For vaporization of water at 1 atmospheric pressure, the values of H and S are 40.63 kJ mol-1 and 108.8 J K-1 mol-1, respectively. The temperature when Gibbs energy change ( G) for this transformation will be zero, is
- 273.4 K
- 393.4 K
- 373.4 K
- 293.4 K
Answer
(3) 373.4 K
Solution
G = H - T S
At G = 0: T = H S
⇒ T = 40.63 103 J mol-1108.8 J K-1 mol-1
∴ T = 373.4 K
Answer: (3)