Question

The standard entropies of $N_2$ (g),$H_2$(g) and $N{H}_3$ (g) are 191.5, 130.5, 192.6 $J{K}^{-1}$ $mo{l}^{-1}$.The value of ${\Delta S}^{\circ }$ of formation of ammonia is?

Answer 1

Formation of ammonia-

$\frac{1}{2}{N_2}_{\left(g\right)}+\frac{3}{2}{H_2}_{\left(g\right)}⟶{N{H}_3}_{\left(g\right)}$

${\Delta S}^{\circ }={{\Delta S}^{\circ }}_{\left(\text{product}\right)}-{{\Delta S}^{\circ }}_{\left(\text{reactant}\right)}$

${\Delta S}^{\circ }=\left(192.6\right)-(\frac{1}{2}\times 191.5+\frac{3}{2}\times 130.5)$

$\Rightarrow {\Delta S}^{\circ }=192.6-(95.75+195.75)$

$\Rightarrow {\Delta S}^{\circ }=192.6-291.5=-98.9J/mol-K$

Hence the value of ${\Delta S}^{\circ }$ of formation of ammonia is $-98.9J/mol-K$.