For vaporization of water at 1 atmospheric pressure, the values of ΔH and ΔS are 40.63 kJ mol⁻¹ and 108.8 J K⁻¹ mol⁻¹, respectively. The temperature when Gibb's energy change (ΔG) for this transformation will be zero, is: 273.4K, 393.4K, 373.4K, 293.4K

According to Gibb's equation,
ΔG = ΔH - TΔS
when ΔG = 0,
ΔH = TΔS
Given,
ΔH = 40.63 kJ mol⁻¹ = 40.63 x 10³ J mol⁻¹
ΔS = 108.8 J K⁻¹ mol⁻¹
∴T = ΔH/ΔS = 40.63 × 10³/108.8 = 373.43 K
The temperature when Gibb's energy change (ΔG) for this transformation will be zero, is 373.4 K
The zero value of the Gibb's free energy change indicates equilibrium state.