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 + Dn_gRT

Where Dn_g=number moles of  product in gaseous state - number moles of reactant in gaseous state

          Dn_g =(n_P-n_R)_g  

case-(1) If  Dn_{g=0   then}   DH = DE

case-(2) If  Dn_{g>0    then}  DH > DE    

case-(3) If  Dn_{g<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

                       ΔH_R= -185 KJ/mole ,ΔU_R= -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) + 12O_2(g) à10CO_2(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) C₄H_10(g) + 6.5O_{2 (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)