Question

The heat produced on combustion of methane is approximately

• A. $ 890\, kJ $ , per $ g $
• B. $ 74.2\, kJ $ , per $ g $
• C. $ 55.6\, kJ $ , per $ g $
• D. $ 49.5\, kJ $ , per $ g $

Answer 1

Answer: (C)

$CH_4 (g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$
Enthalpy change for a given reaction is equal to $=$ (sum of standard enthalpies of formation of products) $-$ (sum of standard enthalpies of formation of reactants).
Given,
enthalpy of $CO_2 = - 393.5 \,kJ / mol$
enthalpy of $H_2O = - 285.8 \,kJ / mol$
enthalpy of $CH_4 = - 74.9 \,kJ / mol$
$\therefore $ Enthalpy of combustion of methane
$= [- 393.5 + 2 \times ( - 285.8)] - [- 74.9 + 0]$
$= -965.1 + 74.9$
$ = - 890.2 \,kj/mol$
$ = \frac{890.2}{16}\,kJ/g$
$ = -55.6\,kJ/g$
The standard enthalpy of formation is taken as zero, because it is most stable form at $298 \,K$ and $1$ atm pressure.