Question

$\Delta U^{\circ }$ of combustion of methane is $-XkJmo{l}^{-1}$ The value of $\Delta H^{\circ }$ is

• A. $=\Delta U^{\circ }$
• B. $>\Delta U^{\circ }$
• C. $<\Delta U^{\circ }$
• D. $=0$

Answer 1

Answer: (C)

$C{H}_{4(g)}+2O_{2(g)}\to C{O}_{2(g)}+2H_2{O}_{(l)}$
$\Delta n_g=1-3=-2$
$\Delta H^{\circ }=\Delta U^{\circ }+\Delta n_{g}RT=-X-2RT$
Thus, $\Delta H^{\circ }<\Delta U^{\circ }$