An Overview of Chemical Reaction Analysis - Bioprocess Engineering: Kinetics, Biosystems, Sustainability and Reactor Design - page 107

We can now return to our reaction system and examine the situation near chemical
equilibrium. Assuming the simple power-law rate expressions and for a reversible reaction
we have

k f Y

N S

k b Y

N S

C O f R j
j

C O b R j
j

0 ¼ r ¼ r f r b ¼ (3.79)

j ¼1 j ¼1

because the rate is zero at chemical equilibrium. This is consistent with the arguments based
on nonequilibrium thermodynamics, Eqn (3.52) , as well. Rearranging this equation, we
obtain

Q N S
j

C O bRj
j

Y

N S

k f
k b ¼

¼1

C O bRj O fRj
j

¼ (3.80)

Q N S
j

C O fRj
j

j ¼1

¼1

Since we just noted that

! C j

C j

! n j

Y

N S

G R
RT

D

K eq ¼ exp ¼ (3.81)

j ¼1

We can immediately identify terms in these equations,

C j n j

K eq Y

N S

k f
k b

¼ (3.82)

j ¼1

and

n j ¼ O bRj O fRj (3.83)

at equilibrium.
From the preceding equations it can be seen that the rate coefficients and the equilibrium
constant are related. Recall from thermodynamics that

DG R ¼ DH R TDS R

(3.84)

0
R

0
R

where DH is the standard state enthalpy change and DS is the standard state entropy

0
R

0
R

change in the reaction. Both DH and DS are only weakly dependent on temperature. We

can therefore write

K eq ¼ exp DG R = RT ¼ exp DS R = R exp DH R = RT

(3.85)

Noting that Arrhenius law can be applied to the reaction rate constants:

E f

E b RT

C j n j C j n j

Y

N S

Y

N S

k f
k b

k f0
k b0

¼ exp (3.86)

j ¼1 j ¼1

Therefore, we can identify

E b ¼ DH R E f (3.87)