For the following reaction:
`2A + B +C rarr A_(2)B + C`
The rate law has been determined to be
Rate `= k[A][B]^(2)` with `k = 2.0 xx 10^(-6) mol^(-2) L^(2) s^(-1)`
For this reaction, determine the initial rate of the reaction with `[A] = 0.1 mol L^(-1), [B] = 0.2 mol L^(-1), C = 0.8 mo, L^(-1)`. Determine the rate after `0.04 mol L^(-1)` of `A` has reacted.

Answer

`8xx10^(-9)" mol L"^(-1)s^(-1),3.9xx10^(-9)" mol L"^(-1)s^(-1)`

Explanation

Initial rate `=(2.0xx10^(-6)" mol"^(-2)L^(2)s^(-1))xx0.1" mol L"^(-1)xx(0.2" mol L"^(-1))^(2)=8xx10^(-9)" mol L"^(-1)s^(-1)`
After `0.04" mol L"^(-1)" of A has reacted. "[A]=0.1-0.04=0.06" mol L"^(-1)`
`[B]=0.2-0.02=0.18" mol L"^(-1)`
Then rate `=(2xx10^(-6))xx(0.06)xx(0.18)^(2)=3.9xx10^(-9)" mol L"^(-1)s^(-1).`