For the reaction
`2A+B +C rarr A_(2)B+C`
the rate law has been found to be
Rate = k[A] `[B]^(2) ` with `k = 2.0xx10^(-6) mol^(-2) L^(2)s^(-1)` .
The initial rate of reaction with
[A] =0.1 mol `L^(-1)`
[B] = 0.2 mol `L^(-1)` and
[C] = 0.8 mol `L^(-1) s^(-1)`

Options

A. `6.4xx10^(-8) "mol L"^(-1)s^(-1)`
B. `4xx10^(-3) "mol L"^(-1)s^(-1)`
C. `8.0xx10^(-9) "mol L"^(-1) s^(-1)`
D. `4.0 xx10^(-7) "mol L"^(-1)s^(-1)`

Answer

C

Explanation

(C ) Rate `= k[A] [B]^(2)`
`= (2.0xx10^(-6))(0.1) (0.2)^(2)`
`=8xx10^(-9)`