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The enthalpy of combustion is always negative. CH4 + 2O2 → CO2 + 2H2O; ΔH = −x kJ CH3OH + 3/2 O2 → CO2 + 2H2O; ΔH = −y kJ CH4 + 1/2 O2 → CH3OH; ΔH = ΔH1 − ΔH2 It is given that ΔH is negative. Thus, −x − (−y) = ve y − x = ve Hence, x > y.

1. (a) x > y
2. (b) x < y
3. (c) x = y
4. (d) x ≥ y

Correct answer: (a) x > y

Solution

The enthalpy of combustion is always negative, and the given equations show that ΔH for combustion of CH4 (x) is more negative than that of CH3OH (y). Thus, x > y.

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