derive the relation delta H =delta U + delta nRT
Answers
Explanation:
When the change in volume is zero i.e., ΔV=0 or when none of the reactants or products are in gaseous states or when the number of moles of reactants is equal to the number of moles of products (Δng=0) , then change in enthalpy is equal to change in internal energy, i.e., ΔH = ΔU .
Answer:
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)
→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
ΔH=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
,
∴ΔH=U
2
+P
2
+P
2
V
2
−(U
1
+P
1
V
1
)
∴ΔH=U
2
+P
2
V
2
−U
1
−P
1
V
1
∴ΔH=U
2
−U
1
+P
2
V
2
−P
1
V
1
Now, ΔU=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
=ΔnRT
where, Δn= [No. of moles of gaseous products] - [No. of moles of gaseous reactants]
∴ΔH=ΔU+ΔnRT
In an isochoric process, the volume remains constant i.e., ΔV=0
Therefore,
ΔH=ΔU