00:01
Hey guys, so in this question, we're given a reaction where a is decomposing into b and c, and we're also given some time and concentration data, and we're asked to find the order of the reaction as well as its rate constant, and the rate of formation of b at t equals 30 minutes.
00:21
So what i have here is what i like to use to find the order of a reaction like this, given the given data.
00:31
So here i have three different scenarios where my blue line represents the concentration graph with versus time in minutes.
00:40
Then i have my red graph, which is the natural log of my concentration versus time.
00:46
And then i have the inverse of my concentration over or with respect to time.
00:52
And these all represent zero first and second order.
00:56
So when we use this method of graphing, we want to make sure that the the graph is as most linear as possible when comparing the graphs to determine the order.
01:09
So i use that.
01:11
To do that, i make trend lines for each of them and compare the r squared values.
01:16
The r squared values just tell me how linear each of these are.
01:21
And you can see we want to have an r square value of as close to one as possible.
01:28
And what we can see here is that our square value for the natural log is 1.
01:36
So that means this line is perfectly linear as compared to the other two orders.
01:44
So that means since the natural log of the concentration is linear, that means this reaction must be first order.
01:56
Now another way to find the orders would be to calculate the k values for each of these scenarios, for each concentration, and when we compare the k values, we want to make sure each k value is constant in the order that we choose.
02:15
And if we were to calculate the k values, we would have found that in the first order, that the k values would have been constant for that one.
02:25
Now moving on to part b, it asks us to find the k constant.
02:30
So i'm going to use my integrated rate law for a first order reaction.
02:35
So that is the natural log of my concentration at some time t is equal to the natural log of my initial concentration minus kt.
02:45
I'm going to go ahead and solve for k.
02:48
So k is going to be equal 1 over t times the natural log of a sub t or a.
02:57
Initial, excuse me, over my concentration at some time t.
03:03
I got this by first adding this over to the left side and subtracting this natural log expression over to the right.
03:17
And remember when i subtract a natural log, where there is a natural log on the same side, i can use my log rules to form a division expression.
03:29
That i have this, i can go ahead and plug in my values that i have.
03:32
So, k is going to be equal to 1 over 8 minutes times the natural log of my concentration value...