1.

A first order reaction was started with a decimolar solution of the reactant , 8 minutes and 20 seconds later its concentration was found to be M/100 . So the rate of the reaction is

Answer»

`2 . 303 xx 10^(-5) sec^(-1)`
`2 . 303 xx 10^(-4) sec^(-1)`
`4 . 606 xx 10^(-3) sec^(-1)`
`2 . 606 xx 10^(-5) sec^(-1)`

Solution :For first order REACTION `K = (2.303)/(t)` log `(a)/(a-x)`
GIVEN : `a = (1)/(10) = 0.1` m , a -x = `(1)/(100) = 0.01` m , t = 500 sec
`thereforek = (2.303)/(500) "log" (0.10)/(0.01)= (2.303)/(500) ` log 10
= `(2.303)/(500) = 0.004606 = 4.6 xx 10^(-3) sec^(-1)`.


Discussion

No Comment Found

Related InterviewSolutions