 ###### Jonathan Fong

U.C.Berkeley
M.Ed.,San Francisco State Univ.

Jonathan has been teaching since 2000 and currently teaches chemistry at a top-ranked high school in San Francisco.

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# Tips for Figuring Out What a Rate Law Means - Concept

Jonathan Fong ###### Jonathan Fong

U.C.Berkeley
M.Ed.,San Francisco State Univ.

Jonathan has been teaching since 2000 and currently teaches chemistry at a top-ranked high school in San Francisco.

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Here's some tips and tricks for figuring what a rate law means. So say if we have an equation where we have A plus B yields C. We have 2 reactants A and B, and so when we write a general equation for a rate law, we would write; basically rate equals. And basically rate equals this would be your initial rate of your reaction and then K is the rate constant. You would just multiply times the concentration of A to whatever the power is and the power would be the order. Then you can multiply it times the concentration of B to whatever the power it is or whatever order it is.

Let me tell you how the orders work out. Say for example, if we just simplify this and wrote rate equals K times the concentration A to the 0 order or 0 power. This will be 0 order.

What happens is if I increase A or what happens, well anything to the 0 power, would equal 1 so basically this would equal 1. What happens is, rate would equal K in this case. 0 order, no matter how much I increase the concentration: increasing concentration has no effect on reaction rate and that's for 0 order only. That's what the order means.

Say for example if we have rate equals K times basically the concentration A, and this would be, since there is no exponent, this would be first order, or like having an exponent of 1. If I increase the concentration, what's going to happen to the rate? Well it's going to increase by that same amount. Say if A is 2, then the rate would equal 2K because that would just multiply times that. Now keep in mind this is 2 Molar because it would have to be concentration. (I'll make a prime here just to show that it's different) rate equals K times, maybe if the concentration is now 4 Molar what happens? Well then it equals to 4K. By doubling the concentration, we also double the rate. So basically increasing concentration has a directly proportional effect on reaction rate, and that's for first order only.

Let's take a look at one more. Let's take a look at second order. We would write, rate equals K times concentration A squared. What happens is if I start off with one Molar, from my concentration, I would square that and add K. Rate equals K times 1 Molar, so that would equal 1k. Now if I can double the concentration, just like we did before when we went from 2 to 4 Molar, so maybe I'll take rate equals to K times 2 Molar and then the square, because that's the order.

The new rate would equal, not 2k but 4k. And so basically increasing concentration basically instead of just doubling the concentration, it increased it by times l to the to the second power. This was double to the first power. And then this would be like times double to the 0 power. So I have those there.

Increasing concentration has a squared effect on reaction rate, and that's for second order. Third order would obviously be cubing it. Fourth order would be basically to the fourth power etcetera. Hopefully, these rate laws will help you in determining the reaction rate and also figuring out how the concentration of one particular reactant might affect your particular reaction rate. Have a good one.