- #1
PFuser1232
- 479
- 20
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?
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?
Last edited: