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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.

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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
$$K_1 \times K_2 \times K_3~=~1$$

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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

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  • $\begingroup$ 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. $\endgroup$ – Osman Mamun Jul 7 '15 at 16:04
  • $\begingroup$ 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. $\endgroup$ – Chet Miller Jul 20 '15 at 14:40
  • $\begingroup$ 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) en.wikipedia.org/wiki/Steady_state_(chemistry) 2)researchgate.net/post/… $\endgroup$ – Osman Mamun Jul 20 '15 at 18:31
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    $\begingroup$ 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. $\endgroup$ – Chet Miller Jul 21 '15 at 15:34

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