00:01
Hello students welcome here in this question we have a reaction.
00:04
So to clo2 like qs is reacting with 2oh -h minus to give clo2 plus clo3 plus h2.
00:18
Okay so for this reaction we have a data is given.
00:23
So the first question they are asking here we have to write the rate law.
00:30
Expression for this reaction and the second one is we had to calculate the rate constant for this reaction okay we will see we have already data so from this we can calculate the rate count rate value so from r1 rate 1 and rate 2 there are three experiments are given so from this one we can calculate 5 .75 multiplied by 10 power minus 2 is r1 is given and r2 is 2 .30 multiplied by 10 power minus 1 is given and with respect to r1 is k so we want k divided by k so both are same so this will get cancelled in both these two b values are same so this one is so rate for this one k c l o2.
01:32
This one is a and oh minus per b.
01:38
So in the first and second there is only o b and concentration is same and clo minus ion concentration is given as 0 .05 0 .0 divided by 0 .10 0 .10.
01:51
Whole power b, a.
01:56
Okay.
01:57
Then we have this one is 0 .25 and this one is 0 .25 and this one is is 0 .5a okay then we have 1 5 4 this one is 1 divided by 2 power a so this can be 1 divided by 2 whole square so a is equal to 2 from this one okay now from experiment 2 and 3 we can write here this is 1 -00 whole power a b is here the concentration of c l -o2 is same that's why directly we are taking this one so k divided by k both will get cancelled and this one is 0 .050 whole power b this one is so here this one 2 .3 multiplied by 10 power minus 1 divided by 1 .1315 multiplied by 10 power minus 1.
03:08
So here the value is equal to 2 this one is so this one become 2 power b and this one become approximately okay so b is equal to 1 now we know b and a so rate law is equal to power k is given a clo2 concentration power 2 and oh minus ion.
03:39
Okay, so this is the equation for rate law expression...