Question

Combustion of 1 mole of benzene is expressed as C$_6$H$_6$(l) + $\frac{15}{2}$O$_2$(g) → 6CO$_2$(g) + 3H$_2$O(l). The standard enthalpy of combustion of 2 mol of benzene is x kJ. x = __________
(1) standard Enthalpy of formation of 1 mol of C$_6$H$_6$(l), for the reaction 6C(graphite) + 3H$_2$(g) → C$_6$H$_6$(l) is 48.5 kJ mol$^{-1}$.
(2) Standard Enthalpy of formation of 1 mol of \(CO_2\)(g),  for the reaction 6(graphite) + \(O_2(g)\) → CO$_2$(g) is -393.5 kJ mol$^{-1}$.
(3) Standard Enthalpy of formation of 1 mol of H$_2$O(l), for the reaction H$_2$(g) + $\frac{1}{2}$O$_2$(g) → H$_2$O(l) is -286 kJ mol$^{-1}$.

Answer 1

Correct Answer: 6535

The enthalpy change for the combustion reaction can be calculated using Hess's law:
\[\Delta H = \sum \Delta H_f (\text{products}) - \sum \Delta H_f (\text{reactants}).\]
Step 1: Write the given reaction
The reaction for 1 mole of benzene is:
\[\text{C}_6\text{H}_6(l) + \frac{15}{2} \text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 3\text{H}_2\text{O}(l).\]
Step 2: Calculate the enthalpy change for 1 mole of benzene
Using the standard enthalpies of formation:
\[\Delta H_f (\text{CO}_2(g)) = -393.5 \, \text{kJ/mol},\]
\[\Delta H_f (\text{H}_2\text{O}(l)) = -286 \, \text{kJ/mol},\]
\[\Delta H_f (\text{C}_6\text{H}_6(l)) = 48.5 \, \text{kJ/mol}.\]
For the products:
\[\Delta H_f (\text{products}) = [6 \times (-393.5)] + [3 \times (-286)].\]
\[\Delta H_f (\text{products}) = -2361 - 858 = -3219 \, \text{kJ/mol}.\]
For the reactants:
\[\Delta H_f (\text{reactants}) = [1 \times 48.5] + \left(\frac{15}{2} \times 0 \right).\]
\[\Delta H_f (\text{reactants}) = 48.5 \, \text{kJ/mol}.\]
The enthalpy change for the combustion of 1 mole of benzene is:
\[\Delta H = \Delta H_f (\text{products}) - \Delta H_f (\text{reactants}),\]
\[\Delta H = -3219 - 48.5 = -3267.5 \, \text{kJ/mol}.\]
Step 3: Calculate for 2 moles of benzene  
For 2 moles of benzene:
\[\Delta H = 2 \times (-3267.5) = -6535 \, \text{kJ}.\]
Final Answer: \(x = 6535 \, \text{kJ}\).