InterviewSolution
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InterviewSolution
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Find the values of A,B and C in the following table for the reaction `X + Y to Z`. The reaction is of first order w.r.t X and zero order w.r.t. Y. `{:("Exp",[X](mol L^(-1)),[Y](mol L^(-1)),"Initial rate (mol L^(-1) s^(-1))),(1,0.1,0.1,2xx10^(-2)),(2,A,0.2,4 xx 10^(-2)),(3,0.4,0.4,B),(4,C,0.2,2 xx 10^(-2)):}`A. `A = 0.2 mol L^(-1), B = 8 xx 10^(-2) mol L^(-1) s^(-1), C = 0.1 mol L^(-1)`B. `A = 0.4 mol L^(-1), B = 4 xx 10^(-2) mol L^(-1) s^(-1), C = 0.02 mol L^(-1)`C. `A = 0.2 mol L^(-1), B = 2 xx 10^(-2) mol L^(-1) s^(-1), C = 0.4 mol L^(-1)`D. `A = 0.4 mol L^(-1), B = 2 xx 10^(2) mol L^(-1) s^(-1), C = 0.4 mol L^(-1)` |
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Answer» Correct Answer - A Rate `= k [X][Y]^(0)` Rate is independent of the conc. Of Y and it depends only on the conc.of X and it is the first order reaction. From exp. (1), `2 xx 10^(-2) = k (0.1)` From exp, (2) `4 xx 10^(-2) = k(A)` Dividing (ii) and (i) , `(4 xx 10^(-2))/(2 xx 10^(-2)) = (k(A))/(k(0.1)) = (A)/(0.1)` `implies 2 xx 0.1 = A` `implies A = 0.2 mol L^(-1)` From exp. (3), `B = k(0.4)` Dividing (iii) and (i), `(B)/(2 xx 10^(-2)) - (k(0.4))/(k(0.1)) = 4` `implies B = 4 xx 2 xx 10^(2) = 8 xx 10^(-2) mol L^(-1) s^(-1)` From exp. (4), `2 xx 10^(2) = k (C )` dividing (iv) and (i), `(2 xx 10^(-2))/(2 xx 10^(-2)) = (k(C ))/(k(0.1)) = (C )/(0.1)` `implies C = 0.1 mol L^(-1)` |
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