In the reaction `aArarrbB` , when the concentration of A is `2.2xx10^(-3)M` the rate is `2.4xx10^(-3)M.s^(-1)` and when the concentration of A is halved , the rate becomes `0.6xx10^(-3)M.s^(-1)` . For what concentration of A will the rate be `1.8xx10^(-3)M.s^(-1)` ?

Explanation

If the order of the reaction be n , then
rate (r) = `k[A]^(n)` [k=rate constant]
when `[A]=2.2xx10^(-3)(M), "rate"=2.4xx10^(-3)M.s^(-1)`
Therefore , `2.4xx10^(-3)=k(2.2xx10^(-3))^(n) " "….[1]`
When `[A]=(1)/(2)xx2.2xx10^(-3)=1.1xx10^(-3)(M)`,
rate `=0.6xx10^(-3)M.s^(-1)`
Therefore , `0.6xx10^(-3)=k(1.1xx10^(-3))^(n)" "...[2]`
Dividing equation [1] by [2] , we get
`(2.4xx10^(-3))/(0.6xx10^(-3))=((2.2xx10^(-3))/(1.1xx10^(-3)))^(n)" or, "2^(n)=4=2^(2) " or, " n=2`
Substituting n = 2 into the equation (1), we get,
`2.4xx10^(-3)=k(2.2xx10^(-3))^(2)" or, " k = 4.95xx10^(2)M^(-1).s^(-1)`
Suppose , when the concentration of A is x (M) , the rate is `1.8xx10^(-3)M.s^(-1)` . This means,
`1.8xx10^(-3)M.s^(-1)=4.95xx10^(2)M^(-1).s^(-1)xx(x)^(2)`
or , ` x^(2)=(1.8xx10^(3))/(4.95xx10^(2)) "or" , x = 1.9xx10^(-3)`
`therefore ` Therefore , the concentration of A is `1.9xx10^(-3)(M)` , then the rate reaction will `1.8xx10^(-3)M^(-1).s^(-1)`