InterviewSolution
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InterviewSolution
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Rate of a reaction `A + B rarr` Product, is given as a function of different initial concentration of `A` and `B`. `|{:([A] (mol L^(-1)),(B) (mol L^(-1)),"Initial rate" (mol L^(-1) min^(-1)),),(0.01,0.01,0.005,),(0.02,0.01,0.010,),(0.01,0.02,0.005,):}|` Determine the order of the reaction with respect to `A` and with respect to `B`. What is the half life of `A` in the reaction ? |
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Answer» Correct Answer - `t_(1//2) = 1.386 min` `OR` w.r.t. `A = 1` `OR` w.r.t. `B = 0` `|{:([A] (mol L^(-1)),[B] (mol L^(-1)),"Initial rate" (mol L^(-1) min^(-1)),),(0.01,0.01,0.005,),(0.02,0.01,0.010,),(0.01,0.02,0.005,):}|` form careful observation of the given data, it can be conclude that doubling the concentration of `A` alone doubles the rate of the reaction, therefore order with respect to `A` is one. Doubling the concentration of `B` alone does not affect the rate of the reaction, therefore order with respect to `B` is zero. Rate `= k[A][B]^(0) = k[A]` or `k = (0.005)/(0.01) = 0.5 min^(-1)` Half life `= (0.693)/(k) = (0.693)/(0.5) = 1.386 min`Correct Answer - `t_(1//2) = 1.386 min` `OR` w.r.t. `A = 1` `OR` w.r.t. `B = 0` `|{:([A] (mol L^(-1)),[B] (mol L^(-1)),"Initial rate" (mol L^(-1) min^(-1)),),(0.01,0.01,0.005,),(0.02,0.01,0.010,),(0.01,0.02,0.005,):}|` form careful observation of the given data, it can be conclude that doubling the concentration of `A` alone doubles the rate of the reaction, therefore order with respect to `A` is one. Doubling the concentration of `B` alone does not affect the rate of the reaction, therefore order with respect to `B` is zero. Rate `= k[A][B]^(0) = k[A]` or `k = (0.005)/(0.01) = 0.5 min^(-1)` Half life `= (0.693)/(k) = (0.693)/(0.5) = 1.386 min` |
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