00:01
We're given the weights of the composition of phosphine, p .s3, we equals to k times the phosphine concentration to power of 1.
00:10
So we have a first order reaction.
00:13
So for we have started with one motor, it would take 120 seconds to get to 0 .25 motor.
00:22
So what would be the time required for 2 motor to get to decrease to 0 .35? motor.
00:31
Okay, first of all, because we have the first other reaction, we are looking at the concentration and time.
00:37
So we're going to use the integrated radar again.
00:40
So the integrator will be the phosphine concentration, let alone of that at any time, equals to the original phosphine concentration, minus kt.
00:55
Okay, minus kt.
00:57
Okay, so if we want to find the time for the second scenario, so we just need to find out the weight constant.
01:09
Okay, so for the weight constant, if you actually plug in number for the scenario, one, so the time will be 0 .250, and then initial will be 1, and then we're minus k, and it will be 120.
01:28
Okay, so we all know that lateral lot of 1 is equal to 0.
01:35
So we can just remove this number over here.
01:40
So we're minus k times 120 equals to a letter lot of 0 .25...