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.




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  

$$K\_1 \times K\_2 \times K\_3~=~1$$


 2

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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