00:02
In this question, decompotion reaction given of n2 .05.
00:08
So the reaction is 4 no2 plus o2.
00:14
So here the given statements and values or k rate constant values which is 4 .8 into 10 to the power minus 4 per second at 45 degrees celsius and the initial concentration of n2 .5 we can denoted as concentration of n2 .5.
00:33
Which is equal to 2 .12 into 10 to the power minus 2 mole per liter.
00:39
Then next we need to find the concentration of n2 05 after after 5 4 second.
00:51
So this reaction follows the first order reaction.
00:55
So the equation which is k is equal to 2 .303 divided by t log of initial concentration which is n205 divided by concentration of n2 .05.
01:11
Then we can substitute these values into this formula, rate constant which is 4 .8 into 10 to the power minus 4, which is equal to 2 .303 divided by the time is 500 .04 log of initial concentration which is 2 .12 into 10 to the power minus 2 .12.
01:33
And we need to find the concentration of n2 .05.
01:37
So here, if we divided these values 0 .0, 456, and this is we can say log of 2 .1 to 10 to the power minus 2, developed by concentration of n205, and this is 4 .8 into 10 to the power minus 4.
01:53
And the next step will be, this one we can write it as 4 .8 into 10 to the power minus 4, and this one 4 .5 into 10 to the power minus 2.
02:05
Then here log of 2 .12 into 10 to the power minus 2 divided by concentration of n to 05.
02:15
And here 10 to the power minus 2 minus 2 cancel each other.
02:18
So 10 to the power minus 2 will be remaining and 4 .8 divided by 4 .5 which is nothing but 0 .0106.
02:27
Then log 2 .12 into 10 to the power minus 2 divided by concentration of n2 o5.
02:36
And the next step, we can inverse this value, which is concentration of n2 05 divided by 2 .12 into 10 to the power minus 2 because we need concentration of n2 o5.
02:51
So that's why we inverse...