For the reaction `A+B to ` product
`[A]=[B]=1 ` M the rate of reaction is `2xx10^(-2 )`mol `dm^(-3) sec^(-1)` , then the rate of reaction at the time of 90 % of each of the reactants are converted into product is `

Options

A. `1.8xx10^(-2) mol dm^(-3) sec^(-1)`
B. `2xx10^(-4)mol dm^(-3)sec^(-1)`
C. `1.62xx10^(-4)mol dm^(-3) sec^(-1)`
D. `4.05xx10^(-1) mol dm ^(-3) sec^(-1)`

Answer

B

Explanation

when `[A]=[B]=1M`
Rate =`r_(1)=k[A][B]`
`2xx10^(-2)=K[1][1]`
when `[A]=[B]=0.1M`
Rate `=R_(2)=k[0.1][0.1]`
dividing I by II
`(2xx10^(-2))/(r_(2))=([1][1])/([0.1][0.1])`
`therefore r_(2) =2xx10^(-4)mol dm^(-3) sec^(-1)`