The enthalpy of vaporisation of a certain liquid at its boiling point of 35° C is 24.64 kJ mol -1. The value of change in entropy for the process is

Question

The enthalpy of vaporisation of a certain liquid at its boiling point of 35° C is 24.64 kJ mol -1. The value of change in entropy for the process is (A) 704 J K-1 mol-1 (B) 80 JK-1 mol-1 (C) 24.64 JK-1 mol-1 (D) 70 JK-1 mol-1

Tardigrade Admin · Accepted Answer

Entropy of vaporisation (Δ S) =( text enthalpy of vaporisation (Δ H)/ text boiling point (in K )) Given, Δ H=24.64 kJ mol -1 and boiling point =35+273=308 K ∴ Δ S=(24.64 × 103 J mol -1/308 K ) =80 JK -1 mol -1

Q. The enthalpy of vaporisation of a certain liquid at its boiling point of $3 5^{\circ} C$ is $24.64 k J m o l^{- 1}$ . The value of change in entropy for the process is

$704 J K^{- 1} m o l^{- 1}$

$80 J K^{- 1} m o l^{- 1}$

$24.64 J K^{- 1} m o l^{- 1}$

$70 J K^{- 1} m o l^{- 1}$

Entropy of vaporisation $(Δ S)$
$= \frac{enthalpy of vaporisation ( Δ H )}{boiling point (in K )}$
Given, $Δ H = 24.64 k J m o l^{- 1}$ and
boiling point $= 35 + 273 = 308 K$
$∴ Δ S = \frac{24.64 \times 1 0 ^{3} J m o l ^{- 1}}{308 K}$
$= 80 J K^{- 1} m o l^{- 1}$

Q. The enthalpy of vaporisation of a certain liquid at its boiling point of 35∘C is 24.64kJmol−1. The value of change in entropy for the process is

704JK−1mol−1

80JK−1mol−1

24.64JK−1mol−1

70JK−1mol−1