Question

$ H _2( g )+\frac{1}{2} O _2( g ) \rightarrow H _2 O (l) $
$ BE ( H - H )=x_1 ; $
$BE ( O = O )=x_2$
$ BE ( O - H )=x_3$
Latent heat of vaporization of water liquid into water vapour $=x_4$, then $\Delta_f H$ (heat of formation of liquid water) is

• A. $x_1+\frac{x_2}{2}-x_3+x_4$
• B. $2 x_3-x_1-\frac{x_2}{2}-x_4$
• C. $x_1+\frac{x_2}{2}-2 x_3-x_4$
• D. $x_1+\frac{x_2}{2}-2 x_3+x_4$

Answer 1

Answer: (C)

$\Delta H=( BE )_{\text {reactant }}-( BE )_{\text {products }}$
[But all the species must be in gaseous state.
In product, $\left[ H _2 O (l) \rightarrow H _2 O ( g )\right] \Delta H$ must be added
Hence, $H _2( g )+\frac{1}{2} O _2( g ) \rightarrow H _2 O (l)$
$\Delta H=\left[( BE )_{ H - H }+\frac{1}{2}( BE )_{ O = 0 }\right]$
$=\left[(\Delta H)_{ vap }+2( BE )_{ O - H }\right]$
$=x_1+\frac{x_2}{2}-\left[x_4+2 x_3\right]$
$=x_1+\frac{x_2}{2}-x_4-2 x_3$