# How to relate equilibrium constant and equilibrium conversion?

For the following gas phase reaction: $$A\leftrightarrow B$$ The concentration equilibrium constant ($$K_c$$) can be expressed as: $$K_c = \frac{c_B}{c_A} = \frac{\frac{F_B}{\vartheta}}{\frac{F_A}{\vartheta}} = \frac{F_{A0}X_{eq}}{F_{A0}(1-X_{eq})} = \frac{X_{eq}}{1-X_{eq}}$$ Where $$c_i$$ is the concentration and $$F_i$$ is the molar flow rate of component i ($$F_{A0}$$ is the initial molar flow rate of A, assuming no B is present initially), $$\vartheta$$ is the volumetric flow rate, and $$X_{eq}$$ is the equilibrium conversion of the reaction.

Following the same methodology for the following gas-phase reaction: $$A \leftrightarrow 2B$$ $$K_c = \frac{c_B^2}{c_A} = \frac{(\frac{F_B}{\vartheta})^2}{\frac{F_A}{\vartheta}} = \frac{(2F_{A0}X_{eq})^2}{\vartheta F_{A0}(1-X_{eq})} = \frac{4F_{A0}X_{eq}^2}{\vartheta (1-X_{eq})}$$ Would this be correct?

## 1 Answer

Well, the second reaction is most likely not chemically possible, but if it was, you'd be correct.
One nice thing is you can phrase the second using the first.
$$K_{c,2}=\frac{4F_{A0}X_{eq}^2}{\vartheta (1-X_{eq})}=\frac{4F_{A0}}{\vartheta}4K_{c,1}$$

• A and B aren't meant to be the same things in the two reactions. Commented Dec 17, 2021 at 14:26
• If so, then the calculations seem correct. Commented Dec 17, 2021 at 15:41