Ratio of rate constants

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.
6

663697

While reviewing my FL I noticed there was a question i was completely stumped on. It gave the dissociation constant of a reaction and the rate constant of the forward reaction, and asked to find the rate constant of the reverse reaction.

The solution says that this can be easily solved using Keq = k(forward)/k(reverse). I don't recall ever learning this formula, and it wasn't in my TPR book either. Can somebody please explain how to derive this formula? I've searched other sources but couldn't find anything.

Members don't see this ad.
 
Consider a reversible chemical reaction expressed as the following: x X + y Y --> c C + d D, where X, Y, C, D are compounds/elements in the reaction and x, y, c, d are stoichiometric coefficients to keep the reaction balanced. Now assume this reaction is in equilibrium.

By definition, the equilibrium constant = K = [products] / [reactants], so for the above reaction, it's K = ([C]^c * [D]^d) / ([X]^x * [Y]^y)

Now a chemical equilibrium means the rate of the forward reaction is equal to the rate of the reverse reaction. Assume that the reaction above involves only a single-step mechanism. This assumption is necessary because in multi-step reaction mechanisms, the rate laws depend on the slowest/rate-limiting step. Having only a single step allows us to treat the reaction as elementary and rate-limiting, thereby allowing stoichiometric coefficients to be used as rate law exponents.

With this assumption, we can see the following:

Rate forward = k(forward) * [reactants] = k(forward) * [X]^x * [Y]^y
Rate reverse = k(reverse) * [products] = k(reverse) * [C]^c * [D]^d

Because we are dealing with a chemical equilibrium, rate forward = rate reverse or:

k(forward) * [X]^x * [Y]^y = = k(reverse) * [C]^c * [D]^d -->
k(forward) / k(reverse) = ([C]^c * [D]^d) / ([X]^x * [Y]^y)

More generally:

Rate forward = Rate reverse -->
k(forward) * [reactants] = k(reverse) * [products] -->
k(forward) / k(reverse) = [products] / [reactants]

Since the equilibrium constant K = [products] / [reactants], we find that K = [products] / [reactants] = k(forward) / k(reverse).
 
  • Like
Reactions: 1 users
Consider a reversible chemical reaction expressed as the following: x X + y Y --> c C + d D, where X, Y, C, D are compounds/elements in the reaction and x, y, c, d are stoichiometric coefficients to keep the reaction balanced. Now assume this reaction is in equilibrium.

By definition, the equilibrium constant = K = [products] / [reactants], so for the above reaction, it's K = ([C]^c * [D]^d) / ([X]^x * [Y]^y)

Now a chemical equilibrium means the rate of the forward reaction is equal to the rate of the reverse reaction. Assume that the reaction above involves only a single-step mechanism. This assumption is necessary because in multi-step reaction mechanisms, the rate laws depend on the slowest/rate-limiting step. Having only a single step allows us to treat the reaction as elementary and rate-limiting, thereby allowing stoichiometric coefficients to be used as rate law exponents.

With this assumption, we can see the following:

Rate forward = k(forward) * [reactants] = k(forward) * [X]^x * [Y]^y
Rate reverse = k(reverse) * [products] = k(reverse) * [C]^c * [D]^d

Because we are dealing with a chemical equilibrium, rate forward = rate reverse or:

k(forward) * [X]^x * [Y]^y = = k(reverse) * [C]^c * [D]^d -->
k(forward) / k(reverse) = ([C]^c * [D]^d) / ([X]^x * [Y]^y)

More generally:

Rate forward = Rate reverse -->
k(forward) * [reactants] = k(reverse) * [products] -->
k(forward) / k(reverse) = [products] / [reactants]

Since the equilibrium constant K = [products] / [reactants], we find that K = [products] / [reactants] = k(forward) / k(reverse).

This totally makes sense. Is this a commonly used formula? Just wondering as I hadn't come across it before.
 
This totally makes sense. Is this a commonly used formula? Just wondering as I hadn't come across it before.

I think I saw it in EK Chemistry manual. It's a useful shortcut but I don't recall seeing it often on practice tests. I usually just stick with K = [products] / [reactants].
 
  • Like
Reactions: 1 user
Consider a reversible chemical reaction expressed as the following: x X + y Y --> c C + d D, where X, Y, C, D are compounds/elements in the reaction and x, y, c, d are stoichiometric coefficients to keep the reaction balanced. Now assume this reaction is in equilibrium.

By definition, the equilibrium constant = K = [products] / [reactants], so for the above reaction, it's K = ([C]^c * [D]^d) / ([X]^x * [Y]^y)

Now a chemical equilibrium means the rate of the forward reaction is equal to the rate of the reverse reaction. Assume that the reaction above involves only a single-step mechanism. This assumption is necessary because in multi-step reaction mechanisms, the rate laws depend on the slowest/rate-limiting step. Having only a single step allows us to treat the reaction as elementary and rate-limiting, thereby allowing stoichiometric coefficients to be used as rate law exponents.

With this assumption, we can see the following:

Rate forward = k(forward) * [reactants] = k(forward) * [X]^x * [Y]^y
Rate reverse = k(reverse) * [products] = k(reverse) * [C]^c * [D]^d

Because we are dealing with a chemical equilibrium, rate forward = rate reverse or:

k(forward) * [X]^x * [Y]^y = = k(reverse) * [C]^c * [D]^d -->
k(forward) / k(reverse) = ([C]^c * [D]^d) / ([X]^x * [Y]^y)

More generally:

Rate forward = Rate reverse -->
k(forward) * [reactants] = k(reverse) * [products] -->
k(forward) / k(reverse) = [products] / [reactants]

Since the equilibrium constant K = [products] / [reactants], we find that K = [products] / [reactants] = k(forward) / k(reverse).

Excellent physical derivation!
 
Top