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For the reaction :2A+BtoA_(2)B the rate =k[A][B]^(2) with k=2.0xx10^(-6) mol^(-2) L^(2)s^(-1) Calculate the initial rate of the reaction when [A]=0.1 mol L^(-1),[B] =0.2 mol L^(-1).Calculate the rate of reaction after [A] is reduced to 0.06 mol L^(-1) |
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Answer» SOLUTION :Given reaction :`2A+BtoA_(2)B` Rate EQUATION r=`k[A][B]^(2)` Where k=`2.0xx10^(-6) mol^(-2) L^(-2)s^(-1)` (i)Initial rate =`r_(1)` at initial [A]=0.1 mol `L^(-1)` [B]=0.2 mol `L^(-1)` `THEREFORE r_(1)(2.0xx10^(-6) mol^(-2) L^(2) s^(-1))` `(0.1 mol L^(-1))(0.2 mol L^(-1))^(2)` `=8xx10^(-9) mol L^(-1)s^(-1)` ..........(i) (ii)When [A]=0.06 mol `L^(-1)` than rate `r_(2)`: Where ,`r_(2)=k[A][B]^(2)` Reaction :`2A+BtoA_(2)B` Initial concentration 0.1M0.2M CHANGE in reaction -2x-x Concentration at NEW 0.06M(0.2-x)M equilibirum =(0.1-2x)M=0.18 M Decrease in concentration of A 0.06=(0.1-2x)M `therefore` (2x)=(0.1-0.06)M `therefore` 2x=0.04M `therefore` x=0.02M `therefore r_(2)=k[A][B]^(2)` `=3.888xx10^(-9) mol L^(-1)s^(-1)` |
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