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## Chemistry library

### Course: Chemistry library>Unit 17

Lesson 1: Reaction rates and rate laws

# Units of the rate constant

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?
• Very keen observation. Yep, that formula will work out for the units of k as the order changes.
• Would the rate constant for third order reactions be 1/M^2 *s?
• Yes, the units of the rate constant are 1/(M²s), although they are more frequently written as M⁻²s⁻¹ or L²mol⁻²s⁻¹.
• Have someone heard about the Arrhenius plot? Can I find it somewhere in the videos here?
• what exactly does # order of reaction mean?
• 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.
• 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"?
• 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!
• What is the general formula for finding the unit of k?
• There is no general formula for the constant K you have to find it based on your units.
• 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
• 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

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?
• 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).