The free energy and entropy change in kJ per mole when liquid water boils at 1 atmosphere are respectively (latent heat of water = 2.0723 kJ g-1)

Question

The free energy and entropy change in kJ per mole when liquid water boils at 1 atmosphere are respectively (latent heat of water = 2.0723 kJ g-1) (A) 0, 0 (B) 0.1, 0.1 (C) 0.1, 0 (D) 0, 0.1

Tardigrade Admin · Accepted Answer

Δ Hvap of water = 2.0723 × 18 = 37.30 kJ mol-1 Δ Svap=(Δ Hvap/Tb)=(37.30 kJ mol-1/373 K)=0.1 kJ mol-1 K-1 Δ Gvap=Δ H(kJ mol-1)-T(K)Δ S(kJ K-1 mol-1) =37.30-373×0.1=0

Q. The free energy and entropy change in $k J$ per mole when liquid water boils at 1 atmosphere are respectively (latent heat of water $= 2.0723 k J g^{- 1}$ )

$0, 0$

$0.1, 0.1$

$0.1, 0$

$0, 0.1$

$Δ H_{v a p}$ of water $= 2.0723 \times 18 = 37.30 k J m o l^{- 1}$
$Δ S_{v a p} = \frac{Δ H _{v a p}}{T _{b}} = \frac{37.30 k J m o l ^{- 1}}{373 K} = 0.1 k J m o l^{- 1} K^{- 1}$
$Δ G_{v a p} = Δ H (k J m o l^{- 1}) - T (K) Δ S (k J K^{- 1} m o l^{- 1})$
$= 37.30 - 373 \times 0.1 = 0$

Q. The free energy and entropy change in kJ per mole when liquid water boils at 1 atmosphere are respectively (latent heat of water =2.0723kJg−1)

0,0

0.1,0.1

0.1,0

0,0.1