Question

Consider the reaction: $N_2+3H_2\to 2N{H}_3$. It is carried out at constant temperature and pressure. If $\Delta H$and $\Delta U$ are the enthalpy and internal energy changes for the reaction respectively, then which of the following expression is true?

• A. $\Delta H>\Delta U$
• B. $\Delta H<\Delta U$
• C. $\Delta H=\Delta U$
• D. $\Delta H=0$

Answer 1

Answer: (B)

$\Delta H<\Delta U$

$\Delta H=\Delta U+\Delta n_{g}RT$

Given that

$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$

$\Delta n_g=2-3-1=-2$

$\therefore \Delta H=\Delta U-2RT$

$\therefore \Delta U=\Delta H+2RT$

$\therefore \Delta U>\Delta H$