Question

$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)⟶H_{2}O\left(l\right)$

$BE(H--H)=x_1;BE(O==O)=x_2$

$BE(O--H)=x_3$

Latent heat of vaporisation of water liquid into water vapour = $x_4$, then ${\Delta }_{f}H$(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. $x_1+\frac{x_2}{2}-x_4-2x_3$
• D. $x_1+\frac{x_2}{2}-2x_3+x_4$

Answer 1

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

$\Delta H=(BE{)}_{reactant}-(BE{)}_{products}$

But all the species must be in gaseous state. In product, $\left[H_{2}O\right(l)⟶H_{2}O(g\left)\right]$ $\Delta H$ must be added.

Hence, $H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)⟶H_{2}O\left(l\right)$

$\Delta H=\left[\right(BE{)}_{H--H}+\frac{1}{2}\left(BE{)}_{O==O}\right]$

$=\left[\right(\Delta H{)}_{vap}+2\left(BE{)}_{O--H}\right]$

$=x_1+\frac{x_2}{2}-[x_4+2x_3]$

$=x_1+\frac{x_2}{2}-x_4-2x_3$