I Equilibrium constant change with stoichiometric doubling (Callen)?

[EE18]

Callen asks us (with respect to an ideal gas)

How is the equilibrium constant of a reaction related to that for the same reaction when written with stoichiometric coefficients twice as large? Note this fact with caution!

I had thought to proceed as follow. We have the definition for the singular reaction:
$$\ln K_s(T) = - \sum_j \nu_j \phi_j(T).$$
Now a reaction which is the sum of this reaction with itself (doubled reaction) has $\nu_j \to 2\nu_j$ so that its equilibrium constant obeys, by definition,
$$\ln K_d(T) = - \sum_j 2\nu_j \phi_j(T) = 2\ln K_s(T) \implies K_d = e^2K_s.$$
But when I look online it says the equilibrium constant should square in this case, $K_d = K_s^2$. Can someone point out what I'm doing wrong?

 

[TSny]

$2 \ln x = \ln(x^2)$

 

[EE18]

$2 \ln x = \ln(x^2)$

You are saying to use
$$\ln K_d(T) = 2\ln K_s(T) = \ln K^2_s(T)\implies K_d = K_s^2$$
which makes sense to me. I'm embarrassed to say I don't know what I'm doing wrong by using
$$\exp(\ln K_d(T)) = K_d = \exp(2\ln K_s(T)) = e^2 exp(\ln K_s(T)) = e^2K_s$$
which is different. What elementary algebra is being slipped under on me here?

 

[TSny]

You are saying to use
$$\ln K_d(T) = 2\ln K_s(T) = \ln K^2_s(T)\implies K_d = K_s^2$$
which makes sense to me. I'm embarrassed to say I don't know what I'm doing wrong by using
$$\exp(\ln K_d(T)) = K_d = \exp(2\ln K_s(T)) = e^2 exp(\ln K_s(T)) = e^2K_s$$
which is different. What elementary algebra is being slipped under on me here?

Note that $\exp(2\ln K_s(T)) \neq e^2 \exp(\ln K_s(T))$.

Instead, $\exp(2\ln K_s(T)) = [\exp(\ln K_s(T)]^2$. This follows from $x^{ab} = (x^a)^b$.

 

[EE18]

Note that $\exp(2\ln K_s(T)) \neq e^2 \exp(\ln K_s(T))$.

Instead, $\exp(2\ln K_s(T)) = [\exp(\ln K_s(T)]^2$. This follows from $x^{ab} = (x^a)^b$.

Oof, of course. $\exp(2+\ln K_s(T)) =e^2K_s$ which is of course not what we have here.

My bad, and thanks for the clarification on this silly error.

 

[TSny]

Oof, of course. $\exp(2+\ln K_s(T)) =e^2K_s$ which is of course not what we have here.

Right.