For vaporization of water at 1 atmospheric pressure, the values of H and S are 40.63 kJmol^-1 and 108.8 JK^-1 mol^-1, respectively The temperature when Gibb’s energy change (G) for this transformatio…

Question

For vaporization of water at 1 atmospheric pressure, the values of H and S are 40.63 kJmol^-1 and 108.8 JK^-1 mol^-1, respectively The temperature when Gibb’s energy change (G) for this transformation will be zero, is

Accepted Answer

Correct answer: C. (c) According to Gibbs equation, G= H-T S when G=0 , H=T S Given, H=40.63 kJmol^-1=40.63 10^3 Jmol^-1 S=108.8JK^-1 mol^-1 T= H S= 40.63 10^3108.8=373.43K

$For vaporization of water at 1 atmospheric pressure, the values of Δ H and Δ S are 40.63 kJmo l^{- 1} and 108.8 J K^{- 1} mo l^{- 1}, respectively The temperature when Gibb’s energy change (G) for this transformation will be zero, is$

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