Question

Using the enthalpies of formation, calculate the energy (kJ) released when $3.00g$ of $N{H}_{3\left(g\right)}$ reacts according to the following equation?
(Atomic weights: $N=14.00,H=1.008$).
$4N{H}_3\left(g\right)+5O_2\left(g\right)⟶4NO\left(g\right)+6H_{2}O\left(g\right)$
$\Delta HN{H}_3\left(g\right)=-46.1kJ/mole$
$\Delta HNO\left(g\right)=+90.2kJ/mole$
$\Delta H{H}_{2}O\left(g\right)=-241.8kJ/mole$

• A. $34.3$
• B. $30.8$
• C. $39.9$
• D. $42.6$

Answer 1

Answer: (C)

$39.9$

Energy released, when four mole of $N{H}_3$ reacts, is given by ${\Delta }_{r}H=H_{Products}-H_{Reactants}$ ${\Delta }_{r}H=(4\times \Delta HNO\left(g\right)+6\times \Delta H{H}_{2}O\left(g\right))-(4\times \Delta HN{H}_3\left(g\right)+5\times \Delta H{O}_2\left(g\right))$

${\Delta }_{r}H=4\times 90.2+6\times (-241.8)-4\times (-46.1)+5\times 0$ (As $\Delta H{O}_2\left(g\right)=0$) ${\Delta }_{r}H=-905.6kJ$ Thus, for four mole of $N{H}_3$, i.e., $(4\times 14=)68g$ of $N{H}_3$ amount of energy released is 905.6 kJ. Therefore, for 3g of $N{H}_3$ amount of energy released will be $\frac{(905.16\times 3)}{68}=39.9kJ$