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For the reaction A+B rarr C. The followi...

For the reaction `A+B rarr C`. The following data were obtained. In the first experiment, when the initial concentrations of both `A` and `B` are `0.1 M`, the observed initial rate of formation of `C` is `1xx10^(-4) mol litre^(-1) minute^(-1)`. In the second experiment when the initial concentrations of `A` and `B` are `0.1 M` and `0.3 M`, the initial rate is `9.0xx10^(-4) mol liter^(-1) "minute"^(-1)`. In the third experiment when the initial concentrations of both `A` and `B` are `0.3M` the initial rate is `2.7xx10^(-3) mol litre^(-1) "minute"^(-1)`.
(a) Write rate law for this reaction.
(b) Calculate the value of specific rate constant for this reaction.

Text Solution

Verified by Experts

Let Rate `= K[A]^(x) [B]^(y)`
`r_(1) = 1 xx 10^(-4) = K ( 0 .1)^(x) ( 0.1) ^(y)` …..(1)
`r_(2) = 9 xx 10^(-4) = K ( 0.1) ^(x) ( 0.3)^(y)` ......(2)
`r_(3) = 2.7 xx 10^(-3) = K(0.3)^(x) ( 0.3) ^(y) ` .....(3)
By equations ( 1) and ( 2) `( r_(1))/( r_(2)) = ( 1 xx 10^(-4))/( 9 xx 10^(-4)) = ((1)/( 3))^(gamma) :. y = 1 `
By equation ( 2) and ( 3)` ( r_(2))/( r_(3)) = ( 9 xx 10^(-4))/( 27 xx 10^(-4)) = ((1)/( 3))^(x) :. x = 1 `
`:. ` Rate `= K [ A] ^(1) [B]^(2)`
Also `1 xx 10^(-4) = K ( 0.1)^(1) ( 0.1)^(2)`
`:. K = 10^(-1) = 0.1 L t^(2) "mol"^(-1) "min"^(-1)`
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