# Is there any relation between reaction quotient and rate constant in the context of order of the reaction?

How many times will the rate of the reaction

$$\ce{2 A + B -> A2B}$$

change if the concentration of substance $$\ce{A}$$ is doubled and that of substance $$\ce{B}$$ is halved?

The given solution takes

$$\mathrm{rate} = k[\ce{A}]^2[\ce{B}]\tag{1}$$

which is absurd to me since the order of the reaction is an experimental value and not necessarily stoichiometric coefficient.

Here is my attempt:

$$K_\mathrm{i} = \frac{[\ce{A_2B}]}{[\ce{A}]^2[\ce{B}]}\tag{2}$$

and doing the same for $$K_\mathrm{f}$$ just replacing $$[\ce{A}]$$ with with $$2[\ce{A}]$$ and $$[\ce{B}]$$ with $$[\ce{B}]/2$$ and then dividing the two equations to get $$2.$$

This just so happens to match up with the answer. Is there any relation and can I use reaction quotients this way?

• Be aware that at equilibrium, forward and backward rates are equal, regardless of the complexity of the overall reaction and particular values of concentrations of involved substances. Jul 15, 2021 at 17:55
• @Poutnik how would affect the order of the reaction though? its an experimental value so I don't see what difference would it make. Jul 15, 2021 at 18:12
• The equilibrium constant is an experimental value as well and must be consistent with the reaction rates. Jul 15, 2021 at 19:06
• @Poutnik but the reaction isnt give at equilibrium in this question Jul 15, 2021 at 20:32
• It does not matter. Why do you ever consider reaction quotient for the forward reaction rate? Jul 16, 2021 at 3:03

1. When you are implying reaction quotient, you should use $$Q$$ instead of $$K$$. As $$Q$$ is defined at any concentrations other than equilibrium.
2. When you are changing $$[\ce{A}]$$ and $$[\ce{B}]$$, $$[\ce{A2B}]$$ may change also, which you have neglected.