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Question

Prove that $$\Delta H = \Delta U+ \Delta n RT$$. What is the condition under which, $$\Delta U = \Delta H$$?


Solution

Let $$H_1, U_1, P_1, V_1$$ and $$H_2, U_2, P_2, V_2$$ represent enthalpies, internal energies, pressures and volumes in the initial and final states respectively.
For a reaction involving $$n_1$$ moles of gaseous reactants in initial state and $$n_2$$ moles of gaseous products at final state,
$$n_1 X_{(g)} \rightarrow n_2 Y_{(g)}$$
If $$H_1$$ and $$H_2$$ are the enthalpies in initial and final states respectively, then the heat of reaction is given by enthalpy change as
$$\Delta H = H_2 - H_1$$
Mathematical definition of 'H' is $$H = U + PV$$
Thus, $$H_1 = U_1 + P_1 V_1$$ and $$H_2 = U_2 + P_2 V_2$$,
$$\therefore \Delta H = U_2 + P_2 + P_2 V_2 - (U_1 + P_1 V_1) $$
$$\therefore \Delta H = U_2 + P_2 V_2 - U_1 - P_1 V_1$$
$$\therefore \Delta H = U_2 - U_1 + P_2 V_2 - P_1 V_1$$
Now, $$\Delta U = U_2 - U_1$$
Since, $$PV = nRT$$
For initial state, $$P_1 V_1 = n_1 RT$$
For final state, $$P_2 V_2 = n_2 RT$$
$$P_2V_2 - P_1 V_1 = n_2 RT - n_1 RT$$
$$= (n_2 - n_1) RT$$
$$= \Delta n RT$$
where, $$\Delta n = $$ [No. of moles of gaseous products] - [No. of moles of gaseous reactants]
$$\therefore \Delta H = \Delta U + \Delta n RT$$
In an isochoric process, the volume remains constant i.e., $$\Delta V = 0$$
Therefore,
$$\Delta H = \Delta U$$

Chemistry

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