For vaporization of water at 1 atmospheric pressure, the values of ∆H a n d ∆S are 40.63 k J m o l - 1 a n d 108.8 J K - 1 m o l - 1 respectively. The temperature when Gibbs energy change ∆ G for this transformation will be zero, is:

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Correct answer: D. 373.4 K

H 2 O ( l ) ⇌ 1 atm H 2 O ( g ) ∆ H = 40630 J m o l - 1 ∆ S = 108.8 J K - 1 m o l - 1 ∆ G = ∆ H - T ∆ S When ∆ G = 0 , ∆ H - T ∆ S = 0 T = Δ H Δ S = 40630 J mol K − 1 108.8 J mol K − 1 = 373.4 K .

For vaporization of water at 1 atmospheric pressure, the values of ∆H a n d ∆S are 40.63 k J m o l - 1 a n d 108.8 J K - 1 m o l - 1 respectively. The temperature when Gibbs energy change ∆ G for this transformation will be zero, is:

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