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The rate of a reaction increases four-fo...

The rate of a reaction increases four-fold when the concentration of reactant is increased `16` times. If the rate of reaction is `4 xx 10^(-6) mol L^(-1) s^(-1)` when the concentration of the reactant is `4 xx 10^(-4) mol L^(-1)`. The rate constant of the reaction will be

A

`2 xx 10^(-4) mol^(1 // 2)L^(-1 // 2)s^(-1)`

B

`1 xx 10^(-2) s^(-1)`

C

`2 xx 10^(-4) mol^(-1 // 2)L^(1 // 2)s^(-1)`

D

`25 mol^(-1) L min^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`"Rate" prop sqrt("Concentration")= k sqrt("Concentration")`
`k= ("Rate ")/("(Concentration)"^(1 // 2))`
`=(4 xx 10^(-6))/((4 xx 10^(-4))^(1 // 2))=(4 xx 10^(-6))/(2 xx 10^(-2))`
`=2 xx 10^(-4) mol^(1 // 2) L^(-1 // 2) s^(-1)`
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