Question

The value of $\Delta E$ for combustion of 16 g of $C{H}_4$ is -885389 J at 298 K. The $\Delta H$ combustion for $C{H}_4$ in J $mo{l}^{-1}$ at this temperature will be: $(Giventhat,R=8.314J{K}^{-1}mo{l}^{-1})$

• A. $-55337$
• B. $-880430$
• C. $-885389$
• D. $-890348$

Answer 1

The correct option is C $-885389$

$△H=△E+△n_{g}RT$

$\underset{\left(g\right)}{C{H}_4}+\underset{\left(g\right)}{2O_2}⟶\underset{\left(g\right)}{C{O}_2}+\underset{\left(g\right)}{2H_{2}O}$

Since all are attained in gaseous form and $△H_{condensation}$ of $H_{2}O$ is not provided

$\therefore △n_g$ =no of moles of gaseous reactants $-$no of moles of gaseous products

$=3-3=0$

$\therefore \Delta H=\Delta E+ORT$

$\therefore \Delta H_{combustion}=-885389Joules$