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In the following reaction, the initial c...

In the following reaction, the initial concentrations of the reactant and initial rates at 298 K are given
`2ArarrC+D`
`{:([A]_(0)", mol "L^(-1),"Initial rate in mol "L^(-1)s^(-1)),(0.01,5.0xx10^(-5)),(0.02,2.0xx10^(-4)):}`
The value of rate constant of this reaction at 298 K is

A

`0.01s^(-1)`

B

`5xx10^(-3)molL^(-1)s^(-1)`

C

`2.0xx10^(-2)mol^(-1)Ls^(-1)`

D

`5xx10^(-1)mol^(-1)Ls^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
Let the order of reaction with respect to A is n.
`therefore` Rate law is given as,
rate, `r=k[A_(0)]^(n)…(i)`
On putting the given values in Eq. (i), we get
`5.0xx10^(-5)=k[0.01]^(n)...(ii)`
`2.0xx10^(-4)=k[0.02]^(n)...(iii)`
Dividing Eqs. (iii) by (ii) gives,
`[(2.0xx10^(-4))/(5.0xx10^(-5))]=[(0.02)/(0.01)]^(n)`
`[(20)/(5)]=(2)^(n)`
or `(4)=(2)^(n)`
`(2)^(2)=(2)^(n)`
`therefore n=2`
On putting the value of n in Eq. (i), we get
rate, `r=k[A_(0)]^(2)`
`therefore 5xx10^(-5)=k(0.01)^(2)`
`k=(5xx10^(-5))/(10^(-4))=5xx10^(-1)mol^(-1)Ls^(-1)`
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