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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

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


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 + 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)

THERMOCHEMISTRY AND ENERGETICS


ENTROPY (S) INTRODUCTION


ENTHALPY (H) INTRODUCTION

We known 1st law of thermodynamics   
                                        dU=dq+dW 
  If Pext=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
                 QP=nCpmdT (Molar Heat capacity at constant Pressure)
                 Cpm= (dQ/dT)P   for 1 mole of gas
                            dQ=dH  at dP=0
   then         Cpm= (dH/dT)P  
         
For an ideal gas: change in enthalpy at constant temperature with change in
pressure is zero. i.e.

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

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 10C was converted into steam at 100C.( Given Cp(H2O)=75.29 JK-1 and dHVap=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 ; CP(H2O,l)= 75 J / mol /K ; CP(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 UA be the energy of a system in its state A and UB 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 =UB  -UA = 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.
(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
     Continue reading.........

HEAT, HEAT EXCHANGE AND HEAT CAPACITY(Q)

Heat is defined as the energy that flow into or out of the system because of a difference in temperature between system and surrounding.

Note; Work is more organised way of energy transfer as compared to hear exchange.

IUPAC  Sign convention of Heat: sign of heat will negative (-Ve) if heat is released by the system given by system while sign of heat will be positive (+Ve) if heat is given to the system.
(i) qV = nCvdT (for constant volume process)
(ii) qp = nCpdT (for constant pressure process)
(iii) Cp,m – Cv,m = R
(iv) Cv & Cp depends on temperature even for an ideal gas.( C = a + bT + cT2 .....)
(v) It is a path function
Cv, Cp are heat capacity of system and Cv,m, Cp,m are heat capacity of one mole system at constant  volume and pressure respectively.

“Exchange of heat and work(P–V) in between system and surrounding always
   occur through boundry of system”

   Note: heat exchange can be measured with the help of Heat Capacity.

 HEAT CAPACITY(Q)
 We know that usually on increasing in temperature is proportional to the heat transfer
                               q=coefficient x ∆T
The coefficient depend on the size, composition and nature of the system so we can also write it as   
                                 q=C∆T
 where C is called Heat capacity
The heat capacity (C):
                                     q=C∆T  or
                                   C= q/∆T      unit- J/K
                            If ΔT=10=1K
                                Then C=q
It is equal to amount of heat needed to raise the temperature of the sample of any substance by one degree Celsius (or Kelvin).

heat Capacity depend  on quantity, nature as well as physical state of the system. And the heat capacity is extensive. it may be made intensive as specific heat capacity
                                  
Specific Heat Capacity(Cs ): 
                                     q=Csm∆T  or
                                   Cs= q/m∆T      
                            If ΔT=10=1K and m = 1g
                                Then Cs=q
It is equal to amount of heat required to raise the temperature of 1gm substance by one degree centigrade .it is intensive properties.

Molar Heat Capacity(Cm ):
                                     q=Cmn∆T  or
                                   Cm= q/n∆T      
                            If ΔT=10=1K and n=1mole

                                Then Cm=q
It is equal to amount of heat required to raise the temperature of 1 Mole substance by one degree centigrade

EXAMPLE(1): The latent heat of fusion of ice at 0ºCis 80 cal/gm the amount of heat needed to convert 200 gm ice into water at 0ºC is ?
 (A) 80 cal (B) 16000 J (C) 16000 cal (D) 1600 cal

SOLUTION: Ans (C)  q = m.L = 200 × 80 = 16000 cal

EXAMPLE(2): Calculate the amount of heat required to raise the temperature of 50 gm water from 25ºC to 55ºC.Specific heat capacity of water = 4.2 J/ºC-gm.
(A) 126 J    (B) 210 J    (C) 6300 J   (D) 1500 J

SOLUTION: Ans. (C)
  q = m.s.ΔT = 50 × 4.2 × (55 – 25) = 6300 J

EXAMPLE (3): Five moles of a monatomic ideal gas is heated from 300K to 400K at constant pressure. the amount of heat absorbed is :
(A) 500 cal   (B) 1500 cal   (C) 2500 cal   (D) 2500 J

SOLUTION: Ans. (C)     
qp = CpΔT = n.Cp,m ΔT
    = 5 × 5/2× (400 – 300) = 2500 cal

EXAMPLE (4):  2 moles of an ideal gas absorbs 720 cal heat when heated from 27ºC to 87ºC, at constant volume. 'ɤ' for the gas is :
(A) 1.5 (B) 1.4 (C) 1.6 (D) 1.33

SOLUTION :Ans. (D)
 qv = n.Cv,m. ΔT
Cv,m =qv/nΔT
       = 720/2x(87 – 27)
       = 6 cal/K-mol
Now, r = 1 +R/Cv,m
           = 1 + 2/6
           = 1.33

EXAMPLE (5): 500 gm ice at 0ºC is added in 2000 gm water at tºC. If the final temperature of system is 0ºC, then the value of 't' is (latent heat of fusion of ice = 80 cal/gm and specific heat capacity of water=cal/gm-ºC)
(A)   20    (B)  40      (C) 10     (D)     2

SOLUTION: Ans. (A)
Heat lost by water = heat gained by ice
or, (m.s. ΔT)water = (m.L)ice
or, 2000 × 1 × (t – 0) = 500 × 800
Δ t = 20ºC

EXAMPLE(6): What is the heat in Joules required to raise the temperature of 25 grams of water from 0 °C to 100 °C? What is the heat in calories?
(Given: specific heat of water = 4.18 J/g·°C)

SOLUTION: Use the formula   q = mcΔT
where
= heat energy
m = mass
c = 
specific heat
ΔT = change in temperature

q = 25gx4.18 J/g·°x(100 °C - 0 °C)
q = 25gx4.18 J/g·°Cx(100 °C)
q = 10450 J
 We know 1 Calorie=4.18 J
So  10450 J in Calorie   = 10450/4.8=2500 calorie 

RELATION BETWEEN Cp AND Cv :


FOR LARGE HEAT CHANGE :

Case –(1)  when Cm is constant
              Q= nCm (T2-T1)
Case- (2) Cm = f(T)
                 Cm = a + bT+ cT2 +……..

Case- (3) The theoretical value of Cvm and Cpm  for Ideal gas can determined by using degree of freedom.

CHARACTERISTIC OF HEAT CAPACITY:

(1): The heat capacity Of any system should depend upon temperature because by increasing    temperature  of system different degree of freedom get excited.
(2): When temperature approaches zero then heat absorbed by the solid mainly converted into vibration potential energy of molecule resulting in very small increase in temperature, hence ‘C’ increases sharply with increase in temperature.
                   Normally C Directly proportional T3
 (3): When the temperature at Melting point of solid ,then heat capacity becomes nearly  constant  for solid elements.
                    Molar heat capacity= 6.4 Cal/K mole
              Or specific heat capacity x atomic weight=6.4 (Dulong and petite’s law)
(4): Exactly at melting point, the heat capacity become infinite as  ΔT=0
(5): the heat capacity of liquid is greater than that of solid because of rotational degree of freedom also excited.
(6): In liquid heat capacity also depend upon temperature and also infinite at boiling point.
(7): the heat capacity of gas become less than liquid because all vibrational and rotational degree of freedom converted into translational degree of freedom.
(8): the  heat capacity of gases depend upon their atomicity.
(9): If the heat capacity depends upon temperature.
(10); As heat (q) is path function, any substance may have infinite heat capacity.
         Example for any substance.
         Isothermal process = infinite
         Adiabatic process   = 0
         Isobaric process     = Cp
         Isochoric process   = Cv

      Normally ,we use Cp and Cv  value  as characteristic of substance.

DEDREE OF FREEDOM :

It is equal to number of modes of energy transfer when a gaseous molecule undergoes collision.                      OR

        It represent the number of independent modes to describe the molecular motion.  

Total degree of freedom = 3N (Where N  is Number of atom in molecule)


1- Translational degree of freedom is 3 (three) always for mono,di and tri atomic molecule.
2- Rotational degree of freedom is zero for mono atomic gas,2 (two) for diatomic molecules and    3 (three) for triatomic molecule

3-Vibrational degree of freedom is also zero for mono atomic gas and 1(one) diatomic gas molecule  and for polyatomic gases VDOF is calculated individually.( fvib= 3N- ftrans+ frot)

Total degree of freedom:=  ftrans + f rot + f vib     and     fvib= 3N- ftrans + f rot    


Molecules                N                      TDF
He                             1                         3
O2                                    2                         6
CO2                           3                         9
NH3                            4                        12
PCl5                           6                        18

Case-1

Monoatomic       Diatomic       Triatomic (linear)       Triatomic (Non linear)
f total =3                ftotal  =6          ftotal =9                      f total=9
f trans=3                  ftrans =3            ftrans=3                     ftrans=3
f rot  =0                      frot     =2           frot    = 2                    frot    =3
f vib  =0                     fvib     =1           fvib    = 4                        fvib   =3

                                  Q =n CmdT

                                QV=n CvmdT

                                Cvm=(dQ/dT)v
 By FLOT dq=dU+dW     and  at constant volume  dW=0   so   dQv=dU

                         Hence    Cvm= (dU/dT)v

LAW OF EQUIPARTIAL OF ENERGY :
Average energy associate with each molecule per degree of freedom is U= 1/2KT  (where K is Boltz’s man constant.

Let degree of freedom is = f   then U is U=1/2fkT


             And U=1/2fkTNper molecule  we know  kNA=R

                     U=1/2fRT  and   dU/dT=1/2fR


             And  dU/dT=Cv      hence  Cv=1/2fR

Cv=1/2ftransR +1/2frotR  (Where Vib degree inactive in chemistry)

       For ideal gas Cpm-Cvm=R  and  Gama= Cpm/Cvm
         

Adiabatic exponent :Adiabatic exponent (Gama) for a mixture of gas with different heat capacity is defined as :
where n1, n2 ........................ are moles of different gases

Example:Calculate change in internal energy of 10 gm of H2 ,when it's state is changed from(300K, 1Atm) to (500 K, 2Atm) ?
Solution: For ideal gas
       
Cv for H2 (diatomic) in low temperature range will be 5R as vibrational part is not included.




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