Question

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

Answer 1

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,
$n_1{X}_{\left(g\right)}\to n_2{Y}_{\left(g\right)}$
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, $P_2{V}_2=n_{2}RT$
$P_2{V}_2-P_1{V}_1=n_{2}RT-n_{1}RT$
$=(n_2-n_1)RT$
$=\Delta nRT$
where, $\Delta n=$ [No. of moles of gaseous products] - [No. of moles of gaseous reactants]
$\therefore \Delta H=\Delta U+\Delta nRT$
In an isochoric process, the volume remains constant i.e., $\Delta V=0$
Therefore,
$\Delta H=\Delta U$