The enthalpy of vaporisation of liquid water using data
`H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l) DeltaH = -285.77 kJ`
`H_(2)(g) + 1/2O_(2)(g) to H_(2)O(g) DeltaH = -241.84 kJ` is

Options

A. `+ 43.93 kJ "mol"^(-1)`
B. `- 43.93 kJ "mol"^(-1)`
C. `+ 527.61 kJ "mol"^(-1)`
D. `-527.61 kJ "mol"^(-1)`

Answer

A

Explanation

The required equation is : `H_(2)O(l) to H_(2)O(g)`
(i) `H_(2)(g) + 1/2 O_(2)(g) to H_(2)O(l)`
`DeltaH = - 285.77 kJ`
(iii) `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(g)`
`DeltaH = -241.84 kJ`
Subtracting eqn (i) from eqn (ii)
The enthalpy of vaporisation = -241.84 - (-285.77) = `+ 43.93 kJ mol^(-1)`