00:01
All right, we're told that for the first order reaction, we have the reaction of a to b, and we're told it's first order.
00:10
What that means is that the rate law is r equals k times the concentration of a.
00:16
And of course, the integrated law is the initial concentration of a divided by the final concentration of a must be equal to the rate constant times the time.
00:27
This is the rate law for a first order.
00:30
But what were we told? we were told that the rate, the half -life, the half -life is 466 .6 at 25 degrees celsius.
00:46
Okay, so t half is 466 .6 second at the temperature of, the temperature was 25 degrees celsius, right? which will mean that the rate constant at this point, the half -life for a first order reaction, t half is actually 0 .693 divided by k.
01:13
So the rate constant would be 0 .693 divided by 466 .6.
01:21
And this will give you 1 .49 times 10 to the power of minus three per second.
01:27
That was at that temperature, right? now, when the temperature, we're told that when the temperature is 25 degrees celsius, right? that's 40 degrees celsius.
01:44
The half -life was what? 433 .3 seconds.
01:53
Again, it means that the rate constant at that point would be 0 .693 divided by 433 .3.
02:02
And this would give you 0 .693 divided by 433 .3.
02:08
And we have 1 .60 times 10 to the power of minus three per second.
02:16
All right, so now the question says, calculate the time required for the concentration of a to be caught by a factor of five, right? that is, for it to decrease by five.
02:28
So the first thing we need to do here is apply arrhenius's equation because we need to know, it says, at a temperature of 75 degrees celsius.
02:38
So we need to know what the rate constant would be at 75 degrees celsius.
02:43
And so the first step is to determine what the activation energy of that reaction is.
02:50
And ln of k2 over k1, i'm going to call this k2, i'm going to call that k1, and this is t1.
03:02
In kelvins, that's 298 kelvins.
03:05
And this in kelvins, right? the temperature here, this one, when the temperature was 440 degrees celsius, you know, add 270 to that, so this was 313 kelvins when the half -life was 433 .3 seconds.
03:33
Okay, so k1 equals ea over r, into bracket 1 over t1 minus 1 over t2.
03:47
This is what we have.
03:49
So we need to find what the activation energy is first.
03:53
So ln of k2, which is, we had that 1 .60 times 10 to the power of minus 3, 1 .60 times 10 to the power of minus 3, divided by k1, which was 1 .49, 1 .49 times 10 to the power minus 3, is equal to ea, divided by 8 .314, multiplied by 1 over 298 minus 1 over 313.
04:33
So when you rearrange this, you will have 0 .0712 equals ea over 8 .314.
04:49
And when you work this out in the bracket, you should have 1 .61 times 10 to the power of minus 4.
04:59
So when you cross multiply, your ea would be 0 .0712 times 8 .314, divided by 1 .61 times 10 to the power of minus 4...