Question

Enthalpy of the reaction : $C{H}_4+\frac{1}{2}{O}_2⟶C{H}_{3}OH$ is negative. If enthalpy of combustion of $C{H}_4$ and $C{H}_{3}OH$ are $x$ and $y$ respectively, which relation is correct?

• A. $x>y$
• B. $x<y$
• C. $x=y$
• D. $x\ge y$

Answer 1

Answer: (A)

$x>y$

$C{H}_4+2O_2⟶C{O}_2+2H_{2}O;\Delta H=-x$ ......(i)

$C{H}_{3}OH+\frac{3}{2}{O}_2⟶C{O}_2+2H_{2}O;\Delta H=-y$ ......(ii)

Subtract equation (ii) from equation (i) to obtain

$C{H}_4+\frac{1}{2}{O}_2⟶C{H}_{3}OH;\Delta H=y-x$

But $y-x$ is negative.

Hence, $x>y$

Thus, option A is the correct answer.