Question

$A{l}_{(}s)+\frac{3}{2}{O}_2(g)\to \frac{1}{2}A{l}_2{O}_3(s)$
For this reaction relate $\Delta H$ and $\Delta U$.

• A. $\Delta H$ =$\Delta U$
• B. $\Delta H$ > $\Delta U$
• C. $\Delta H$ < $\Delta U$
• D. None of these

Answer 1

The correct option is C $\Delta H$ < $\Delta U$

As we know,
$\Delta H$ =$\Delta U+\Delta n_{g}RT$
here,
$A{l}_s+\frac{3}{2}{O}_2\left(g\right)\to \frac{1}{2}A{l}_2{O}_3\left(s\right)$
$\Delta n_g=-3/2$
so,
$\Delta H$ =$\Delta U-3/2RT$
$\Delta H$ < $\Delta U$