For the reaction `2A +B to A_2B`, the rate `=k[A][B]^2` with `k = 2.0 xx 10^(-6) M^(-2) s^(-1)` . Calculate the initial rate of the reaction when [A] = 0.1M, [B] = 0.2 M. If the rate of reverse reaction is negligible then calculate the rate of reaction after [A] is reduced to 0.06 M.

Explanation

According to the given rate law, `r=k[A] [B]^2`
Initial rate `=k[A] [B]^2 = 2.0 xx 10^(-6) M^(-2) s^(-1) [0.1 M][0.2 M]^2`
`= 8.0 xx 10^(-9) Ms^(-1)`
From the equation, `2A +B to A_2`,B we find that when two moles of A are used then 1 mol of B is also used in the same time. Amount of A consumed = 0.1-0.06 = 0.04 M
` therefore `Amount of B consumed `=(0.04 )/(2) = 0.02 M`
Conc. of A left, [ A ]=0.06M
Conc. of B left, [B] = [0.2-0.02 ]M=0.18 M
Rate `=k [A] [B]^2`
`= 2 xx 10^(-6) s^(-1) [0.06 M] [0.18 M]^2`
`= 3.89 xx 10^(-9) Ms^(-1)`