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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?

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

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

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