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We know that equilibrium in a chemical system is attained when forward and backward reaction rates are equal. What if the reaction mixtures involve more than one reaction?

For example, consider three inter-reacting species A, B and C. The rate constants for the forward and backward reactions of $\ce{A -> B}$ are $k_1$ and $k'_1$, that of $\ce{B -> C}$ are $k_2$ and $k'_2$ and that of $\ce{C -> A}$ are $k_3$ and $k'_3$. What is the relation between the six rate constants so that the system remains in equilibrium? Assume first order kinetics for all the reactions. Thanks in advance.

pentavalentcarbon
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Sagnik
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  • See also http://chemistry.stackexchange.com/questions/15511/can-a-multi-species-system-oscillate-around-equilibrium – Curt F. Jul 06 '15 at 20:46

2 Answers2

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At equilibrium $~r_1~=r_2~=r_3~=~0$.
So, $$k_1[A]-k_{-1}[B]=0$$
$$k_2[B]-k_{-2}[C]=0$$
$$k_3[C]-k_{-3}[A]=0$$
So, The criteria for equilibrium is
1) $$k_1[A]=k_{-1}[B]$$
2) $$k_2[B]=k_{-2}[C]$$
3) $$k_3[C]=k_{-3}[A]$$
Now defining equilibrium constant as $K_i=\frac{k_i}{k_{-i}}$
We can derive a formula
$$K_1 \times K_2 \times K_3~=~1$$

Osman Mamun
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The previous answer was not done correctly.

These are the balances on the three species:

$$\frac{dA}{dt}=-(k_1+k_{-3})A+k_{-1}B+k_3C$$ $$\frac{dB}{dt}=k_1A-(k_{-1}+k_2)B+k_{-2}C$$ $$\frac{C}{dt}=k_{-3}A+k_2B-(k_3+k_{-2})C$$

If the system is at equilibrium, the three time derivatives must be equal to zero. So one must have: $$-(k_1+k_{-3})A+k_{-1}B+k_3C=0$$ $$k_1A-(k_{-1}+k_2)B+k_{-2}C=0$$ $$k_{-3}A+k_2B-(k_3+k_{-2})C=0$$

These relationships represent 3 simultaneous homogeneous linear algebraic equations in three unknowns, A, B, and C. The condition for these equations to be satisfied for all values of A, B, and C is that the determinant must be equal zero.

Chet

Chet Miller
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  • Your mathematical explanation is for steady state solution not for equilibrium. It's more applicable for flug flow reactor catalytic system. At equilibrium, you must have equal forward and reverse rate. – Osman Mamun Jul 07 '15 at 16:04
  • mamun comment is incorrect. At equilibrium of a system involving multiple species and reactions, each individual reaction does not have equal forward and reverse rate. The condition for equilibrium of such a system is that the free energy of the reaction mixture is minimum. That condition is achieved when the relationships presented in my previous post are satisfied. – Chet Miller Jul 20 '15 at 14:40
  • That's a common mistakes to mix up the concept of equilibrium and steady state. I would suggest you to please read the following link: 1) https://en.wikipedia.org/wiki/Steady_state_(chemistry) 2)https://www.researchgate.net/post/Can_anyone_explain_briefly_the_difference_between_the_equilibrium_and_steady_state_system_in_chemistry – Osman Mamun Jul 20 '15 at 18:31
  • 3)http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Chemical_Equilibria/Principles_of_Chemical_Equilibria/Principles_of_Chemical_Equilibrium/Characteristics_Of_The_Equilibrium_State 4) https://en.wikipedia.org/wiki/Chemical_equilibrium -Thanks – Osman Mamun Jul 20 '15 at 18:33
  • In a batch reactor, how do you determine when a system at steady state is not at equilibrium, given that the concentrations of all the species are not changing with time? If the system is not at equilibrium, then it should be moving spontaneously toward equilibrium, and the concentrations should be changing. – Chet Miller Jul 21 '15 at 02:39
  • Ah man! Equilibrium concept is only applicable for closed system. Now if you use a close batch reactor with no input and output, the system will slowly approach equilibrium and the net rate will reach zero. But you don't want that because that's not economical. To disturb the equilibrium, you are withdrawing the product and giving more reactant input. So, the system is at steady state and the driving force towards equilibrium is the highest. – Osman Mamun Jul 21 '15 at 02:47
  • In Chemical Engineering we study thermodynamics to know the boundary and then we study heat and mass transfer to utilize those boundaries as much as possible. - Thanks! – Osman Mamun Jul 21 '15 at 02:47
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    Hi Mamun, I just realized that your 3 equations are a linear combination of my three equations, and visa versa, so we must both be correct. I'm sure that if I evaluate the determinant of my 3 homogeneous linear algebraic equations, it will lead to your final result. So I unjustifiably criticized you in my original post. Sorry about that. Incidentally, I am a chemical engineer also (1967 PhD) with over 50 years of experience, so I fully understand what you are saying. – Chet Miller Jul 21 '15 at 15:34