# Equilibrium constant change with stoichiometric doubling (Callen)?

• I
• EE18
In summary: So, in summary, the equilibrium constant for a reaction which is the sum of this reaction with itself (doubled reaction) is 2 times the equilibrium constant for the same reaction when written with stoichiometric coefficients twice as large.
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?

Last edited:
##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.

## What is the equilibrium constant in the context of stoichiometric doubling?

The equilibrium constant (K) is a dimensionless number that expresses the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction. When the stoichiometry of the reaction is doubled, the equilibrium constant changes because it depends on the stoichiometric coefficients of the balanced chemical equation.

## How does the equilibrium constant change when the stoichiometric coefficients are doubled?

When the stoichiometric coefficients of a balanced chemical equation are doubled, the equilibrium constant is squared. For example, if the original reaction is A + B ⇌ C with an equilibrium constant K, then the doubled reaction 2A + 2B ⇌ 2C will have an equilibrium constant K^2.

## Why does doubling the stoichiometric coefficients affect the equilibrium constant?

Doubling the stoichiometric coefficients affects the equilibrium constant because the equilibrium expression depends on the concentrations of reactants and products raised to the power of their coefficients. When the coefficients are doubled, the exponents in the equilibrium expression are also doubled, leading to a change in the value of the equilibrium constant.

## Can you provide an example to illustrate how the equilibrium constant changes with stoichiometric doubling?

Consider the reaction N2 + 3H2 ⇌ 2NH3 with an equilibrium constant K. If we double the stoichiometric coefficients, we get 2N2 + 6H2 ⇌ 4NH3. The equilibrium constant for this new reaction will be K^2 because the exponents in the equilibrium expression are doubled.

## Is the change in the equilibrium constant due to stoichiometric doubling a common topic in thermodynamics?

Yes, the change in the equilibrium constant due to stoichiometric doubling is a common topic in thermodynamics and chemical kinetics. It is important for understanding how changes in the stoichiometry of a reaction can affect the position of equilibrium and the concentrations of reactants and products at equilibrium.

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