Question

Enthalpy of formation of \(CO(g)\), \(CO_2(g)\), \(N_2O(g)\) and \(N_2O_4(g)\) are \(-110\), \(-393\), \(81\), \(9.7\ \text{kJ mol}^{-1}\) respectively. Calculate \(\Delta_rH\) for the following reaction: \[ N_2O_4(g)+3CO(g)\rightarrow N_2O(g)+3CO_2(g) \]

• A. \(-569\ \text{kJ mol}^{-1}\)
• B. \(+569\ \text{kJ mol}^{-1}\)
• C. \(+778\ \text{kJ mol}^{-1}\)
• D. \(-778\ \text{kJ mol}^{-1}\)

Hint:

Hess's law: \[ \Delta H_{\text{reaction}} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}) \] Always multiply each enthalpy of formation by its stoichiometric coefficient before substitution.

Answer 1

The Correct Option is D

Step 1: Write the target reaction.
We need to find the standard enthalpy change for: \[ \text{N}_2\text{O}_4(g) + 3\text{CO}(g) \rightarrow \text{N}_2\text{O}(g) + 3\text{CO}_2(g) \]

Step 2: List the given enthalpies of formation.
\(\Delta_f H^\circ[\text{CO}(g)] = -110\) kJ/mol
\(\Delta_f H^\circ[\text{CO}_2(g)] = -393\) kJ/mol
\(\Delta_f H^\circ[\text{N}_2\text{O}(g)] = +81\) kJ/mol
\(\Delta_f H^\circ[\text{N}_2\text{O}_4(g)] = +9.7\) kJ/mol

Step 3: Apply Hess's Law using the standard formula.
\[ \Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants}) \]
This principle states that the enthalpy of a reaction equals the sum of enthalpies of formation of products minus that of reactants.

Step 4: Calculate the sum for products.
Products: N2O(g) and 3 CO2(g)
\[ \Delta_f H^\circ(\text{products}) = 1 \times (+81) + 3 \times (-393) = 81 - 1179 = -1098 \text{ kJ} \]

Step 5: Calculate the sum for reactants.
Reactants: N2O4(g) and 3 CO(g)
\[ \Delta_f H^\circ(\text{reactants}) = 1 \times (+9.7) + 3 \times (-110) = 9.7 - 330 = -320.3 \text{ kJ} \]

Step 6: Find the reaction enthalpy.
\[ \Delta_r H^\circ = -1098 - (-320.3) = -1098 + 320.3 = -777.7 \approx -778 \text{ kJ/mol} \]
The reaction is highly exothermic because three CO molecules are oxidised to CO2, releasing large amounts of energy.
\[ \boxed{-778 \text{ kJ/mol}} \]