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

Options

A. `+43.93kJ//mol`
B. `-43.93kJ//mol`
C. `527.61kJ//mol`
D. `-527.61kJ//mol`

Answer

A

Explanation

(I) `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),DeltaH=-285.77kJ//mol`
(II) `H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(g),DeltaH=-241.84kJ//mol`
`H_(2)O(l)toH_(2)O(g),dDeltaH=?`
Subtracting the Eqs. (I) from (II), we get
`H_(2)O(l)toH_(2)O(g)`
`DeltaH=-241.84-(-285.77)`
`DeltaH=+43.93kJ//mol`