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RELATIONSHIP BETWEEN DH and DU

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    

EXAMPLE (1): 1 mole of a real gas is subjected to a process from (2 bar, 40 lit.,300K) to (4 bar, 30 lit., 400 K). If change in internal energy is 35 kJ then calculate enthalpy change for the process.
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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