I have a few questions regarding rate and rate constant. I am wondering what effect certain concentrations, temperature, etc., has on rate and rate constant. Would be of great help if you can clarify this for me!

If a concentration was halved (and all other factors remained the same), does that mean rate would decrease? If instead the concentration was doubled, I would assume that it would also double. For rate, if the temperature of the solution decreased, does that mean that it would also decrease since it means it would take less time?

For rate constant, if the concentration was halved, I think there would be no effect (because concentration effects rate only). But, what if the concentration was doubled, does that also mean that there's no effect or would the rate constant decrease? For rate constant, let's say the ionic strength of the solution increased. Does that mean that the rate constant decreases?

Thanks for clarifying!


1 Answer 1


Assuming you are speaking about an elementary reaction, and by that I mean a reaction on the form $\ce{A -> B}$ the rate of the reaction is

$$r = \frac{\mathrm d[\ce{B}]}{\mathrm dt} = -\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{A}]$$

Most of the time $k$ can be express following the Arrhenius law:

$$k = A\exp\left(\dfrac{-E_\mathrm{a}}{k_\mathrm{B}T}\right)$$

Where $A$ is the pre-exponential factor (the constant rate when them temperature is really high), and $E_\mathrm{a}$ is the activation energy of the reaction.

In this case the initial concentration has no affect on the rate constant, only the temperature. Looking at the first equation the initial concentration has effect on the rate, this effect will depend on the order of the reaction…

$$r = k[\ce{A}]^2$$

$$r = k[\ce{A}]^3$$

Here a second and third-order reaction. If you study a complex mechanism, it can happen that initial concentration will affect the global rate constant, it can also happen with some approximation:

$$\ce{A + B -> C} \qquad r = k[\ce{A}][\ce{B}]$$

If $\ce{A},$ for example, is in really large quantity compared to $\ce{B},$ you can admit that for every time the concentration will approximately stay the same $[\ce{A}] = [\ce{A}]_0$ and then put it in the constant

$$k' = k[\ce{A}]_0$$

I hope that helped! If not, you need to study more.

  • $\begingroup$ A is (collision frequency) times (orientation factor), shouldn't collision frequency increase with increase in concentration. Or is it a constant, perhaps calculated when temperature is high? $\endgroup$ Jun 2, 2022 at 3:48

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