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Comprehension
Entropy is a state function and its value depends on two or three variables temperature (T), pressure (P) and volume (V). Entropy change for an ideal gas having number of moles (n) can be determined by the following equation.
\[
\begin{array}{l}
\Delta \mathrm{S}=2.303 \mathrm{nC}_{\mathrm{V}} \log \left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)+2.303 \mathrm{nR} \log \left(\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}\right) \\
\Delta \mathrm{S}=2.303 \mathrm{nC}_{\mathrm{p}} \log \left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)+2.303 \mathrm{nR} \log \left(\frac{\mathrm{P}_{1}}{\mathrm{P}_{2}}\right)
\end{array}
\]
Since free energy change for a process or a chemical equation is a deciding factor for spontaneity, which can be obtained by using entropy change \( (\Delta \mathrm{S}) \) according to the expression, \( \Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \) at a temperature \( \mathrm{T} \).

For a reaction \( \mathrm{M}_{2} \mathrm{O}(\mathrm{s}) \longrightarrow 2 \mathrm{M}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) ; \Delta \mathrm{H}=30 \mathrm{~kJ} / \) \( \mathrm{mol} \) and \( \Delta \mathrm{S}=0.07 \mathrm{~kJ} / \mathrm{K} \)-mol at \( 1 \mathrm{~atm} \). Calculate upto which temperature the reaction would not be spontaneous?
(b) \( \mathrm{T}300.8 \mathrm{~K} \)
(d) \( \mathrm{T}428.6 \mathrm{~K} \)
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