`H_(2)(g) + 1/2 O_(2)(g) rightarrow H_(2)O(l)`
`H_(2)O(f) rightarrow H_(2)O(g), Delta H = x_(4)`
Given
`E_(H-H) = x_(1)`
`E_(O=O) = x_(2)`
`E(O-H) = x_(3)`
`Delta H_(F) of H_(2)O` vapour is

Options

A. `x_(1) + (x_2)/(2) - x_(3) + x_(4)`
B. `2x_(3) - x_(1) - (x_2)/(2) - x_(4)`
C. `x_(1) + (x_2)/(2) - 2x_(3) - x_(4)`
D. `x_(1) + (x_2)/(2) - 2x_(3) + x_(4)`

Answer

C

Explanation

`H_(2)(g) + 1/2 O_(2)(g) rarr H_(2)O(g)`
`Delta H_(1) = x_(1) + (x_2)/(2) - 2x_(3)`
`DeltaH_(1) = x_(1) + (x_2)/(2) - 2x_(3)`
`H_(2)O(l) rarr H_(2)O(g) Delta H_(2) = x_(4)`
Now `Delta H_(1) - Delta H_(2)`
Gives `H_(2(g)) + 1/2 O_(2_(g)) rarr H_(2)O_(l) implies Delta H_("net") i= x_(1) + (x_2)/(2) - 2x_(3) - x_(4)`.