Question

The difference in $\Delta H$ and $\Delta E$ for the combustion of ethane at ${27}^o$ C would be_________.

• A. $\Delta H>\Delta E$
• B. $\Delta H<\Delta E$
• C. $\Delta H=\Delta E$
• D. no relation

Answer 1

The correct option is B $\Delta H<\Delta E$

$C_2{H}_6\left(g\right)+\frac{7}{2}{O}_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$
$\Delta H=\Delta E+\Delta n_{g}RT$
$\Delta H-\Delta E=\Delta n_{g}RT=(2-\frac{7}{2}-1)RT=\frac{-5}{2}\times RT$
$\therefore \Delta E>\Delta H$