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Ch.14 - Chemical Kinetics

Chapter 14, Problem 42

Consider the first-order reaction AS B in which A molecules (red spheres) are converted to B molecules (blue spheres).

(a) Given the pictures at t = 0 min and t = 1 min, draw pictures that show the number of A and B molecules present at t = 2 min and t = 3 min.

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Video transcript

Hello Everyone in this video is trying to determine the number of X and Y as molecules present at 20 to 60 seconds. So when time is at 60 seconds and when time is at 90 seconds were given these two images below this is when nothing happens though, t goes 20 seconds and this is what happens after 30 seconds. So we're being told in this problem that we have our grace fears being our X molecules and then these red spheres being ry molecules, ross being told that this reaction is a first order reaction. So we can see that half of these X molecules are converted into why molecules in seconds. So since the reaction is first order, the X molecules will be halved every 30 seconds. So what we can see in this diagram here is that we currently at seconds have eight X and its Y molecules. So then when our X molecules are at 60 seconds time, this means again we're going to go ahead and have this. So eight divided by two as we go to four. So we have four X molecules. And then for why we originally have these eight in 30 seconds. But then we have these eight already for our X. So those remaining four will be added to the eight that we have in our seconds. So eight plus four is going to be equal to 12. And now for our x molecules given the time, 90 seconds we're gonna go ahead and have this. So now we have to And then for the y molecules at 90 seconds again, we're just gonna go ahead and add those two that we have to hear onto here. So now 12 plus two is a go to 14. So when T goes to 60 seconds, we have four X molecules and 12 molecules. And then when the time is equal to 90 seconds, when when T is equal to 90 seconds are X molecules, we all have to and then why molecules will have 14? And this right here is going to be my final answer for this problem.