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Step-by-Step Solution
Step 1: Write Down the Balanced Chemical Equation
The combustion reaction of methanol can be written as:
\mathrm{CH_{3}OH(l)} + \frac{3}{2}\,\mathrm{O_{2}(g)} \rightarrow \mathrm{CO_{2}(g)} + 2\,\mathrm{H_{2}O(l)}
Step 2: Identify the Known Data
Internal energy change, \Delta U = -726\,\mathrm{kJ\,mol^{-1}} (obtained from bomb calorimetry)
Temperature, T = 27^\circ\mathrm{C} = 300\,\mathrm{K}
Gas constant, R = 8.3\,\mathrm{J\,K^{-1}\,mol^{-1}} = 0.0083\,\mathrm{kJ\,K^{-1}\,mol^{-1}}
Step 3: Determine the Change in the Amount of Gaseous Moles
Count the moles of gas molecules on both sides of the reaction:
Reactants: \frac{3}{2} \,\mathrm{O_2(g)} \implies 1.5 \,\text{moles of gas} (methanol is liquid, so it does not contribute to moles of gas)
Products: 1\,\mathrm{CO_2(g)} \implies 1 \,\text{mole of gas} (water is liquid, so it does not contribute to moles of gas)
Hence, \Delta n_{g} = 1 - 1.5 = -0.5
Step 4: Apply the Relationship Between \Delta H and \Delta U
The relation is:
\Delta H = \Delta U + \Delta n_{g}\,R\,T
Substitute the values:
\Delta H = -726\,\mathrm{kJ\,mol^{-1}} + (-0.5) \times 0.0083\,\mathrm{kJ\,K^{-1}\,mol^{-1}} \times 300\,\mathrm{K}
Calculate the correction term:
\Delta n_{g}\,R\,T = (-0.5) \times 0.0083 \times 300
= -1.245 \,\mathrm{kJ\,mol^{-1}} \approx -1.25\,\mathrm{kJ\,mol^{-1}}
Step 5: Calculate the Enthalpy of Combustion
\Delta H = -726\,\mathrm{kJ\,mol^{-1}} - 1.25\,\mathrm{kJ\,mol^{-1}}
= -727.25\,\mathrm{kJ\,mol^{-1}}
\approx -727\,\mathrm{kJ\,mol^{-1}}
The enthalpy of combustion (nearest integer) is thus:
-727\,\mathrm{kJ\,mol^{-1}}
