If the enthalpy of vaporization of water is 186.5 kJ mol-1 , the entropy if its vaporization will be :

Question

If the enthalpy of vaporization of water is 186.5 kJ mol-1 , the entropy if its vaporization will be : (A) 0.5JK-1 mol-1 (B) 1.0 JK-1 mol-1 (C) 1.5JK-1 mol-1 (D) 2.0 JK-1 mol-1

Tardigrade Admin · Accepted Answer

Given enthalpy of vapourization Δ H = 186.5 kJ mol-1 Boiling point of water = 100° C = 100 + 273 = 373 K Δ S = (Δ H/T) = (186.5 kJ mol-1/373 K) = 0.5 kJ mol-1 k-1

Q. If the enthalpy of vaporization of water is $186.5 k J m o l^{- 1}$ , the entropy if its vaporization will be :

$0.5 J K^{- 1} m o l^{- 1}$

$1.0 J K^{- 1} m o l^{- 1}$

$1.5 J K^{- 1} m o l^{- 1}$

$2.0 J K^{- 1} m o l^{- 1}$

Given enthalpy of vapourization
$Δ H = 186.5 k J m o l^{- 1}$
Boiling point of water
$= 10 0^{\circ} C = 100 + 273 = 373 K$
$Δ S = \frac{Δ H}{T} = \frac{186.5 k J m o l ^{- 1}}{373 K}$
$= 0.5 k J m o l^{- 1} k^{- 1}$

Q. If the enthalpy of vaporization of water is 186.5kJmol−1, the entropy if its vaporization will be :

0.5JK−1mol−1

1.0JK−1mol−1

1.5JK−1mol−1

2.0JK−1mol−1