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H (2 (g)) + (1)/(2) O (2 (g)) to H (2) O...

`H _(2 (g)) + (1)/(2) O _(2 (g)) to H _(2) O _((l)) , BE (H-H) = x_(1), BE (O=O)= x _(2), BE (O-H) = x _(3)`
Latent heat of vaporisation of water liquid into water vapour `=x_(4),` then heat of formation of liquid water is

A

`x _(1) = (x _(2))/(2 ) - x _(3) + x _(4)`

B

`2 x _(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)`

Text Solution

Verified by Experts

The correct Answer is:
C
`Delta H = (BE) _("reactant") - (BE) _("products") ` [But all the species must be in gaseous state. In product, ]
`[ H_(2) O (l) to H _(2) O (g) ] Delta H ` must be added, Hence, `H _(2) (g)+ (1)/(2) O_(2) (g) to H _(2) O (1)`
`Delta H = [ (BE) _(H - H) + (1)/(2) (BE) _(O=O) ] = [(Delta H ) _(vap) + 2 ( BE) _( O - H) ] = x _(2) + (x _(2))/(2) = [ x _(4) + 2x _(3)] = x _(1) + (x _(2))/(2) x _(4) - 2x _(3)`
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