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 Gibbs energy change (ΔG) for this transformation will be zero, is:

Question

Accepted Answer

373.4 K. For ΔG = 0, the equation ΔG = ΔH - TΔS simplifies to T = ΔH/ΔS. Substituting ΔH = 40.63 kJ mol⁻¹ (40630 J mol⁻¹) and ΔS = 108.8 J K⁻¹ mol⁻¹, T = 40630 / 108.8 = 373.4 K.

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 Gibbs energy change (ΔG) for this transformation will be zero, is:

Solution

Related NEET Chemistry questions