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The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table. |
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Answer» Solution :the rate EQUATION for the reaction is : r`=k[A]^(1)[B]^(0)` i) Comparing EXPERIMENTS I and II. `2.0 xx 10^(-2) = k[0.1]^(1)[0.1]^(0)` `4.0 xx 10^(-2) = k[x]^(1)[0.2]^(0)` Dividing eqn.(ii) by eqn. (i), `(4.0 xx 10^(-2))/(2.0 xx 10^(-2)) = (k[x]^(1)[0.2]^(0))/(k[0.1]^(1)[0.1]^(0)) , 2 = x/0.1 or x = 2xx 0.1= 0.2 M` Thus, concentration of A in experiment II is 0.2 M ii) Comparing experiments II and III. When the concentration of A is made double, the reaction rate will also becomes twice. `therefore` Rate of reaction in experiment III is `8.0 xx 10^(-2)` iii) Comparing experiments I and IV, Since, the reaction rates are the same in both the experiments, the molar concentration of A in experiment IV must be the same as in experiment I. i.e., it must be 0.1 M. |
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