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## Chemistry library

### Course: Chemistry library > Unit 17

Lesson 1: Reaction rates and rate laws# Units of the rate constant

AP.Chem:

TRA‑3 (EU)

, TRA‑3.B (LO)

, TRA‑3.B.4 (EK)

The units of the rate constant,

*k*, depend on the overall reaction order. The units of*k*for a zero-order reaction are*M*/s, the units of*k*for a first-order reaction are 1/s, and the units of*k*for a second-order reaction are 1/(*M*·s). Created by Yuki Jung.## Want to join the conversation?

- It seems that there is a general formula for this.

For n th order, the unit of k will be M/((M^n)*s), right?(11 votes)- Very keen observation. Yep, that formula will work out for the units of k as the order changes.(9 votes)

- Would the rate constant for third order reactions be 1/M^2 *s?(6 votes)
- Yes, the units of the rate constant are 1/(M²s), although they are more frequently written as M⁻²s⁻¹ or L²mol⁻²s⁻¹.(14 votes)

- Have someone heard about the Arrhenius plot? Can I find it somewhere in the videos here?(2 votes)
- Have you tried searching for "Arrhenius".

An Arrhenius plot is based on the Arrhenius equation, which you can find several videos for on this site.

For example:

https://www.khanacademy.org/science/chemistry/chem-kinetics/arrhenius-equation/v/forms-of-the-arrhenius-equation

You may also find this wikipedia page helpful:

https://en.wikipedia.org/wiki/Arrhenius_plot

Does that help?(7 votes)

- What if the concentration is increased 3 times and the rate of reaction is increased 6 times? In this case, do we need to take 3 to the power of 1.63504 to get to 6 or do we need simplify by dividing 3 and 6 by 3 and turn it to a first order reaction? If the first solution is correct (i.e. taking 3 to the power of 1.63094), what would be the order of this reaction and what would be the unit of "k"?(3 votes)
- The rate law usually gets rounded in those cases - probably a rate of 2 in this case.

Rate law isn't exact but it helps when you're trying to figure how long a reaction will take!(3 votes)

- what exactly does # order of reaction mean?(2 votes)
- It's just the sum of the exponents of the chemical species in the rate law (usually the reactants). The order matters because the equations and graphs for the integrated rate law and half life are different depending on the overall order of the reaction. Hope that helps.(4 votes)

- What is the general formula for finding the unit of k?(2 votes)
- There is no general formula for the constant K you have to find it based on your units.(4 votes)

- How to predict the order of reaction and rate law eq when a rraction take place in two steps eg A+B gives E annd second step E+A gives C where first eq is reversible with const k1 & k1- and second reaction const k2 .. ( overall reaction is 2A+B gives C(3 votes)
- A reaction that has more than 1 step generally has a slow step (The step that is the slowest)

So when we want to derive the rate law of a multi-step reaction, we usually consider only the slow step (Since the slowest step is most likely to affect the rate of the reaction as a whole)

For example, consider a multi-step reaction :-

A + B → C + D

Step 1 (Slow Step):- A + A → C + E (Rate constant, K1 )

Step 2 (Fast Step) :- E + B → A + D (Rate constant, K2 )

Here E is an intermediate, the product in step 1 and a reactant in step 2 that does not show up in the overall reaction. This is because when steps 1 and 2 are added, intermediate E cancels out, along with the extra reactant A from step 1 (The extra reactant was added to make sure that when we add the 2 equations, we get the original equation).

So the rate law is

Rate = K1 [A] [A] = K1 [A] ^ 2

I found this answer at:-

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/Rate_Laws/Reaction_Mechanisms/Rate-Determining_Step(1 vote)

- What is the equation for pseudo third order rate constant and can you provide few solved examples for the same?(3 votes)
- what would be the solution for fractional-order reaction?(1 vote)
- Let the reaction be

aA +bB --> Products,

And the rate the equation be,

R = k * [A]^x * [B]^y

Then by comparing units,

M/s = units(k) * M^x * M^y,

or, M/s = units(k) * M^(x+y)

or, M^(1-(x+y))/s = units(k) <--(1)

If x and/or y are/is fractional, then x+y shall be evaluated (may be a fraction) and the units of k can be found by (1).(2 votes)

- Aren't constants and rates always the same and/or similiar? And is the constant k, kelvin?(1 vote)
- Not necessarily, the rate of a reaction represents how quickly a reaction can proceed (so how fast does it go to create the product) while the constant K (which is determined experimentally) takes into account some factors that make a reaction happen that we wouldn't necessarily be able to calculate (like the form of the molecules, how they act at a certain temperature, etc..). It is important to take into account that the constant K is unique to 1 specific reaction at 1 specific temperature, and will be different if you change the temperature or have a different reaction.

Also, the constant K is not in kelvin since it is not a temperature, but it does have units (sorry I confused k with the equilibrium constant).(2 votes)

## Video transcript

- [Voiceover] In this video,
we're going to be talking about how you can find the units
for your rate constant k. So the two things you should
know before we get started are that, one, rate constant k has units. So this isn't always true
of constants in chemistry, but it is true of k. The second thing to remember is that your rate constant, the units of k depend on your rate law. And so we're going to
use this second point to use the rate law to
derive the units of k. And this is really
handy because that means you don't have to memorize
what the units of k are for different orders of reactions. So we're going to focus on the
three most common rate laws that you see in chemistry class. So we're going to talk about zeroth, first, and second order reactions. And we will derive their units. So first, let's look at zeroth order. So zeroth order reactions have a rate law that look like this. So the rate is equal to k times the concentration of your reactant A to the zeroth power. And anything to the
zeroth power is just one. So our rate is equal
to the rate constant k. The units of rate are
always going to be the same. So the units of rate are
always molar per second and you can also just think
of units almost like numbers. If you have an equal sign,
the units on both sides of your equal sign have to be the same and they have to match. So here, since we have rate equal to k, that means k must also have
units of molars per second. So this tell us that the units
for a zeroth order reaction are molar per second. We can use that same idea
to figure out the units of k for first and second order reactions, too. So for a first order reaction, so for first order, a first order reaction rate law is rate is equal to our rate constant k times the concentration of our reactant raised to the first power. Units of rate are molar per second, and the units of concentration are always going to be molar. So now we know that the
units of k times molar equals molar per second. So we have molar on both sides, so we don't have to worry about that, but we're missing a
one over a second term. So that tells us that the units
of k are one over seconds. The other way that we can try
to figure out the units here, if you're not comfortable
with back-calculating what the units are, is we can actually rearrange this rate law, So if we just put k on one side and everything else on the other side, we get that k is equal to rate divided by the concentration of A. So all I did was divide both sides here by the concentration of A. And since we know that the units on both sides of the equal
sign have to be the same, then we can figure out the units of k by dividing the units of rate by the units of our concentration. So that's just molar per
second, for the rate, divided by molar, for the concentration. And then the molar cancels out, and we're left with one over seconds. So that's an even more
straight-forward way to find the units of k. But the idea is the same. You can treat units the
same way you treat numbers, and you just have to make sure they match on both sides of your equal sign. The last example we're going to go through is going to be for second order reactions. So second order reactions, or second order rate laws have the form rate is equal to our rate constant k times the concentration of our reactant to the second power. So on one side, we have
molar per second for the rate and on the other side, now, since our concentration is squared, we have molar squared. So molar squared times something is equal to molar per second. We need to add a one over seconds in our units for k because
we need to make sure when we multiple these, we
get the seconds on the bottom, and we need to cancel out one
of these concentration terms, so we need to put molar in
the denominator as well. So the units of k for
a second order reaction are one over molar, molar-seconds. so these are the three
most common molecularities that you might see in a chemistry class. And sometimes, you have reactions that aren't zeroth,
first, or second order, and whenever that happens, you can always use the rate law to find the units of the rate constant k.