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)

THERMOCHEMISTRY AND ENERGETICS

ENTROPY (S) INTRODUCTION

ENTHALPY (H) INTRODUCTION

We known 1^st law of thermodynamics   

                                        dU=dq+dW 

  If P_ext=constant

                Then dW= -PdV

     and     dU=dq- PdV

                dq=dU-PdV

                dq=(U₂-U₁)+P(V₂-V₁)

                dq=(U2+PV2)-(U1+PV1)

                dQ=H2-H1                    ( H₂=U₂+PV₂  and H₁=U₁+PV₁)

                dq=dH

hence

The enthalpy of a system is defined as:

                  H = U + PV

                  DH = DU + D(PV)

Where

H is the enthalpy of the system

U is the internal energy of the system

P is the pressure at the boundary of the system and its environment.

(1) In thermodynamics the quantity U + PV is a new state function and known as the    enthalpy of the system and is denoted by H=U+PV. It represents the total energy stored in the system.

(2) It may be noted that change in enthalpy is equal to heat exchange at constant pressure.(3) Enthalpy is also an extensive property as well as a state function.                                 (4)The absolute value of enthalpy cannot be determined, however the change in enthalpy can  be experimentally determined.

               DH = DU + D(PV)

(5) Change in enthalpy is a more useful quantity than its absolute value.

(6) The unit of measurement for enthalpy (SI) is joule.

(7)The enthalpy is the preferred expression of system energy changes in many chemical and physical measurements, because it simplifies certain descriptions of energy transfer. This is because a change in enthalpy takes account of energy transferred to the environment through the expansion of the system under study.

(8)The change dH is positive in endothermic reactions, and negative in exothermic processes. dH of a system is equal to the sum of non-mechanical work done on it and the heat supplied to it.

(9) For quasi static processes under constant pressure, dH is equal to the change in the internal energy of the system, plus the work that the system has done on its surroundings. This means that the change in enthalpy under such conditions is the heat absorbed (or released) by a chemical reaction.

NOTE:

Transfer of heat at constant volume brings about a change in the internal energy(DU) of the system whereas that at constant pressure brings about a change in the enthalpy (DH) of the system.

For Ideal gas

                     H=U+PV   and U=f(T)

                     PV=nRT

                    H=U+nRT   and H=f(T) only for ideal gas

For other substance and real gas

                     H=U+PV  

                     U=f(P,V,T) and H=f(,PV,T)

       So         H=f(P,T)/ f(V,T)/ f(P,V)

H=f(T,P)

           dH=(dH/dT)_p dT+(dH/dP)_T dP------------------------------------- (1)

H=f(V,T)

           dH=(dH/dV)_T dV+(dH/dT)_V dT------------------------------------- (2)

H=f(P,V)

           dH=(dH/dV)_P dV+(dH/dP)_V dP------------------------------------- (3)

Out of the above three relation H as function on of (T,P) Has a greater significance. The above differential equation simplified for different substance for different condition.

For isobaric process : dP = 0 (Molar Heat capacity at constant Pressure)

We known

                 Q_P=nC_pmdT (Molar Heat capacity at constant Pressure)

                 C_pm= (dQ/dT)_P   for 1 mole of gas

                            dQ=dH  at dP=0

   then         C_pm= (dH/dT)_P  

         

For an ideal gas: change in enthalpy at constant temperature with change in

pressure is zero. i.e.

Continue...........

RELATIONSHIP BETWEEN DH AND DU