Question

Calculate ΔH (in kJ/mol) for the reaction described by the equation.

6 NH₃(g) + 5 O₃(g) → 6 NO(g) + 9 H₂O(l)

Answer 1

Ans. Following Hess’s law, dH formation of the reaction is given by-

dH_rxn = (Sum of dH⁰_f of products) – (sum of dH⁰_f of reactants)

Or, d_Hrxn = (6 x dH⁰_f of NO + 9 x dH⁰_f of H₂O, l) - (6 x dH⁰_f of NH₃ + 5 x dH⁰_f of O₃)

Or, d_Hrxn = [6 x 90.4 kJ mol^-1 + 9 x (- 285.8 kJ mol^-1)] –

[6 x (- 46.2 kJ mol^-1) + 5 x 143.0 kJ mol^-1]

Or, d_Hrxn = (542.4 kJ mol^-1 – 2572.2 kJ mol^-1) – (-261.6 kJ mol^-1 + 715.0 kJ mol^-1)

Or, d_Hrxn = -2029.8 kJ mol^-1 + 453.4 kJ mol^-1

Hence, d_Hrxn = - 1576.4 kJ mol^-1