Q. The value of $\mathrm{\Delta}E$ for combustion of 16 g of $CH_{4}$ is -885389 J at 298 K. The $\mathrm{\Delta}H$ combustion for $CH_{4}$ in $J\ mol^{- 1}$ at this temperature will be (Given that, $R = 8.314\ JK^{- 1}\ mol^{- 1})$
A) $- 55337$
B) $- 880430$
C) $- 885389$
D) $- 890348$
Explanation
(b)
$CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l)$
$\mathrm{\Delta}n_{g} = 1 - 3 = - 2$
We know that,
$\mathrm{\Delta}E = \mathrm{\Delta}H + \mathrm{\Delta}n_{g}RT$
$\therefore$ $\mathrm{\Delta}H = ( - 885389) - ( - 2) \times 8.314 \times 298$
= $-$885389+4955.1440
= $-$880433.86 J $mol^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.