Question
Derive a relation between $\Delta H$ and $\Delta U$ for a chemical reaction.
✓
Answer
Relation between $\triangle H$ and $\triangle U$
The heat of reaction is given by enthalpy change
$\triangle H=H_2 - H_1$
by defination $H= U+PV$
$H_1=U_1+P_1V_1$
$H_2=U_2+P_2V_2$
$\triangle H = (U_1+P1V_1) - (U_2+P_2V_2)$
$=(U_2-U_1)+(P_2V_2-P_1V_1)$
$=\triangle U+(P_2V_2-P_1V_1)(\because \triangle U=U_2-U_1)$
since $PV=nRT$
for initial state $P_1V_1=n_1RT$
for final state $P_2V_2=n_2RT$
$P_2V_2-P_1V_1=n_2RT-n_1RT$
$=(n2-n)RT$
$=\triangle nRT$
where $\triangle n =$ number of moles of gaseous product $-$ number of moles of gaseous reactant.
$\triangle H=\triangle U+\triangle nRT$
The heat of reaction is given by enthalpy change
$\triangle H=H_2 - H_1$
by defination $H= U+PV$
$H_1=U_1+P_1V_1$
$H_2=U_2+P_2V_2$
$\triangle H = (U_1+P1V_1) - (U_2+P_2V_2)$
$=(U_2-U_1)+(P_2V_2-P_1V_1)$
$=\triangle U+(P_2V_2-P_1V_1)(\because \triangle U=U_2-U_1)$
since $PV=nRT$
for initial state $P_1V_1=n_1RT$
for final state $P_2V_2=n_2RT$
$P_2V_2-P_1V_1=n_2RT-n_1RT$
$=(n2-n)RT$
$=\triangle nRT$
where $\triangle n =$ number of moles of gaseous product $-$ number of moles of gaseous reactant.
$\triangle H=\triangle U+\triangle nRT$
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