Question

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

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

Answer 1

The correct option is A $x>y$

$C{H}_4\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to C{H}_{3}OH\left(l\right)$

$\therefore \Delta H=-\left[\right(\Delta HofcombustionofC{H}_{3}OH\left)\right]-\left[\right(\Delta HofcombustionofC{H}_4\left)\right]$

$=-\left[\right(-y)-(-x\left)\right]$

$=-[-y+x]=y-x$

$\Delta H$ = negative

$y-x$ = negative

$\therefore x>y$