InterviewSolution
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InterviewSolution
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For the non-stoichiometric reaction `2A+BrarrC+D` The following kinetic data were obtained in theee separate experiment, all at `98 K` `|{:("Initial concentration (A)","Initial concentration (B)","Initial rate of formation of C" (molL^(-1) s^(-1))),(0.01 M,0.1 M,1.2 xx 10^(-3)),(0.1 M,0.2 M,1.2 xx 10^(-3)),(0.2 M,0.1 M,2.4 xx 10^(-3)):}|` The rate law for the formation of `C` is:A. `(dc)/(dt)=K[A][B]^(2)`B. `(dc)/(dt)=k[A]`C. `(dc)/(dt)=k[A][B]`D. `(dc)/(dt)=k[A]^(2)[B]` |
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Answer» Correct Answer - B (I) `R = k[A]^(x)[B]^(y)` `1.2 xx 10^(-3) = k[0.1]^(x)[0.1]^(y)` …(i) `1.2 xx 10^(-3) = k[0.1]^(x)[0.2]^(y)` …(ii) Divide Eq. (ii) by Eq. (i) `1 = (2)^(y)` `:. (2)^(0) = (2)^(y) :. y = 0` (II) `1.2 xx 10^(-3) = k[0.1]^(x)[0.1]^(y)` `= k[0.1]^(x)` , `(because y = 0)` ...(iii) `2.4 xx 10^(-3) = k[0.2]^(x)[0.1]^(0)` ...(iv) Divide Eq. (iv) by Eq. (iii) `(2)^(1) = (2)^(x) :. x = 1`. `R = k[A]` Hence answer (b) is correct.Correct Answer - B (I) `R = k[A]^(x)[B]^(y)` `1.2 xx 10^(-3) = k[0.1]^(x)[0.1]^(y)` …(i) `1.2 xx 10^(-3) = k[0.1]^(x)[0.2]^(y)` …(ii) Divide Eq. (ii) by Eq. (i) `1 = (2)^(y)` `:. (2)^(0) = (2)^(y) :. y = 0` (II) `1.2 xx 10^(-3) = k[0.1]^(x)[0.1]^(y)` `= k[0.1]^(x)` , `(because y = 0)` ...(iii) `2.4 xx 10^(-3) = k[0.2]^(x)[0.1]^(0)` ...(iv) Divide Eq. (iv) by Eq. (iii) `(2)^(1) = (2)^(x) :. x = 1`. `R = k[A]` Hence answer (b) is correct. |
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