InterviewSolution
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InterviewSolution
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In the following reaction , the initial concentration of the reactant and initial rate at 298 K are given `2 A to C + D ` `{:([A]_(0) "mol" L^(-1) ,, "Initial rate in mol" L^(-1) s^(-1)), (0.01 ,, 5. 0 xx 10^(-5)), (0.02 ,, 2xx 10^(-4)):}` The value of rate constant of this reaction at 298 K isA. `0.01 s^(-1)`B. `5 xx 10^(-3) mol L^(-1) s^(-1)`C. `2.0 xx 10^(-2) mol^(-1) L s^(-1)`D. `5 xx 10^(-1) mol^(-1) L s^(-1)` |
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Answer» Correct Answer - d Let the order with respect to A is `alpha` . Rate = k `[A]^(alpha)` `5 xx 10^(-5) = k[0.01]^(alpha) " " … (i)` `2 xx 10^(-4) = k[0.02]^(alpha) " "… (ii)` By dividing eqn. (ii) by (i) , `(2xx 10^(-4))/(5 xx 10^(-5)) = (k[0.02]^(alpha))/(k[0.01]^(alpha)) implies 4 (2)^(alpha)` i.e., `alpha = 2` Substituting the values of `alpha` in eq. (i) , `5 xx 10^(-5) = k[0.1]^(2) implies k = (5 xx 10^(-5))/(10^(-4)) = 5 xx 10^(-1) mol^(-1) L s^(-1)` |
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