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In a reaction , `A + B rarr` Product, rate is doubled when the concentration of `B` is doubled, and rate increases by a factor of `8` when the concentration of both the reactants (A and B) are doubled, rate law for the reaction can be written asA. `Rate = k[A][B]^(2)`B. `Rate = k[A]^(2)[B]^(2)`C. `Rate = k[A][B]`D. `Rate = k[A]^(2)[B]` |
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Answer» Correct Answer - D Let the rate law be `Rate=k[A]^(x)[B]^(y)` Let the initial concentrations `A` and `B` be `a` and `b` respectively. According to first observation `R_(1)k[a]^(x)[b]^(y)` `R_(2)=2R_(1)=k[a]^(x)[2b]^(y)` Dividing the `2^(nd)` equation by `1^(st)` , we have `2=2^(y)` or `2^(1)=2^(2)` Hence, `y=1` According to the second observation ltbr. `R_*(1)=k[a]^(x)[b]^(y)` `R_(2)8R_(1)=k[2a]^(x)[2b]^(y)` Dividing the `2^(nd)` equation by `1^(st)` , we get `8=2^(x).2^(y)` `2^(3)=2^(x+y)` or `x+y=3` Substituting the value of `y=1` , we have `x+1=3` or `x=3-1=2` Thus, the correct rate law is `Rate=K[A]^(2)[B]` |
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