The following pictures represent the progress of a reaction in which two A molecules combine to give a more complex molecule A2, 2 AS A2.

(b) What is the rate law?

Video transcript

Hi everyone. This problem reads the image below shows the reaction of to be molecules that combine to produce a complex molecule. B. two determine the rate law of the reaction. And were given three images here. T equals zero minutes, two equals one minute, and t equals two minutes. So the first thing that we're going to need to do is recall what the different orders are for this. Okay. And so one of these could be one option. Could be zero order. So if we have zero order, what that means is the rate of conversion of B is constant. If this is a first order, what this means is half of the B molecules per minute are transformed to be two molecules so one half of B is transformed to be too per minute. And if this is a second order, what this means is B molecule concentration is inversely proportional to half life. Okay, so it's going to be inversely proportional. Two half life. Alright, so let's take a look at our images. Okay, so at T equals zero minutes. So at T equals zero minutes. Our concentration of B is equal to At T equals or excuse me, it's equal to At T equals one minute concentration of B Is equal to 12. And at T equals two minutes. The concentration of B is equal to six. So here we can see that the half life is inversely proportional to the concentration of the molecules. So the rate of the reaction is second order. Okay, so we have a second order reaction. So what that means is as a second order reaction. When we write this rate, we're going to have rate is equal to K times the concentration of B squared. That is going to be the rate law for this reaction, and that is it for this problem. I hope this was helpful.