# How can equilibrium constant be derived by irreversible reaction?

I am solving the example from the book [Essentials of Chemical Reaction Engineering, Chapter 3, P3-12B] and finding out a contradictive point.

"The rate law for the A->B reaction was obtained at low temperatures. The reaction is highly exothermic and, therefore, reversible at high temperatures. Suggest a rate law for the reaction at high temperature."

The reaction A->B is irreversible at low temperatures and the rate law is $$-r_A=k C_A^2$$

The answer given in the book is:

However, if I look back at the definition of the equilibrium constant

The answer should be $$K_c=\frac{C_B}{C_A}$$.

If I use the irreversible rxn to derive the answer, it will be $$K_c=\frac{C_B^2}{C_A^2}$$

I am confused by two different answers.

• Only elementary reactions satisfy the condition that the order of a species is equal to its molar coefficient: $\alpha=a$; $\beta=b$. If the reaction is not elementary, $\alpha\neq a$; $\beta\neq b$. Mar 29 at 0:58
• Note that using photos/screenshots of text instead of typing text itself is highly discouraged. The image text content cannot be indexed nor searched for, nor can be reused in answers. Specifically handwritten scripts can be difficult to decipher. Consider copy/pasting or rewriting of essential parts. // Optional: Formatting guides for texts and formulas/equations/expressions. Mar 29 at 2:45

No matter what the kinetics, the equilibrium constant for the reaction $$\ce{A <=> B}$$ is $$K = \frac{[\mathrm{B}]}{[\mathrm{A}]}$$
On the other hand, the equilibrium constant for the reaction $$\ce{2A <=> 2B}$$ is $$K = \frac{[\mathrm{B}]^2}{[\mathrm{A}]^2}$$