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Question
Calculate the standard enthalpy of combustion of methane if the standard enthalpy of formation of methane, carbon dioxide and water are −74.8, −393.5 and −285.8 kJmol−1 respectively.
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Solution
Standard enthalpy of formation of methane
\[\ce{\Delta_fH^\circ(CH4)}\] = −74.8 kJ mol−1
\[\ce{\Delta_fH^\circ(CO2)}\] = −393.5 kJ mol−1
\[\ce{\Delta_fH^\circ(H2O)}\] = −285.8 kJ mol−1
To find:
Standard enthalpy of combustion of \[\ce{CH4}\] = ?
The equation for the combustion of \[\ce{CH4}\] is
\[\ce{CH4_{(g)} + 2O2_{(g)} -> CO2_{(g)} + 2H2O_{(l)}; \Delta_cH^\circ}\] = ?
\[\ce{\Delta_cH^\circ = [\Delta_fH^\circ(CO2) + 2\Delta_fH^\circ (H2O)] - [\Delta_fH^\circ](CH4) + 2\Delta_fH^\circ(O2)}\]
= [1 × (−393.5) + 2 × (−285.8)] − [1 × (−74.8) + 2 × 0]
= −965.1 + 74.8
= −890.3 k J
∴ Standard enthalpy of combustion of methane
\[\ce{\Delta_cH^\circ(CH4)}\] = −890.3 k J
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