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?


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.
$K_{c,2}=\frac{4F_{A0}X_{eq}^2}{\vartheta (1-X_{eq})}=\frac{4F_{A0}}{\vartheta}4K_{c,1}$

  • 1
    $\begingroup$ A and B aren't meant to be the same things in the two reactions. $\endgroup$
    – orthocresol
    Dec 17 '21 at 14:26
  • $\begingroup$ If so, then the calculations seem correct. $\endgroup$
    – razivo
    Dec 17 '21 at 15:41

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