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For a first-order reaction A rarr Produ...

For a first-order reaction A `rarr` Products, the concentration of A changes from 0.1 M to 0.025 M in 40 min. The rate of the reaction when the concentration of A is 0.01 M is

A

`1.73 xx 10^(-5) M "min"^(-1)`

B

`3.47 xx 10^(-4) M "min"^(-1)`

C

`3.47 xx 10^(-5) M "min"^(-1)`

D

`1.73 xx 10^(-4) M "min"^(-1)`

Text Solution

Verified by Experts

Using the expression of rate constant for first -order reaction, we have `k = (2.303)/(t) log""([A]_(0))/([A])`
Given that t = 40 min, `[A]_(0) = 0.1 M` and [A] = 0.025 M.
Therefore, `k = (2.303)/(40)log""(0.1)/(0.025) = 0.03465 "min"^(-1)`
Now, rate = k[A] (for first-order reaction) `= 0.03465 xx 0.01 = 3.47 xx 10^(-4) M "min"^(-1)`
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