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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?

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    $\begingroup$ 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. $\endgroup$
    – Poutnik
    Jul 15 at 17:55
  • $\begingroup$ @Poutnik how would affect the order of the reaction though? its an experimental value so I don't see what difference would it make. $\endgroup$ Jul 15 at 18:12
  • $\begingroup$ The equilibrium constant is an experimental value as well and must be consistent with the reaction rates. $\endgroup$
    – Poutnik
    Jul 15 at 19:06
  • $\begingroup$ @Poutnik but the reaction isnt give at equilibrium in this question $\endgroup$ Jul 15 at 20:32
  • $\begingroup$ It does not matter. Why do you ever consider reaction quotient for the forward reaction rate? $\endgroup$
    – Poutnik
    Jul 16 at 3:03
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  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.

  3. From the given data, the change in rate of reaction cannot be calculated for the reason you have pointed out in the beginning.

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  • $\begingroup$ @andselisk About the edit: If A and B denote a particular atom then they can be represented upright, if A and B can be any atom then they can be in italic? $\endgroup$
    – Apurvium
    Jul 16 at 15:35
  • $\begingroup$ Symbols for chemical elements and symbols/abbreviations referring to them are always upright. The only exception I can come up with from the top of my head are locants depicted with italicized Roman letters, e.g. N,N-dimethylaniline. $\endgroup$
    – andselisk
    Jul 16 at 15:39

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