Ncert Exercises & Solutions

The value of $∆_{f} H^{⊝}$ for ${NH}_{3}$ is $- 91.8 kJ {mol}^{- 1}$ . Calculate enthalpy change for the following reaction.

$2 {NH}_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$

Given,

\frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) \to {NH}_{3} (g); ∆_{f} H^{⊝} = - 91.8 kJ {mol}^{- 1}

(

∆_{f} H^{⊝}

means enthalpy of formation of 1 mole of NH₃)

∴

Enthalpy change for the formation of 2 moles of NH₃

N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g); ∆_{f} H^{⊝} = 2 \times - 91.8 = - 183.6 kJ {mol}^{- 1}

And for the reverse reaction,

2 {NH}_{3} (g) \to N_{2} (g) + 3 H_{2} (g); ∆_{f} H^{⊝} = + 183.6 kJ {mol}^{- 1}

Hence, the value of

∆_{f} H^{⊝}

for NH₃ is

+ 183.6 kJ {mol}^{- 1}