Question

$\Delta \, U^{\circ}$ of combustion of methane is $ -X\,kJ\,mol^{-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)

$CH_{4(g)}+2O_{2(g)} \to CO_{2(g)}+2H_{2}O_{(l)}$
$\Delta\,n_{g} = 1-3 =-2$
$\Delta\,H^{\circ} = \Delta\,U^{\circ}+\Delta\,n_{g}RT=-X-2RT$
Thus, $\Delta\,H^{\circ} <\, \Delta\,U^{\circ}$