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Electrochemistry

Question

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For vaporisation of water at 1 atm pressure, the values of ΔH and ΔS are 40.63 kJ mol^- and 108.8 JK^-1 mol^-1, respectively. The temperature when Gibbs energy change (ΔG) for this transformation will be zero, is

273.14 K

393.4 K

373.4 K

293.4 K

Solution

C.

373.4 K

ΔG = ΔH - TΔS
ΔG = 0 (given)
ΔH = TΔS,
T = 40.63 x 10³ / 108.8 = 373.4 K

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