# Equilibrium of more than two inter reacting species

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.

• Commented Jul 6, 2015 at 20:46

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

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

• 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. Commented Jul 7, 2015 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. Commented Jul 20, 2015 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) en.wikipedia.org/wiki/Steady_state_(chemistry) 2)researchgate.net/post/… Commented Jul 20, 2015 at 18:31
• Commented Jul 20, 2015 at 18:33
• 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. Commented Jul 21, 2015 at 15:34