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Question
Calculate the standard enthalpy of combustion of CH4(g) if ΔfH°(CH4) = – 74.8 kJ mol–1, ΔfH°(CO2) = – 393.5 kJ mol–1 and ΔfH°(H2O) = – 285.8 kJ mol–1.
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Solution
Given: ΔfH° (CO2) = – 393.5 kJ mol–1
ΔfH° (H2O)= – 285.8 kJ mol–1
ΔfH°(CH4) = – 74.8 kJ mol–1
To find: Standard enthalpy of combustion (ΔcH°)
Formula: ΔrH° = ∑ΔfH°(products) − ∑ΔfH° (reactants)
Calculation: The equation for the combustion of CH4 is
\[\ce{CH_{4(g)} + 2O_{2(g)} -> CO_{2(g)} + 2H2O_{(l)}}\]
ΔrH° = ∑ΔfH°(products) − ∑ΔfH° (reactants)
= [ΔfH°(CO2) + 2ΔfH°(H2O)] – [ΔfH°(CH4) + 2ΔfH°(O2)]
= [1 mol × (– 393.5 kJ mol–1) + 2 mol × (– 285.8 kJ mol–1)] – [1 mol × (– 74.8 kJ mol–1) + 0]
= – 890.3 kJ
ΔcH°(CH4) = – 890.3 kJ
The standard enthalpy of combustion is – 890.3 kJ.
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