Question

The standard entropies of $N_2\left(g\right),H_2\left(g\right)and{NH}_3\left(g\right)$ are 191.5, 130.5, 192.6 ${JK}^{-1}{mol}^{-1}$. The value of $△S$ during the formation of one mole of ammonia is:

• A. $-98.9{KJ}^{-1}{mol}^{-1}$
• B. $-9.78{JK}^{-1}{mol}^{-1}$
• C. ${+129.4JK}^{-1}{mol}^{-1}$
• D. zero

Answer 1

The correct option is A $-98.9{KJ}^{-1}{mol}^{-1}$

Solution:- (A) $-98.9kJ/mol$

Formation of ammonia-

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

$\Delta S$ for the given reaction-

$\Delta S={\sum \Delta S}_{product}-{\sum \Delta S}_{reactant}$

$\Delta S=(2\times 192.6)-(191.5+(3\times 130.5))$

$\Delta S=385.2-583=-197.8kJ/mol$

$\Delta S$ for the formation of $2$ mole of ammonia $=-197.8kJ/mol$

$\Delta S$ for the formation of $1$ mole of ammonia $=\frac{-197.8}{2}=-98.9kJ/mol$

Hence the value of $\Delta S$ during the formation of one mole of ammonia is $-98.9kJ/mol$.