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: Use Hess's law.
The enthalpy of reaction is \[ \Delta_rH = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}) \]

Step 2: Calculate total enthalpy of products.
Products: \[ N_2O(g)+3CO_2(g) \] Therefore, \[ \sum \Delta H_f^\circ(\text{products}) = 81+3(-393) \] \[ =81-1179 \] \[ =-1098\ \text{kJ mol}^{-1} \]

Step 3: Calculate total enthalpy of reactants.
Reactants: \[ N_2O_4(g)+3CO(g) \] Therefore, \[ \sum \Delta H_f^\circ(\text{reactants}) = 9.7+3(-110) \] \[ =9.7-330 \] \[ =-320.3\ \text{kJ mol}^{-1} \]

Step 4: Calculate \(\Delta_rH\).
\[ \Delta_rH = (-1098)-(-320.3) \] \[ =-1098+320.3 \] \[ =-777.7 \] \[ \approx -778\ \text{kJ mol}^{-1} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\Delta_rH=-778\ \text{kJ mol}^{-1}} \] Hence, the correct option is \[ \boxed{(4)} \]