Chemical Thermodynamics — Chemistry STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryChemical Thermodynamics2 Marks
Question
Derive the relation $\Delta H-\Delta U=\Delta n R T$.
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Answer
Derivation for the relation $\triangle H − \triangle U = \triangle nRT:$
$1) \triangle H$ and $\triangle U$ at constant pressure are related as,
$\triangle H = \triangle U + P\triangle V$
For reactions involving solids and liquids, $\triangle V$ is usually very small because solids and liquids do not expand or contract significantly as pressure changes. For such reactions neglecting $P\triangle V, \triangle H = \triangle U$
However, for reactions involving gases, $\triangle V$ cannot be neglected. The equation
$\triangle H = \triangle U + P\triangle V$
$= \triangle U + P (V_2 − V_1)$
$= \triangle U + PV_2 − PV_1 ….. (1)$
where, $V1$ is the volume of gas$-$phase reactants $($initial state$)$ and $V2$ is the volume
of gas$-$phase products $($final$-$state$).$
If we assume that reactant and product gases are ideal, we can apply ideal gas equation, $PV = nRT.$ Suppose that $n1$ moles of gaseous reactants produce $n_2$ moles of gaseous products. Then,
$PV_1 = n_1RT$ and $PV_2 = n_2RT ….. (2)$
Substitution of equation $(2)$ into equation $(1)$ gives,
$\triangle H = \triangle U + n_2RT − n_1RT$
$= \triangle U + (n_2 − n_1)RT$
$= \triangle U + \triangle nRT$
where, $\triangle n$ is the difference between the number of moles of gaseous products and that of
gaseous reactants
$\therefore \triangle H − \triangle U = \triangle n RT$
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