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
Messages
112
Reaction score
13
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?
 
Last edited:
Science news on Phys.org
##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.
 
I need to calculate the amount of water condensed from a DX cooling coil per hour given the size of the expansion coil (the total condensing surface area), the incoming air temperature, the amount of air flow from the fan, the BTU capacity of the compressor and the incoming air humidity. There are lots of condenser calculators around but they all need the air flow and incoming and outgoing humidity and then give a total volume of condensed water but I need more than that. The size of the...
Thread 'Why work is PdV and not (P+dP)dV in an isothermal process?'
Let's say we have a cylinder of volume V1 with a frictionless movable piston and some gas trapped inside with pressure P1 and temperature T1. On top of the piston lay some small pebbles that add weight and essentially create the pressure P1. Also the system is inside a reservoir of water that keeps its temperature constant at T1. The system is in equilibrium at V1, P1, T1. Now let's say i put another very small pebble on top of the piston (0,00001kg) and after some seconds the system...
Back
Top