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$^{-1}$.
(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$^{-1}$.
Correct Answer: 6535
Approach Solution - 1
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: \(x = 6535 \, \text{kJ}\).
Approach Solution -2
\[ C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l) \]
Step 2: Write the standard enthalpy of combustion formula.
The standard enthalpy of combustion (ΔH°₍comb₎) = Σ ΔH°₍f,products₎ – Σ ΔH°₍f,reactants₎
Step 3: Substitute the given data.
ΔH°₍f,CO₂₎ = –393.5 kJ mol⁻¹
ΔH°₍f,H₂O(l)₎ = –286 kJ mol⁻¹
ΔH°₍f,C₆H₆(l)₎ = +48.5 kJ mol⁻¹
Step 4: Apply the enthalpy formula for 1 mol of benzene.
\[ ΔH°_{comb} = [6(–393.5) + 3(–286)] – [48.5] \]
\[ ΔH°_{comb} = [–2361 – 858] – 48.5 \]
\[ ΔH°_{comb} = –3219 – 48.5 = –3267.5 \, \text{kJ mol}^{–1} \]
Step 5: Calculate for 2 moles of benzene.
\[ x = 2 × 3267.5 = 6535 \, \text{kJ} \]
Step 6: Final Answer.
The standard enthalpy of combustion of 2 mol of benzene is:
\[ \boxed{x = 6535 \, \text{kJ}} \]
Final Answer: 6535 kJ
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