The difference between dH and dU becomes significant only when gases are involved (insignificant in solids and liquids)
DH = DU + D(PV)
If substance is not undergoing chemical reaction or phase change.
we know PV=nRT
PDV=DnRT
hence
DH = DE + DnRT
In case of chemical reaction
DH = DE + DngRT
Where Dng=number moles of product in gaseous state - number moles of reactant in gaseous state
Dng =(nP-nR)g
case-(1) If Dng=0 then DH = DE
case-(2) If Dng>0 then DH > DE
case-(3) If Dng<0 then DH < DE
SOLUTION: DH = DU + D(PV)
D(PV) = P2V2 – P1V1
= 4 × 30 – 2 × 40
= 40 liter -bar = 4 kJ
so DH = 35 + 4 = 24 kJ
EXAMPLE (2).: What is the relation between DH and DE in this reaction?
CH4(g) + 2O2(g) ---------> CO2(g) + 2H2O(l)
SOLUTION: DH = DE + DnRT
Dn = no. of mole of products - no. of moles of reactants = 1– 3 = –2
DH = DE – 2RT
EXAMPLE(3):
Consider the chemical
reaction at 300 K H2 (g)+Cl2
à2HCl(g) ΔH= -185KJ/mole calculate ΔU if
3 mole of H2 completely react
with 3 mole of Cl2(g) to form HCl.
SOLUTION: H2
(g)+Cl2 à2HCl(g)
ΔH= -185KJ/mole
Δng=0
ΔH= ΔU+
ΔngRT
ΔH= ΔU
ΔHR= -185 KJ/mole ,ΔUR=
-185 KJ/mole
H2 (g)+Cl2
à2HCl(g) ΔH= -185KJ/mole
3 mole 3 mole
Hence ΔU= -185 X 3
KJ/Mole
EXAMPLE (4): The heat of combustion of naphthalene
(C10H8(s)) at constant volume was measured to be . 5133 kJ mol.1 at 298K.
Calculate the value of enthalpy change (Given R = 8.314 JK.1 mol.1).
SOLUTION: The combustion reaction of naphthalene.
C10H8(s) + 12O2(g) à10CO2(g) + 4H2O(l)
ΔE = -5133kJ
Δn = 10 -12 = -2 mol.
Now applying the relation.
ΔH = ΔE + (Δn) RT
= -5133 × 103 + (-2) (8. 314) (298)
= -5133000J - 4955.14J
= -5137955. 14 Joule
EXAMPLE(5) What is the true regarding complete
combustion of gaseous isobutene –
(A) ΔH = ΔE (B) ΔH > ΔE (C) ΔH = ΔE = O (D) ΔH < ΔE
SOLUTION: (D) C4H10(g) + 6.5O2
(g) à4CO2(g)
+ 5H2O(l)
Δn = [4 -7.5] = -3.5
ΔH = ΔE + ΔngRT
Δ H < ΔE
EXAMPLE (6): For a gaseous reaction: 2A2 (g) +
5B2(g) à2A2B5(g)
at 27ºC the heat change at constant pressure is found to be .50160J. Calculate
the value of internal energy change (ΔE). Given that R = 8.314 J/Kmol.
(A) -34689 J (B) -37689 J (C) -27689 J (D) -38689 J
SOLUTION : 2A2(g) + 5B2(g) à 2A2B5 (g); ΔH= -50160 J
Δ n = 2-(5 + 2) = -5 mol.
ΔH = ΔE + (Δn) RT
-50160 = ΔE + (Δn) RT
Δ E = -50160- (-5) (8.314) (300)
= -50160 + 12471 = -37689 J
The answer is (B)
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