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Step-by-Step Solution
Step 1: Identify the Given Quantities
• Heat of combustion at constant volume, $ \Delta U $, is given as $ -3263.9 $ kJ mol–1.
• Temperature, $ T $, is 25 °C which is $ 25 + 273 = 298 $ K.
• Gas constant, $ R $, is $ 8.314 \, \text{J K}^{-1}\text{mol}^{-1} $ or equivalently $ 8.314 \times 10^{-3} \, \text{kJ K}^{-1}\text{mol}^{-1} $.
Step 2: Write Down the Combustion Reaction
The balanced combustion reaction of benzene is:
$ C_{6}H_{6} (l) + \frac{15}{2} \, O_{2}(g) \rightarrow 6 \, CO_{2}(g) + 3 \, H_{2}O(l). $
Step 3: Determine the Change in Moles of Gaseous Species ($ \Delta n_{g} $)
Only gaseous species are counted when calculating $ \Delta n_{g} $.
• Moles of gas on product side: 6 (from 6 CO2 (g))
• Moles of gas on reactant side: $ \frac{15}{2} = 7.5 $ (from $ \frac{15}{2} $ O2)
Therefore,
$ \Delta n_{g} = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 6 - 7.5 = -1.5. $
Step 4: Use the Relation Between $ \Delta U $ and $ \Delta H $
The formula connecting internal energy change ($ \Delta U $) and enthalpy change ($ \Delta H $) is:
$ \Delta H = \Delta U + \Delta n_{g} \, R \, T. $
Step 5: Calculate the Correction Term ($ \Delta n_{g} \, R \, T $)
Substitute $ \Delta n_{g} = -1.5 $, $ R = 8.314 \times 10^{-3} \, \text{kJ K}^{-1}\text{mol}^{-1} $, and $ T = 298 \, \text{K} $:
$ \Delta n_{g} \, R \, T = (-1.5) \times (8.314 \times 10^{-3}) \times 298. $
Step 6: Compute $ \Delta H $
From the given data,
$ \Delta U = -3263.9 \, \text{kJ mol}^{-1}. $
So,
$
\Delta H
= -3263.9
+ \Big[ (-1.5) \times (8.314 \times 10^{-3}) \times 298 \Big].
$
Numerically, this term is approximately $ -3.7 $ kJ mol–1 (negative sign arises from $ \Delta n_{g} $ being negative), yielding:
$
\Delta H \approx -3263.9 - 3.7 = -3267.6 \, \text{kJ mol}^{-1}.
$
Step 7: Final Answer
Therefore, the heat of combustion of benzene at constant pressure is $ \boxed{-3267.6 \, \text{kJ mol}^{-1}} $.
