Rutger's University
M.Ed., Columbia Teachers College
Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.
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Rutger's University
M.Ed., Columbia Teachers College
Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.
Tips on finding the units of K instantly. The K I'm speaking off is the K in Kinetics. It's the one in the Differential Rate Law, that you typically have to find. I have the basic back bone of the rate law here, and this is the k that I'm referring to. Not capital K in equilibrium, but lower case K in Kinetics.
In order to find the units of K, we have to understand the rest to the equation. We want to make sure all the units cross out so we get the units we need to have. Actually let's go back and I'll show you the first order, rate law. What that means remember is that the exponents add up to 1.
So in this case, rate=k[A] (concentration of A) and we can assume that the exponent is 1. If we were to look at the units for everything, rate is Molarity over some unit of time, equals K. And that's what we're trying to find, times concentration. And concentration is also Molarity.
So we want to figure out the units for K. I'm going to isolate K and divide by Molarity on both sides and that these are going to cross out. and also these are going to cross out. K in the first order of reaction is 1/time and time is typically in seconds, or in minutes. Depending on, get to actually look at your data to see what time is, in this particular problem.
For a first order reaction, this is going to be the units for k, 1/time.
For our second order reaction, second order rate law, I'm going to say rate, the exponents add up to 2. I'm going to make it simple on myself and rate equals k[A]². My rate again is Molarity over some unit of time equals K times Molarity, and this time Molarity is squared. I isolate K, so I'm going to divide the Molarity squared. And that means, this is going to cross out with one of these. So K is going to be equal 1 over Molarity times time.
Now don't forget this can also be viewed, since Molarity is Moles over a Litre, this can also be, k equals Litre over Mole times time, because Molarity is in the bottom I can flip this so that this for a second order, create the equation.
Now there is a pattern to this. I'll just tell you the pattern straight out. The pattern is K is going to equal 1 over Molarity. The exponent of Molarity is going to be the order minus 1 times time. Let's look back and see if that works.
For our first order reaction rate, k equals 1 over molarity, 1 minus 1 times time. This is going to be Molarity to the 0 power, which is 1, which you can eliminate, 1 over time.
Let's do for our second order rate equation. k is equal to 1 over Molarity, 2 minus 1 times time. 2 minus 1 is 1. So it's 1 over molarity times time, the second order which checks off.
For our third order rate equation, k equals 1 over Molarity to the third minus 1 times time 1 over a Molarity squared times time. Which also can be Litres squared over Moles squared times time equals K, for third order.
I know your teacher is going to ask you about units of K. So this right here will tell you the units of K for any order reaction rate law that you have. This can actually give it to you instantly rather than having to actually fiddle with units and possibly make a mistake. This 1 over Molarity to the order minus 1, is the exponent times whatever time this is. You can't just write time, I want you to write seconds or minutes. You will always get the correct unit for K, which will give you some extra points on your test. Hopefully this helps.
Unit
Kinetic-Molecular Theory