Thermo Chemical Reaction Equilibria

CHAPTER 13 CHEMICAL-REACTION EQUILIBRIA

13.1 THE REACTION COORDINATE  The general chemical reaction :

v1 A1  v2 A2  ...  v3 A3  v4 A4  ... where | vi | is a stoichiometric coefficient and Ai stands for a chemical formula.  For vi : Positive ( + ) for product

Negative ( - ) for reactants

Example : CH4 + H2O → CO + 3H2 The stoichiometric numbers are :

vCH 4  1

vH 2 O  1

vCO  1

vH 2  3

 The stoichiometric number for an inert species is zero.  Since

dn1 dn2 dn3 dn4     ...  d v1 v2 v3 v4

 The general relation between a differential change dni in the number of moles of a reacting species and dε is :

dni = vi dε

( i = 1 , 2 ,….N)

 This variable ε called the reaction coordinate, characterizes the extent or degree to which a reaction has taken place.

or



ni

ni0



dni  vi  d

ni  ni0  vi

0

( i = 1 , 2 ,….N)

 Summation over all species yields :

n   ni   ni0    vi

or where

n = n0 + vε

n   ni i

i

i

n0   ni0 i

i

v   vi i

 Thus the mole fractions yi of the species present are related to ε by :

ni ni0  vi  yi   n n0  v

Example 13.2 Consider a vessel which initially contains only no mol of water vapor. If decomposition occurs according to the reaction, 1 H 2 O  H 2  O2 2 find expressions which relate the number of moles and the mole fraction of each chemical species to the reaction coordinate .

Solution 13.2

1 1 For the given reaction v  1  1   2 2 Application of Eqs. (13.4) and(13.5) yields:

n H 2O  n0   y H 2O

n0    1 n0   2

nH 2   yH2 

 1 n0   2

nO2 y O2

1   2 1   2 1 n0   2

The fractional decomposition of water vapor is:

n0  n H 2O n0

n0   n0       n0 n0

Thus when no = 1,  is identified with the fractional decomposition of the water vapor.

dni   vi , j d j

( i = 1 , 2 ,….N)

j

 After integration :

ni  ni0   vi , j  j j

 Summing over all species yields :



n   ni0   vi , j  j  n0    i

i

j

j

 For the mole fraction: ni0   vi , j  j

yi 

j

n0   v j  j j



v i

 i, j



j

13.2 APPLICATION OF EQUILIBRIUM CRITERIA TO CHEMICAL REACTIONS  At the equilibrium state : ( dGt )T , P = 0 The total Gibbs energy Gt is a minimum Its differential is zero.

13.3 THE STANDARD GIBBS-ENERGY CHANGE AND THE EQUILIBRIUM CONSTANT

The fundamental property relation for single-phase systems, provides an expression for the total differential of the Gibbs energy: d  nG    nV  dP   S  dT   dn



i

i

i result of a single  If changes in mole numbers ni occur as the chemical reaction in a closed system , then each dni may be replaced by the product vi dε .

d  nG    nV  dP   S  dT   vi  i d

 Since nG is a state function , the right sidei of this an exact differential expression : t      nG  G   i vi  i         T ,P   T ,P

 

 For chemical-reaction equilibrium :

v  i

i

0

i

 Recall the definition of the fugacity of a species in solution :

 i  i  T   RT ln fˆi  For a pure species i in its standard start at the same temperature :

Gi  i  T   RT ln f i 

 The difference between these two equations is : fˆi   i  Gi  RT ln  fi

 For the equilibrium state of a chemical reaction :



 RT  ln  fˆ

 

 ˆ f 0 v G  RT ln f i i i i i

v G i

 i

i

i



ln  fˆi f i  i

ln  i

where





i

vi

fˆi f i 



 vi

fi



 i vi Gi RT

K

  G    K  exp  RT 

vi

0

 Alternative expression for K :

 G  ln K  RT  Also by definition : G  

  G  i i

K is a function of temperature only . In spite of its dependence on temperature , K is called the equilibrium constant for the reaction ;

 v G i i i , represented by ΔG˚ , is called the

standard Gibbs-energy change of reaction ,

13.4 EFFECT OF TEMPERATURE ON THE EQUILIBRIUM CONSTANT  The dependence of ΔG˚ on T :



 Then become:



d G  RT  H   dT RT 2

d ln K H   dT RT 2  Equation above gives the effect of temperature on equilibrium constant and hence on the equilibrium conversion.

 When ΔH˚ negative = exothermic reaction positive = endothermic reaction  If ΔH˚ is assumed independent of T, then: K H   1 1  ln     K R  T T

 This approximate equation implies that a plot of ln K vs the reciprocal of absolute temperature is a straight line. Figure 13.2 in textbook, shows a plot of ln K vs 1/T for a number of common reactions, illustrates this near linearity.

 The rigorous development of the effect of temperature on the equilibrium constant is based on the definition of the Gibbs energy , written for a chemical species in its standard state: G˚i = H˚i -TS˚i  Multiplication by vi and summation over all species gives:   v G  v H  T v S i i i i i i i

i

i

 As a result of the definition of standard property change of reaction, this reduces to: ΔG˚=ΔH˚ - TΔS˚  Where the standard heat of reaction and standard entropy change is related to temperature:   C P H   H 0  R  dT T0 R T

  C   P dT S  S0  R  T0 R T T

 However whence

   H   G 0 0 S 0  T0

G  G0  H 0 H 0o 1    RT RT0 RT T and

 G  ln K  RT

 T C dT C P P dT  T0 R T0 R T T

 The preceding equation may be reorganized so as to factor K into three terms , each representing a basic contribution to its value : K = K0 K1 K2  K0 represents the equilibrium constant at reference temperature T0 :   G   K 0  





0

RT0 

 H 0  T0 K 1  exp   1 T  RT0 

     

 1 K 2  exp   T

 T C dT  C P P dT  T0 R T0 R T  T

with heat capacities given : 

  1 K 2  exp A ln          





2 2 2  1   1 1 1 D   1  2   1   2      2 BT0   6 CT0 2 2  2 T   0 



13.5 EVALUATION OF EQUILIBRIUM CONSTANTS Example 13.4 Calculate the equilibrium constant for the vapor-phase hydration of ethylene at 145 and at 320°C from data given in App. C. Solution 13.4 First determine values for ΔA, ΔB, ΔC, and ΔD for the reaction: C2H4 (g) + H2O (g) → C2H5OH (g)

The meaning of Δ is indicated by: A = (C2H5 OH) - (C2H4) (H2O). Thus, from the heat-capacity data of Table C. 1: ΔA = 3.518 - 1.424 - 3.470 = - 1.376 ΔB = (20.001 - 14.394 - 1.450) x 10-3 = 4.157 x l0-3 ΔC = (-6.002 + 4.392 - 0.000) x 10-6 = -1.610 x 10-6 ΔD = (-0.000 - 0.000 - 0.121) x 105 = -0.121 x 105

Values of ΔH°298 and ΔG°298 at 298.15 K for the hydration reaction are found from the heat-of-formation and Gibbs-energy-of-formation data from Table C.4: ΔH°298 = -235,100 - 52,510 - (-241,818) = -45,792 J mol-1 ΔG°298 = -168,490 - 68,460 - (-228,572) -8,378 J mol-1 For T = 145 + 273.15 = 418.15 K, values of the integrals in Eq. (13.18) are: IDCPH(298.15,418.15;-1.376,4.157E-3,-1.610E-6,-0.121 E+5) = -23.121 IDCPS(298.15,418.15;-1.376,4.157E-3,-1.610E-6,-0.121 E+5) = -0.06924

Substitution of values into Eq. (13.18) for a reference temperature of 298.15 gives:  G418  8,378  45,792  45,792  23.121     0.06924  1.9356  8.314 298.15  8.314 418.15 418.15 RT

For T = 320 + 273.15 = 593.15 K, IDCPH(298.15,593.15;-1.376,4.157E-3,-1.610E-6,-0.121 E+5) = 22.632 IDCPS(298.15,593.15;-1.376,4.157E-3,-1.610E-6,-0.121 E+5) = 0.01731 Whence,  G593  8,378  45,792  45,792 22.632     0.01731  5.8286  8.314 298.15  8.314 593.15 593.15 RT

Finally, @418.15K: ln K = -1.9356 and K = 1.443 x 10-1 @593.15K: ln K = -5.8286 and K = 2.942x 10-3 Application of Eqs. (13.21), (13.22), and (13.24) provides an alternative solution to this example. By Eq. (13.21),  8,378  H  45,792 0 K 0  exp  29.366   18.473  8.314 298.15  8.314 298.15 RT0

Moreover, With these values, the following results are readily obtained: T/ K

τ

K0

K1

K2

K

298.15

1

29.366

1

1

29.366

418.15

1.4025

29.366

4.985x10-3

0.9860

1.443x10-1

593.15

1.9894

29.366

1.023x10-4

0.9794

2.942x10-3

Clearly, the influence of K1, is far greater than that of K2. This is a typical result, and accounts for the fact that the lines on Fig. 13.2 are nearly linear.

13.6 REELATION OF EQUILIBRIUM CONSTANTS TO COMPOSITION The standard state for gas is the ideal gas-state of the pure gas at the standard-state pressure P˚ of 1 bar. Since for ideal gas : f˚i = P˚ Thus :

fˆi

fi





fˆi

and

P

 fˆi  i  P    

vi

K

where the constant K is a function of temperature only .

For a fixed temperature the composition at equilibrium must change with pressure in such a way that ˆf P  v  K remains constant .





i

i

i

The fugacity is related to the fugacity coefficient :

fˆi  ˆi y i P

Substitution of this equation :

  y ˆ  i i

i

vi

 P     P 

v

K

where v  i vi and P˚ is the standard –state pressure if 1 bar , expressed in the same units used for P .

 For ideal solution : v   y   ii

i

i

 P     P 

v



K

  i for a pure species can be evaluated from a generalized correlation once the equilibrium T and P are specified .  For pressure sufficiently low or temperature sufficiently high, the equilibrium behaves essentially as an ideal gas v   y  i i

i

 P     P 

v

K

For reaction occurring in liquid phase :  fˆi  i  f    i 

Since

vi

K

fˆi   i xi f i

The fugacity ratio expresses :

 fi  fˆi  i xi f i    i xi      fi fi  fi  the fugacities of liquids are weak function of pressure .

For reaction that take place in high pressure

  xi  i 

vi

i



 P  P  K exp   RT

 v V    i

i

i



For low pressure :

v   x   i i

K

i

i

If the equilibrium mixture is an ideal solution , then γi is unity : v   x  i

i

i

K

This simple relation is known as the law of mass action. Since liquids often form nonideal solutions, equation above can be expected in many instances to yield poor result .

For species at low concentration in aqueous solution, Henry’s law can be use :

fˆi  k i mi where ki is a constant which dependent on temperature and mi is molality .

13.7 EQUILIBRIUM CONVERSIONS FOR SINGLE REACTIONS Example 13.5 The water-gas-shift reaction, CO(g) + H2O(g) → CO2(g) + H2(g) is carried out under the different sets of conditions described below. Calculate the fraction of steam reacted in each case. Assume the mixture behaves as an ideal gas. (a)The reactants consist of 1 mol of H2O vapor and 1 mol of CO. The temperature is 1,100 K and the pressure is 1 bar. (b)Same as (a) except that the pressure is 10 bar. (b)Same as (a) except that 2 mol of N2 is included in the reactants.

(d)The reactants are 2 mol of H2O and 1 mol of CO. Other conditions are the same as in (a). (e)The reactants are 1 mol of H2O and 2 mol of CO. Other conditions are the same as in (a). (f) The initial mixture consists of 1 mol of H2O, 1 mol of CO, and 1 mol of CO2. Other conditions are the same as in (a). (g) Same as (a) except that the temperature is 1,650 K.

Solution 13.5 (a) For the given reaction at 1,100 K, 104/T = 9.05, and Fig. 13.2 provides the value. ln K0 or K = 1. For this reaction v   vi  1  1  1  1  0. Since the reaction mixture is an ideal i

gas, Eq. (13.28) applies, and here becomes: y H 2 y CO2  K 1 y CO y H 2O By Eq. (13.5):

y CO

1 e  2

y H 2O

1 e  2

y CO2

e  2

yH2

e  2

Substitution of these values into Eq. (A) gives:  e2 1 2 or εe = 0.5 1   e  Therefore the fraction of the steam that reacts is 0.5. (b) Since v = 0 , the increase in pressure has no effect on the ideal-gas reaction, and εe is still 0.5. (c) The N2 does not take part in the reaction, and serves only as a diluent. It does increase the initial number of moles no from 2 to 4, and the mole fractions are all reduced by a factor of 2. However, Eq. (A) is unchanged and reduces to the same expression as before. Therefore, εe is again 0.5.

(d) In this case the mole fractions at equilibrium are:

y CO

1 e  3

y H 2O

2 e  3

y CO2

e  3

yH2

e  3

and Eq. (A) becomes:

 e2 1 1   e  2   e 

or

εe = 0.667

The fraction of steam that reacts is then 0.667/2 = 0.333

(e)Here the expressions for yCO and yH2O are interchanged , but this leaves the equilibrium equation the same as in (d). Therefore εe = 0.667, and the fraction of steam that reacts is 0.667. (f) In this case Eq. (A) becomes:

 e 1   e 

1   e 

2

1

or

εe = 0.333

The fraction of steam reacted is 0.333.

(g) At 1,650 K, 104/T = 6.06, and from Fig. 13.2, in K = -1.15 or K = 0.316. Therefore Eq. (A) becomes:  e2  0.316 1   e 

or

εe = 0.36

The reaction is exothermic, and conversion decreases with increasing temperature.

13.8 PHASE RULE AND DUHEM’S THEOREM FOR REACTING SYSTEM  This is phase rule for reacting systems. F = 2 – π + N –r where π is number of phases , N number of chemical species and r is number of independent chemical reactions at equilibrium within the system .

13.9 MULTIREACTION EQUILIBRIA  fˆi  i  f    i 

vi , j

 Kj

where j is the reaction index. For gas phase reaction :  fˆi  i  P    

vi , j

 Kj

For the equilibrium mixture is an ideal-gas, v   y  i i

i, j

 P     P 



v j

Kj