I Equilibrium constant change with stoichiometric doubling (Callen)?

AI Thread Summary
The discussion centers on the relationship between the equilibrium constants of a reaction and its doubled stoichiometric coefficients. Callen initially derives that the equilibrium constant for the doubled reaction should be expressed as K_d = e^2K_s, but this contradicts the established relationship K_d = K_s^2. The confusion arises from a misunderstanding of logarithmic properties, specifically that exp(2ln K_s) equals (exp(ln K_s))^2, not e^2 * exp(ln K_s). This clarification resolves the algebraic error, affirming that K_d = K_s^2 is indeed correct. The exchange highlights the importance of accurately applying logarithmic identities in chemical equilibrium calculations.
EE18
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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?
 
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##2 \ln x = \ln(x^2)##
 
TSny said:
##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?
 
EE18 said:
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##.
 
TSny said:
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.
 
EE18 said:
Oof, of course. ##\exp(2+\ln K_s(T)) =e^2K_s## which is of course not what we have here.
Right.
 
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