Question
Derive relationship between $\Delta H$ and $\Delta U$ for gaseous reaction.
✓
Answer
$i.$ At constant pressure, $\triangle H$ and $\triangle U$ are related as
$\triangle H = \triangle U + P\triangle V ...(1)$
$ii.$ For reactions involving gases, $\triangle V$ cannot be neglected.
Therefore, $\triangle H = \triangle U + P\triangle V$
$= \triangle U + P(V_2 – V_1)$
$\triangle H = \triangle U + PV_2 – PV_1 ...(2)$
where, $V_1$ is the volume of gas-phase reactants and $V_2$ that of the gaseous products.
$iii.$ We assume reactant and product behave ideally. Applying an ideal gas equation, $PV = nRT$. Suppose that $n_1$ moles of gaseous reactants produce $n_2$ moles of gaseous products. Then,
$PV_1 = n_1RT$ and $PV_2 = n_2RT ...(3)$
$iv.$ Substitution of equation $(3)$ into equation $(2)$ yields
$\triangle H = \triangle U + n_2RT – n_1RT$
$= \triangle U + (n_2 – n_1) RT$
$= \triangle U + \triangle n_g RT ...(4)$
where, $\triangle n_g$ is the difference between the number of moles of products and those of reactants.
$\triangle H = \triangle U + P\triangle V ...(1)$
$ii.$ For reactions involving gases, $\triangle V$ cannot be neglected.
Therefore, $\triangle H = \triangle U + P\triangle V$
$= \triangle U + P(V_2 – V_1)$
$\triangle H = \triangle U + PV_2 – PV_1 ...(2)$
where, $V_1$ is the volume of gas-phase reactants and $V_2$ that of the gaseous products.
$iii.$ We assume reactant and product behave ideally. Applying an ideal gas equation, $PV = nRT$. Suppose that $n_1$ moles of gaseous reactants produce $n_2$ moles of gaseous products. Then,
$PV_1 = n_1RT$ and $PV_2 = n_2RT ...(3)$
$iv.$ Substitution of equation $(3)$ into equation $(2)$ yields
$\triangle H = \triangle U + n_2RT – n_1RT$
$= \triangle U + (n_2 – n_1) RT$
$= \triangle U + \triangle n_g RT ...(4)$
where, $\triangle n_g$ is the difference between the number of moles of products and those of reactants.
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