Exploring Equilibrium Constants & Kinetics: A Closer Look

In summary, the conversation discusses the concept of equilibrium constants and the ability to define them using different coefficients. The participants also touch on the issue of uniqueness and how multiplying through the stoichiometric equation by a constant can affect the equilibrium constant. The conversation also brings up a similar issue in kinetics and the use of rate laws. Ultimately, it is a matter of convention and as long as calculations are done consistently, the final results should be the same.
  • #1
PFuser1232
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Consider the following generic equilibrium:

aM + bN cO + dP

An equilibrium constant, K, can be defined as:

$$K = \frac{[O]^c [P]^d}{[M]^a [N]^b}$$

But couldn't we also define another equilibrium constant similarly with coefficients that are in the same ratio as our original equation? For instance, α = 2a, β = 2b, and so on. We can then do the following:

$$K' = \frac{[O]^γ [P]^δ}{[M]^α [N]^β}$$

Clearly, those are two conflicting results.

How exactly do we define a unique equilibrium constant and simultaneously grant ourselves the liberty of multiplying through the stoichiometric equation by some constant (which, in this case, was 2)?
Where exactly am I going wrong?

A similar conundrum arises in kinetics. Consider a similar reaction to the one I wrote above, except that now it is a one-way reaction rather than a reversible one. I will use the letter ##v## to denote rate.

$$v = -\frac{1}{a} \frac{d[M]}{dt}$$

To make the problem less abstract, let's consider a real reaction. The bromination of ethene (ethylene).

$$C_2 H_4 + Br_2 → C_2 H_4 Br_2$$

If we consider the equation in that form, then a = 1 in the above equation for rate. If we multiply both sides of the equation by 2 (and I don't see why such an action would be erroneous), then a = 2. Shouldn't the value of a in our definition of rate (not unlike our definition of equilibrium constant) be unique?

Could someone please clarify this for me?
 
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  • #2
I already told you it is a matter of convention. Using other than the lowest integers set of coefficients yields a different value of the equilibrium constant - but as it is defined differently, that's to be expected. What is important is the fact, that these systems are equivalent, and when you use them consistently for calculations, final result should be the same.

[tex]K = \frac {A^aB^b}{C^cD^d}[/tex]

[tex]\log K = a \log A + b \log B - c \log C - d \log D[/tex]

Multiply by any constant you want, equation still holds.
 
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  • #3
With regard to the equilibrium constant, remember that the equilibrium constant of a reaction is related to the change in free energy associated with the reaciton: K = exp(-ΔG/RT)

Clearly, the free energy change associated with the reaction of two moles of a substance will be greater than the free energy change associated with the reaction of one mole of a substance, so the equilibrium constant must be different.
 
  • #4
Borek said:
I already told you it is a matter of convention. Using other than the lowest integers set of coefficients yields a different value of the equilibrium constant - but as it is defined differently, that's to be expected. What is important is the fact, that these systems are equivalent, and when you use them consistently for calculations, final result should be the same.

[tex]K = \frac {A^aB^b}{C^cD^d}[/tex]

[tex]\log K = a \log A + b \log B - c \log C - d \log D[/tex]

Multiply by any constant you want, equation still holds.

Thanks!
I think you've mistaken me for someone else, though. I don't recall ever discussing this topic on PF. o_O
 

1. What is equilibrium in chemistry?

Equilibrium in chemistry refers to the state in which the forward and reverse reactions of a chemical reaction occur at equal rates, resulting in a constant concentration of reactants and products.

2. How is equilibrium constant (K) calculated?

The equilibrium constant is calculated by dividing the concentration of products by the concentration of reactants, with each concentration raised to its respective stoichiometric coefficient in the balanced chemical equation.

3. What factors can affect equilibrium constants?

Temperature, pressure, and the concentrations of reactants and products can all affect the value of the equilibrium constant. Changes in these factors can shift the position of the equilibrium and change the value of K.

4. What is the relationship between kinetics and equilibrium?

Kinetics refers to the study of the rates of chemical reactions, while equilibrium refers to the state in which the rates of the forward and reverse reactions are equal. The kinetics of a reaction can influence the position of the equilibrium and the value of the equilibrium constant.

5. How does Le Chatelier's principle relate to equilibrium?

Le Chatelier's principle states that when a stress is applied to a system at equilibrium, the system will shift in the direction that relieves the stress. This can affect the position of the equilibrium and the value of the equilibrium constant.

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