00:01
Hello students, in the first question we are given that it is a first order reaction and a is getting converted to b.
00:08
And we are given that 46 % of the reaction gets completed in 68 minutes.
00:16
So we need to find a value of rate constant key.
00:20
So we know that the value of p in first order reaction is given as 1 divided by rate constant k, ln of if we are given in terms of percentage, so 100 divided by 100 minus x, where x is the amount of reaction complete or the percentage of reaction completed.
00:42
So if we substitute the values here, we will get k is equals to 1 divided by t, which is given as 60a, multiplied by ln of 100 divided by 100 minus 46.
00:57
And from here if we solve this, we will get rate constant which is k is equal to 1 .14 multiplied by 10 power minus 2 minute inverse.
01:16
And this is given in option b.
01:21
So option b will be the correct answer for the first question.
01:27
Now coming to second question, we are given a second order reaction where a is getting converted to products and we are given that initially we had one molar a and after 5 .82 hours, 50 % of the reaction got completed.
01:51
So amount of a remaining will be 0 .5 motor, that is half of the reaction get got completed...