The heat of combustion of napthalene (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).

 

A sample of gas is compressed by an average pressure of 0.50 atmosphere so as to decrease its volume from 400cm^3 to 200cm^3. During the process 8.00 J of heat flows out to surroundings. Calculate the change in internal energy of the system.

 

SPONTANEOUS AND NON SPONTANEOUS PROCESS

SPONTANEOUS PROCESS: Those process which have natural tendency to take place ,they may or may not require initiation.
For example
(1) Flow of water from higher level to lower level .
(2) Flow of heat from high temperature to lower temperature.
(3) Radioactivity
(4) Cooling of cup of tea .
(5) Evaporation
(6) Condensation 
(7) Sublimation
(8) Burning of candle
(9) Dissolution of salt and sugar in water
(10) Burning of fuel
(11) melting of ice at room temperature.
NON SPONTANEOUS PROCESS: All spontaneous processes are non spontaneous processes in reverse direction it requires spot of external energy for their progress.
CRITERIA FOR SPONTANEITY:
(1) If dS universe is greater than Zero  then process is spontaneous (reversible)
(2) If dS is Zero then process is in reversible state of equilibrium
(3) If dS is lower than Zero then  process is non spontaneous
For  a spontaneous process entropy of univers is greater than Zero . In order to use entropy has a sole criteria , we need to have information about system as well as surrounding.
dS(system) + dS(surr) is greater than  Zero
.
.
.
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In order to explain the spontaneous behaviour by using the parameters of system we can established  two criteria.
(1) Randomness
(2) Criteria of energy
A spontaneous process is one in which energy ( enthalpy) decreasing and Randomness of system increasing.
For spontaneous proceDH=-ve and dS=+ve

GIBB'S FREE ENERGY FUCTION (G)

This is another thermodynamic quantity that helps in predicting the spontaneity of a process, is called Gibbs energy (G).

It is defined mathematically by the equation.

G = H - TS

Where H = heat content, S = entropy of the system, T = absolute temperature

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)

Enthalpy (H) Introduction : Heat change in isochoric process and isobaric process :

The thermal changes taking place at a constant volume (Isochoric Process) are conventionally expressed in terms of internal energy. 

Whereas the thermal changes taking place at a constant pressure (Isobaric Process)  are expressed in terms of  another function ‘H’ called the heat content of the system or enthalpy.

 

Enthalpy is expressed as  H = E + pV.

Enthalpy is a state function  and  Extensive properties 

Molar Enthalpy: Heat absorbed by the mole substance at constant pressure  and it is Intensive properties.  (H = E+PV ) 

Absolute value of Enthalpy can not calculated but we can calculate only change in enthalpy.

We known 1^st law of thermodynamics (FLOT)

                                        dU=dq+dW 

  If P_ext = constant (Isobaric Process)

  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

THIRD LAW OF THERMODYNAMICS (TLOT)

We know that on increasing the entropy of a pure crystalline substance increases because molecular motion increases with increase of temperature and decreases on decreasing temperature. Or we can say that the entropy of a perfectly crystalline solid approaches zero as the absolute zero of temperature is approached. Which means that at absolute zero every crystalline solid is in a state of perfect order and its entropy should be zero. This is third law of thermodynamics.

We can calculate absolute value of entropy of any substance in any state (solid ,liquid, gas) at any temperature by calculating dS for the processes in going from the initial state to the state of the substance for which entropy is to be calculated.

EXAMPLE: Find the entropy of change when 90 g of H₂O at 10^C was converted into steam at 100^C.( Given Cp(H₂O)=75.29 JK^-1 and dH_Vap=43.932 k JK^-1  mol^-1)

SOLUTION:  

EXAMPLE:  Calculate dG for

 (i) H₂O (l, 1 atm, 300 K ) ------à H₂O (g, 1 atm, 300 K)

 (ii) H₂O (l, 2 atm, 373 K ) -----à H₂O (g, 2 atm, 373 K)

 Given: dH₃₇₃ = 40 kJ ; C_P(H₂O,l)= 75 J / mol /K ; C_P(H₂O,g) = 35 J / mol / K

SOLUTION:  (1)

(2)

WORK DONE (PV-WORK ANALYSIS )

Energy that is transmitted from one system to another in such a way that difference of temperature is not directly involved. This definition is consistent with our understanding of work as dw= Fdx. The force (F) can arise from electrical, magnetic, gravitational & other sources. It is a path function.

PV-Work analysis: Consider a cylinder fitted with a friction less piston, which enclosed no more of an ideal gas. Let an external force F pushes the piston inside producing displacement in piston.

Let distance of piston from a fixed point is x and distance of bottom of piston at the same fixed point is l. This means the volume of cylinder = (l – x) A where A is area of cross section of piston.

For a small displacement dx due to force F, work done on the system.

                        dw = F.dx

                                                  Also  F = PA 

                        dW = PA.dx

                                                    V = (l – x)A

                         dV = –A . dx

                        dW = –Pext. dV

             

Note :

(1): Litre atmosphere term is unit of energy. It is useful to remember the conversion: 

         1 litre atm= 101.3 Joules = 24.206 Cal.

(2): During expansion dV is positive and hence sign of w is negative since work is done by the system and negative sign representing decease in energy content of system. During compression, the sign of dV is negative which gives positive value of w representing the increase in energy content of system during compression.

EXAMPLE.1 mole of ideal monatomic gas at 27°C expands adiabatically against a constant external pressure of 1.5 atm from a volume of 4dm³ to 16 dm³.            Calculate (i) q (ii) w and (iii) DU

SOLUTION:   (i) Since process is adiabatic  \ q = 0

                       (ii) As the gas expands against the constant external pressure.

                             W = - PVd=-P(V2-V1)

                             W =-1.5(16-4)

                             W= - 18 dm³

                            (iii) DU = q + w = 0 + (-18) = -18 atm dm³

(1) Work done in Isothermal Irreversible Process:
(2) Work done in Isothermal Reversible Process:
(3) Work done in Adiabatic Irreversible Process:
(4) Work done in Adiabatic Reversible Process:

SECOND LAW OF THERMODYNAMICS (SLOT)

(1) FLOT is law of conservation of energy, and according to it all chemical and physical processes take place in a such way that energy remain constant

(2) In FLOT we introduce enthalpy and internal energy.

(3)When two body at different temperatures are kept closed toeach other there will be transfer of heat while the transfer of heat take place called not explain by FLOT.

(4) we can not explain spontaneous and irreversible behaviour of processes by using FLOT.

In 2nd law of thermodynamics we will introduce entropy as criteria of spontaneity.

STATEMENT OF 2ND LAW :. "It is impossible to construct the heat engine which can take heat from a source and completely convert into work without creating any disturbance in the surrounding.

FIRST LAW OF THERMODYNAMICS (FLOT)

(1) It is also known as law of energy conservation.

(2)The first law of thermodynamics states that energy can neither be created nor destroyed,      although it can be transformed from one form to another.

(3) Another Statement is “total energy of universe is remain constant”

(4) In FLOT we introduce the term ENTHALY and INTERNAL ENERGY.

        

MATHEMATICAL EXPRESSION OF FIRST LAW

Let U_A be the energy of a system in its state A and U_B be the energy in its state B. Suppose the system while undergoing change from state A to state B absorbs heat q from the surroundings and also  performs some work (mechanical or electrical), equal to w. The absorption of heat by the system tends to raise the energy of the system. The performance of work by the system, on the other hand, tends to lower the energy of the system because performance of work requires expenditure of energy. Hence the change of internal energy, DU, accompanying the above process will be given by

                                  dU =U_{B  -}U_{A =} q + w

In general, if in a given process the quantity of heat transferred from the surrounding to the system is q and work done in the process is w, then the change in internal energy,

                                  dU = q + w

This is the mathematical statement of the first law of thermodynamics.

ACCORDINGTO IUPAC:

(1) If work is done by the surroundings on the system (compression of a gas), w is taken as positive so that dU = q + w.

(2) If however work is done by the system on the surroundings ( expansion of a gas), w is taken as negative so that dU = q – w.

(3) (q + w) it will be independent of the way the change is carried out, it only depend on initial and final state. and  and hence it is a state function (dE= q+ w)

(4) If their is no transfer of energy as heat or work (Isolated System) 

      ie, if w=0 and q=0.    then dU=0 

EXAMPLE: A system gives out 25 J of heat and also does 35 J of work. What is the internal energy change ? 

SOLUTION:   According to FLOT dU = q + w

                                                          dU =-25 J+(-35 J)

                                                          dU = -60 J 

Note:

(1) dU = q + w is invalid for open system.

(2) 1st law of T.D. is applicable for closed system in which system is at rest or moving with   constant velocity and in absence of external fields.

(3) The macroscopic energy changes with velocity and elevation of the system are not considered in internal energy change of system.

EXAMPLE:  The pressure of a fluid is a linear function of volume P=a+bV and internal energy of fluid is U=34+3PV in SI unit. find  a ,b, W ,dU for change in state (100 Pascal ,3m cube ) to (400 Pascal ,6m cube). (given Pascal =1J) .

SOLUTION:  We know by FLOT DE=q+W

LIMITATIONS OF FIRST LAW THERMODYNAMICS:

A major limitation of the first law of thermodynamics is that it’s merely indicates that in any process there is an exact equivalence between the various forms of energies involved, but it provides no information concerning the spontaneity or feasibility of the process.

For example, the first law does not indicate whether heat can flow from a cold end to a hot end or not.

The answers to the above questions are provided by the second law of thermodynamics. 

PARAMETER INVOLVE IN  FIRST LAW THERMODYNAMICS

(1) Heat (q)

(2) Work done (w)

(3) Internal Energy (dU) 

(4) Enthalpy (dH)

Note:

(1)  q and w are path functions while internal energy and Enthalpy are state functions.