# 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 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 at 18:12
• The equilibrium constant is an experimental value as well and must be consistent with the reaction rates. Jul 15 at 19:06
• @Poutnik but the reaction isnt give at equilibrium in this question Jul 15 at 20:32
• It does not matter. Why do you ever consider reaction quotient for the forward reaction rate? Jul 16 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.