Question

$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)⟶H_{2}O\left(l\right)$
$H_2{O}_{\left(l\right)}\to H_2{O}_{\left(g\right)};△H=x_4$
Given,
$E_{H-H}=x_1$
$E_{O=O}=x_2$
$E_{O-H}=x_3$
$△H_f$ of $H_{2}O$ Vapour 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

The correct option is D $x_1+\frac{x_2}{2}-2x_3+x_4$

$H_{2\left(g\right)}+\frac{1}{2}{O}_{2\left(g\right)}⟶{H_2}_{\left(l\right)}$

$△H$ for this reaction $=$ (Bond energy of reactant) $-$ (Bond energy of product)

$△H_1=(x_1+\frac{x_2}{2})-\left(2x_3\right)$

$H_2{O}_{\left(l\right)}⟶H_2{O}_{\left(g\right)},△H_2=x_4$

For formation of $H_{2}O$ vapor,

$△H_f=△H_1+△H_2=x_1+\frac{x_2}{2}-2x_3+x_4$