InterviewSolution
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InterviewSolution
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During the kinetic study of the reaction `2A +B rarr C + D` following results were obtained. `{:(, Run[A],[B] i n M,"Initial rate of fo rmation of D in "m s^(-1),),(I,0.1,0.1,6.0 xx 10^(-3),),(II,0.3,0.2,7.2xx10^(-2),),(III,0.3,0.4,2.88xx10^(-1),),(IV,0.4,0.1,2.40xx10^(-2),):}` On the basis of above data which one is correct ?A. `Rate =k[A][B]^(2)`B. `Rate = k[A]^(2)[B]`C. `Rate = k[A][B]`D. `Rate = k[A]^(2)[B]^(2)` |
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Answer» Correct Answer - A From the run `(I)~ and `(IV)` , we observe that when the concentration of `A` is increased `4` times keeping the times. Thus, the reaction is of first order with respect to `A` . From the run `(II)` and `(III)` , we motice that when the concentration od `B` is doubbled, the rate is increased four times Thus this reaction is of second order with respect to `B` i.e. Rate `=K[A][B]^(2)` Alternatively Rate `=k[A]^(x)[B]^(y)` Considering the run `(I)` and `(IV)` , we have `(2.40xx10^(-2))/(6.0xx10^(-3))=((0.4)^(x)(0.1)^(y))/((0.1)^(x)(1.0)^(y))` `4=4^(x)` implies `4^(1)=4^(x)` or `x=1` Considering the run `(II)` and `(III)` , we have `(2.88xx10^(-1))/(7.2xx10^(-2))=((0.3)^(x)(0.4)^(y))/((0.3)^(x)(0.2)^(y))` `4=2^(y)` `2^(2)=2^(y)` Hence, `y=2` |
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