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

Question

For vaporization of water at 1 atmospheric pressure, the values of Δ H and Δ S are 40.63 kJ mol-1 and 108.8 JK-1 mol-1, respectively. The temperature when Gibbs energy change (Δ G) for this transformation will be zero, is (A) 393.4 K (B) 373.4 K (C) 293.4 K (D) 273.4 K

Tardigrade Admin · Accepted Answer

Δ G = Δ H - T Δ S Δ G =0 Given Δ H = T Δ S, T = (40.63 × 103/108.8 ) = 373.4 K

Q. For vaporization of water at 1 atmospheric pressure, the values of $Δ H$ and $Δ S$ are $40.63 k J m o l^{- 1}$ and $108.8 J K^{- 1} m o l^{- 1}$ , respectively. The temperature when Gibbs energy change $(Δ G)$ for this transformation will be zero, is

393.4 K

373.4 K

293.4 K

273.4 K

$Δ G = Δ H - T Δ S Δ G = 0$
Given $Δ H = T Δ S,$
$T = \frac{40.63 \times 1 0 ^{3}}{108.8} = 373.4 K$

Q. For vaporization of water at 1 atmospheric pressure, the values of ΔH and ΔS are 40.63kJmol−1 and 108.8JK−1mol−1, respectively. The temperature when Gibbs energy change (ΔG) for this transformation will be zero, is