Molar heat of vaporisation of a liquid is 6 kJ mole -1. If the entropy change is 16 J mole -1 K -1, the boiling point of the liquid is

Question

Molar heat of vaporisation of a liquid is 6 kJ mole -1. If the entropy change is 16 J mole -1 K -1, the boiling point of the liquid is (A) 375°C (B) 375 K (C) 273 K (D) 102°C

Tardigrade Admin · Accepted Answer

Δ S=16 J mol -1 K -1, Δ Hv=6 kJ mol -1 T b . p .=(Δ H text vapour /Δ S text vapour )=(6 × 1000/16)=375 K

Q. Molar heat of vaporisation of a liquid is $6 k J m o l e^{- 1}$ . If the entropy change is $16 J m o l e^{- 1} K^{- 1}$ , the boiling point of the liquid is

375°C

375 K

273 K

102°C

$Δ S = 16 J m o l^{- 1} K^{- 1}, Δ H_{v} = 6 k J m o l^{- 1}$

$T_{b . p .} = \frac{Δ H _{vapour}}{Δ S _{vapour}} = \frac{6 \times 1000}{16} = 375 K$

Q. Molar heat of vaporisation of a liquid is 6kJmole−1. If the entropy change is 16Jmole−1K−1, the boiling point of the liquid is