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) 0.5 k JK-1 mol-1 (B) 1.0 k JK-1 mol-1 (C) 1.5 k JK-1 mol-1 (D) 2.0 k JK-1 mol-1

Tardigrade Admin · Accepted Answer

Given enthalpy of vaporization, Δ H =186 ⋅ 5 kJ mol -1 Boiling point of water =100° C =100+273=373 K Entropy change, 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 :

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

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

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

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

Given enthalpy of vaporization,
$Δ H = 186 \cdot 5 k J m o l^{- 1}$
Boiling point of water
$= 10 0^{\circ} C = 100 + 273 = 373 K$
Entropy change,
$Δ S = \frac{Δ H}{T} = \frac{186 \cdot 5 k J m o l ^{- 1}}{373 k}$
$= 0 \cdot 5 k J m o l^{- 1} K^{- 1}$

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

0.5kJK−1mol−1

1.0kJK−1mol−1

1.5kJK−1mol−1

2.0kJK−1mol−1