1}K^{-1}}$

For spontaeity,

$\mathrm{\Delta H-T\Delta S< 0}$

$\mathrm{\Rightarrow T> \frac{\Delta H}{\Delta S}}$

For minimum temperature,

$\mathrm{T= \frac{\Delta H}{\Delta S}= \frac{155.8\times10^{3}}{161.65} = 963.8\ K}$

Hence the answer is 964

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