CALCULATION OF HEAT (q ) WORK (W), dU AND dH IN ISOTHERMAL PROCESS

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)