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) 293.4 K (B) 273.4 k (C) 393.4 K (D) 373.4 K

Tardigrade Admin · Accepted Answer

H 2 O (ℓ) stackrel text 1 atm leftharpoons H 2 O ( g ) Δ H =40630 J mol -1 Δ S =108 ⋅ 8 JK -1 mol -1 Δ G =Δ H - T Δ S when Δ G=0 Δ H-T Δ S=0 T =(Δ H /Δ S )=(40630 J mol -1/108 ⋅ 8 J mol -1) =373 ⋅ 4 K

Q. For vaporization of water at $1$ atmospheric pressure, the values of $Δ H$ and $Δ S$ are $40 \cdot 63 k J m o l^{- 1}$ and $108 \cdot 8 J K^{- 1} m o l^{- 1}$ , respectively. The temperature when Gibbs energy change $(Δ G)$ for this transformation will be zero, is:

293.4 K

273.4 k

393.4 K

373.4 K

$H_{2} O (ℓ) ⇌ 1 atm H_{2} O (g)$
$Δ H = 40630 J m o l^{- 1}$
$Δ S = 108 \cdot 8 J K^{- 1} m o l^{- 1}$
$Δ G = Δ H - T Δ S$ when $Δ G = 0$
$Δ H - T Δ S = 0$
$T = \frac{Δ H}{Δ S} = \frac{40630 J m o l ^{- 1}}{108 \cdot 8 J m o l ^{- 1}}$
$= 373 \cdot 4 K$

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: