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Q. $H_{2}\left(g\right)+\frac{1}{2}O_{2}\left(g\right) \rightarrow H_{2}O\left(l\right)$ ; Bond energy of $H-H=x_{1};\left(\Delta \right)_{a}H_{\left(O_{2} , g\right)}=x_{2}$ and Bond energy of $O-H=x_{3}$ . The latent heat of vaporisation of liquid water into water vapour $=x_{4}$ , then heat of formation of liquid water is

NTA AbhyasNTA Abhyas 2020

Solution:

$H_{2}\left(g\right)+\frac{1}{2}O_{2}\left(g\right) \rightarrow H_{2}O\left(l\right)$
$\left(\Delta \right)_{r}H=\left[\left(BE\right)_{\left(H - H\right)} + \frac{1}{2} \left(BE\right)_{\left(O - O\right)}\right]-\left[2 \left(BE\right)_{\left(O - H\right)}\right]$
$=x_{1}+\frac{x_{2}}{2}-2x_{3}$
$\because H_{2}O\left(l\right) \rightarrow H_{2}O\left(g\right)ΔH=x_{4}$
$\Rightarrow H_{2}O\left(g\right) \rightarrow H_{2}O\left(l\right)ΔH=-x_{4}$
so final, $\Delta _{r}H=x_{1}+\frac{x_{2}}{2}-2x_{3}-x_{4}$