Question

$H_2(g)+\frac{1}{2}{O}_2(g)\to H_{2}O(l)$
$BE(H-H)=x_1;$
$BE(O=O)=x_2$
$BE(O-H)=x_3$
Latent heat of 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. $2x_3-x_1-\frac{x_2}{2}-x_4$
• C. $x_1+\frac{x_2}{2}-2x_3-x_4$
• D. $x_1+\frac{x_2}{2}-2x_3+x_4$

Answer 1

Answer: (C)

$\Delta H=(BE{)}_{\text{reactant }}-(BE{)}_{\text{products }}$
[But all the species must be in gaseous state.
In product, $\left[H_{2}O(l)\to H_{2}O(g)\right]\Delta H$ must be added
Hence, $H_2(g)+\frac{1}{2}{O}_2(g)\to H_{2}O(l)$
$\Delta H=\left[(BE{)}_{H-H}+\frac{1}{2}(BE{)}_{O=0}\right]$
$=\left[(\Delta H{)}_{vap}+2(BE{)}_{O-H}\right]$
$=x_1+\frac{x_2}{2}-\left[x_4+2x_3\right]$
$=x_1+\frac{x_2}{2}-x_4-2x_3$