JEE MAIN - Chemistry (2005 - No. 3)
Consider the reaction: N2 + 3H2 $$\to$$ 2NH3 carried out at constant temperature and
pressure. If $$\Delta H$$ and $$\Delta U$$ are the enthalpy and internal energy changes for the
reaction, which of the following expressions is true?
$$\Delta H$$ > $$\Delta U$$
$$\Delta H$$ < $$\Delta U$$
$$\Delta H$$ = $$\Delta U$$
$$\Delta H$$ = 0
Explanation
$$\Delta H = \Delta U + \Delta nRT$$
for $$\,\,\,{N_2} + 3{H_2} \to 2N{H_3}$$
$$\,\,\,\Delta {n_g} = 2 - 4 = - 2$$
$$\therefore$$ $$\,\,\,\Delta H = \Delta U - 2RT\,\,\,$$
or $$\,\,\,\Delta U = \Delta H + 2RT$$
$$\therefore$$ $$\,\,\,\Delta U > \Delta H$$
for $$\,\,\,{N_2} + 3{H_2} \to 2N{H_3}$$
$$\,\,\,\Delta {n_g} = 2 - 4 = - 2$$
$$\therefore$$ $$\,\,\,\Delta H = \Delta U - 2RT\,\,\,$$
or $$\,\,\,\Delta U = \Delta H + 2RT$$
$$\therefore$$ $$\,\,\,\Delta U > \Delta H$$
Comments (0)
