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.

Explanation

a. `2A +B + C rarr A_(2)B + C`
Rate `= k[A][B]^(2)`
`r = 2.0 xx 10^(-6) xx [0.1] xx [0.2]^(2)`
`= 8 xx 10^(-9) mol . L^(-1) s^(-1)`
b. `(-d[A])/(2 dt) = (-d[B])/(dt) = (-d[C])/(dt) = (d[A_(2)B])/(dt) = (d[C])/(dt)`
`[A]_("initial") = 0.1 M, [A]_("reacted"] = 0.04 M`
`[A]_("left") = 0.1-0.04 = 0.06 mol L^(-1)`
`(-d[A])/(2 dt) = (-d[B])/(dt)`
`:. (-d[B])/(dt) = (1)/(2) xx (-d[A])/(dt) = (1)/(2) xx 0.04 = 0.02 M s^(-1)`
`[B]_("initial") = 0.2M` , `[B]_("reacted") = 0.02 M`
`[B]_("left") = 0.2-0.02 = 0.18 mol L^(-1)`
`:. r = k[A][B]^(2) = 2.0 xx 10^(-6)[0.06][0.18]^(2)`
`= 3.88 xx 10^(-9) mol . L^(-1) s^(-1)`