Question

Enthalpies of formation of CO(g), CO₂(g), N₂O(g) and N₂O₄(g) are –110, – 393, 81 and 9.7 kJ mol^-1 respectively. Find the value of ∆_rH for the reaction.
\(N_2O_4(g) + 3CO(g) → N_2O(g) + 3CO_2(g)\)

Answer 1

Reaction:

N2O4(g) + 3 CO(g) -> N2O(g) + 3 CO2(g)}

Given standard enthalpies of formation (in kJ mol\(^{-1}\)):

• \(\Delta_f H^\circ\big(\ce{CO(g)}\big) = -110\)
• \(\Delta_f H^\circ\big(\ce{CO2(g)}\big) = -393\)
• \(\Delta_f H^\circ\big(\ce{N2O(g)}\big) = 81\)
• \(\Delta_f H^\circ\big(\ce{N2O4(g)}\big) = 9.7\)

Use:

\[ \Delta_r H^\circ = \sum \nu_p \Delta_f H^\circ(\text{products}) - \sum \nu_r \Delta_f H^\circ(\text{reactants}) \]

Sum for products:

\[ \Delta_f H^\circ(\ce{N2O}) + 3\,\Delta_f H^\circ(\ce{CO2}) = 81 + 3(-393) = 81 - 1179 = -1098\ \text{kJ} \]

Sum for reactants:

\[ \Delta_f H^\circ(\ce{N2O4}) + 3\,\Delta_f H^\circ(\ce

\(∆_rH\) for a reaction is defined as the difference between \(∆_fH\) value of products and \(∆_fH\) value of reactants.
\(△_rH\) =\(Σ△_fH\) (product) - \(Σ△_fH\) (reactants)
For the given reaction,
\(N_2O_4(g) + 3CO(g) → N_2O(g) + 3CO_2(g)\)
\(△_rH\) = \(△_fH (NO_2)\)+ \(3△_fH (CO_2)\) - \(△_fH (N_2O_4)\) + \(3△_fH (CO)\)
Substituting the values of \(∆_fH\) for \(N_2O, CO_2, N_2O_4,\) and \(CO\) from the question, we get:
\(△_rH = [{81\ kJ mol^{-1} + 3(-393)\ kJ mol^{-1}} - {9.7\ kJ mol^{-1} + 3(-110)\ kJ mol^{-1}}]\)
\(△_rH = -777.7\ kJ mol^{-1}\)

Hence, the value of \(∆_r H\) for the reaction is \(-777.7\ kJ mol^{-1}\).

{CO}) = 9.7 + 3(-110) = 9.7 - 330 = -320.3\ \text{kJ} \]

 

Reaction enthalpy:

\[ \Delta_r H^\circ = -1098 - (-320.3) = -1098 + 320.3 = -777.7\ \text{kJ} \approx -7.78 \times 10^{2}\ \text{kJ} \]

\[ \boxed{\Delta_r H^\circ \approx -7.8 \times 10^{2}\ \text{kJ}} \] (The reaction is highly exothermic.)