JEE MAIN - Chemistry (2010 - No. 8)
For a particular reversible reaction at temperature T, ∆H and ∆S were found to be both +ve. If Te is
the temperature at equilibrium, the reaction would be spontaneous when :
Te > T
T > Te
Te is 5 times T
T = Te
Explanation
At equilibrium $$\Delta G = 0$$
Hence, $$\Delta G = \Delta H - {T_e}\Delta S = 0$$
$$\therefore$$ $$\,\,\,\,\,\,\Delta H = {T_e}\Delta S\,\,\,$$
or $$\,\,\,\,{T_e} = {{\Delta H} \over {\Delta S}}\,\,$$
For a spontaneous reaction
$$\Delta G$$ must be negative
which is possible only if $$\Delta H < T\Delta S$$
or $$\,\,\,T > {{\Delta H} \over {\Delta S}};{T_e} < T$$
Hence, $$\Delta G = \Delta H - {T_e}\Delta S = 0$$
$$\therefore$$ $$\,\,\,\,\,\,\Delta H = {T_e}\Delta S\,\,\,$$
or $$\,\,\,\,{T_e} = {{\Delta H} \over {\Delta S}}\,\,$$
For a spontaneous reaction
$$\Delta G$$ must be negative
which is possible only if $$\Delta H < T\Delta S$$
or $$\,\,\,T > {{\Delta H} \over {\Delta S}};{T_e} < T$$
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