00:01
In this question, we want to look into the disoption of butane, which is c4h10.
00:07
Now looking at this, for the first part of the question, we want to determine the half -life.
00:11
And t -half, we know this to be 0 .693, divided by the right constant of the disruption process or the reaction.
00:20
And this is going to be equal to 0 .693 divided by 0 .128.
00:26
And this gives us a t half, the half -life being equal to 5 .414 per second.
00:33
Now moving on to the next part, we know that this is a first order reaction.
00:41
So, lean concentration of butane at a time t divided by the initial concentration of butane, this is going to be equal to negative kt.
00:57
So this is the rate low for a first order reaction.
01:02
Now that we have this information, we are saying for time t, this is going to be limit the concentration of c4h10 at time t divided by the concentration, the initial concentration of our b -10.
01:23
And we have to divide all of this by negative k -t.
01:30
And this is going to be equal to lynn 0 .75 divided by, rather this is just k divided by negative 0 .128.
01:45
So our time, this is going to be equal to 2 .2 and this is in second.
01:50
Now moving on at 7, at 50 % disruption, we're going to have this is a similar expression.
01:57
We're just using this expression.
01:58
Let's just use this expression in all cases so at 50 % this option we're going to have our t being equal to lynn 0 .5 divided by negative k t which is negative 0 .1 to 8 so our t here is going to be equal to 5 .4 seconds now moving on to the next part we want to determine the friction that remains and the fraction that remains is the concentration of c4 h10 at a certain time divided by the initial concentration of c4 h10...