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Question

△H and △S for the equilibrium reaction, H2O(l)⇌H2O(g) at 1 atmospheric pressure are 40.63 kJmol−1 and 108.8 JK−1mol−1 respectively. Calculate the temperature at which the reaction occur. Also, predict the sign of free energy for this transformation above this temperature.

A
373.4K, negative
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B
373.4K, positive
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C
463.4K, negative
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D
463.4K, positive
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Solution

The correct option is A 373.4K, negative Given: △H=40.63 kJ mol−1 △S=108.8 JK−1mol−1=0.1088 kJK−1mol−1 △G=0 (as the system is in equilibrium) Applying Gibb's free energy equation: △G=△H−T △S 0=40.63−T×0.1088 T=40.630.1088=373.4K The sign of △G above this temperature will become negative as magnitude of (TΔS>ΔH) and the reaction will be spontaneous.

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