Question

Enthalpy of $CH_{4} + \frac{1}{2} O_{2} \rightarrow CH_{3}OH$ is negative. If enthalpy of combustion of $CH_{4}$ and $CH_{3}OH$ 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

Answer: (A)

$ CH _4( g )+\frac{1}{2} O _2( g ) \rightarrow CH _3 OH ( l ) $
$\therefore \Delta H =-\left[\left(\Delta H\right.\right.$ of combustion of
$\left.\left.CH _3 OH \right)\right]-\left[\left(\Delta H\right.\right.$ of
combustion of $\left.\left.CH _4\right)\right]$
$=-[(-y)-(-x)] $
$=-[-y+x]=y-x$
$\Delta H =$ negative
$y-x=$ negative
$ \therefore x > y $