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Enthalpy of $CH_4 + \dfrac{1}{2}O_2 \rightarrow CH_3OH$ is negative. If enthalpies of combustion of $CH_4$ and $CH_3OH$ are x and y respectively, then which relation is correct : -
A
x > y
B
x < y
C
x = y
D
x $\geq$ y
Detailed Solution
Using combustion enthalpies: $\Delta H = \sum\Delta H_c(\text{reactants}) - \sum\Delta H_c(\text{products})$
For $CH_4 + \dfrac{1}{2}O_2 \rightarrow CH_3OH$: $\Delta H = x - y$ (combustion enthalpy of $O_2$ is zero)
Given $\Delta H$ is negative: $x - y \lt 0$
Hence $x \lt y$ (x is more negative than y; methane releases more heat on combustion than methanol).
For $CH_4 + \dfrac{1}{2}O_2 \rightarrow CH_3OH$: $\Delta H = x - y$ (combustion enthalpy of $O_2$ is zero)
Given $\Delta H$ is negative: $x - y \lt 0$
Hence $x \lt y$ (x is more negative than y; methane releases more heat on combustion than methanol).