00:01
All right, so we're told that this reaction is second order.
00:04
So if we want concentration as a function of time, we can use the integrated rate law, which for second order is going to be one over the concentration at any time is equal to one over the initial concentration, plus the rate constant times the half -life t1 over 2, or times the time, sorry.
00:29
If we wanted to solve for k, we can use the half -life.
00:33
Would be the t1 over 2.
00:35
We know that 1 over k times the initial concentration is equal to the half -life t 1 over 2.
00:45
So to solve for k, just rearranging this equation, we can basically just flip the positions of k in the half -life.
00:51
So k is equal to 1 over the half -life times the initial concentration.
00:57
And we're told, let's see, the half -life is 4 .23 times 10 to the second.
01:02
So 423 seconds.
01:06
And the initial concentration is .136 molar.
01:14
So 423, so 1 divided by 423 divided by 0 .13, comes out to 0 .174, and that's inverse seconds times inverse molar, which is consistent with the units for a second order rate constant.
01:35
Now, if we want the concentration to decrease by 76 .9%, then a concentration is going to be a 76 .9 % decrease means that if we subtract that from 100, we'll have 23 .1 % left.
01:57
So it's going to be 0 .231 times the initial concentration.
02:03
So we know this, we know this, and we know this, and then we can solve for the time...
