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In many textbooks, it is written that:

The value of rate constant depends on the nature of the reactants, temperature and catalyst. It is independent of the concentration of the reactants.

However, the unit of the rate constant is

$$\left(\frac{\mathrm{mol}}{\mathrm{l}}\right)^{1-n} \mathrm{s}^{-1}$$

where $n$ is the order of reaction.

Therefore, except for first order reactions, the unit of the rate constant will have a concentration term $\displaystyle\frac{\textrm{mol}}{\textrm{l}}$ for all other reactions. Does this contradict the statement given in books?

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  • $\begingroup$ So what if it will have a concentration term? $\endgroup$
    – Ivan Neretin
    Sep 7, 2019 at 9:29
  • $\begingroup$ Think of it like the equilibrium coefficient, the units don't realllly matter. Still use them to check your units and stuff. $\endgroup$
    – Eashaan Godbole
    Sep 7, 2019 at 12:35
  • $\begingroup$ @EashaanGodbole I think units matter. However, you can't tell from the units what a quantity depends on. Let's say I drive with a constant speed of 100 km/h on the freeway. Just because the units have the length-unit km in them does not mean my speed depends on how far I've already driven (or how long, for that matter). The idea of constant speed is that the ratio of distance and time is constant, even as time passes and you move along. $\endgroup$
    – Karsten ♦
    Sep 9, 2019 at 2:33
  • $\begingroup$ @Ivan Neretin So, rate constant does not depend on the concentration of reactants for all type of reactions including pseudo first order reactions? $\endgroup$
    – Apurvium
    Sep 9, 2019 at 4:12
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    $\begingroup$ @KarstenTheis I wasn't saying units don't matter in general--Units are very important and non-neglectable for most cases. I'm just saying they may not hold as much importance in this specific scenario. $\endgroup$
    – Eashaan Godbole
    Sep 10, 2019 at 13:28

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As an example if the reaction is $\ce{A + B -> C}$ the rate expression for the appearance of $\ce{C}$ is

$$\frac{\mathrm d[\ce{C}]}{\mathrm dt} = +k[\ce{A}][\ce{B}].$$

The units have to be the same on both sides of this equation, these are concentration/time on the left and to be the same on the right the rate constant $k$ has to have units $(1/\text{concentration})(1/\text{time})$ usually expressed as $\mathrm{dm^3\,mol^{-1}\,s^{-1}}.$

It is true that the numerical value of the rate constant usually depends of several things such as temperature, but the units remain the same for each type of rate constant, 0th, 1st, 2nd, 3rd simply because the rate equation units have to balance.

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    $\begingroup$ No, units does not remain the same for each type of reaction. Here you have assumed the order of the reaction w.r.t. both 'A' and 'B' as 1. So net order n= 1+1= 2. $\endgroup$
    – Apurvium
    Sep 9, 2019 at 4:08
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    $\begingroup$ @Apurvium You have probably mis-understood, I wrote that units have to be the same on both sides the the equation because the equation has to balance. By type of rate constant it is clear from what I wrote that all first order have the same units etc. etc. $\endgroup$
    – porphyrin
    Sep 9, 2019 at 17:57
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    $\begingroup$ No, all reactions of 1 st order have the same units for the rate constant and similarly for other orders, which is what 'same for each type of rate constant' means. $\endgroup$
    – porphyrin
    Sep 10, 2019 at 12:56
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    $\begingroup$ No, you can see from the equation I give that $k$ is a constant independent of any concentration. The same is true for all other rate expressions excluding the pseudo rate constant that is sometimes used in special cases. $\endgroup$
    – porphyrin
    Sep 10, 2019 at 16:44
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    $\begingroup$ yes and only because one concentration is vastly in excess of the other and does not change by any realistic amount while the reaction occurs. $\endgroup$
    – porphyrin
    Sep 11, 2019 at 10:54

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