The enthalpy of evaporation of water at 373 K is 40.67 kJ `mol^(-1)`. What will be the enthalpy of evaporation at 353 K and 393 K. If average molar heats at constant pressure in this range for water in liquid and vapour states are 75.312 and `33.89 JK^(-1) mol^(-1)` respectively?

Explanation

`K^(+)(g) + Cl^(-)(g) to KCl`(s)
Equation (6) can be obtained by the following manipulations.
Eq. (1) - Eq. (5) - Eq(2) `-1/2 Eq.(4) - Eq.(3)`
Accordingly, we have
`DetlaH_(6) = DeltaH_(1) - DeltaH_(5) -DeltaH_(2) -1/2DeltaH_(4) - DeltaH_(3)`
`DeltaH_(6) =(-435.9 - 89.8 - 418.9 + 1/2 xx 242.0 + 348.6) kJ mol^(-1) = -475.0 kJ mol^(-1)`