For the non-equilibrium process, `A + B rarr` Product, the rate is first-order w.r.t. `A` and second order w.r.t. `B`. If `1.0 mol` each of `A` and `B` were inrofuced into `1.0 L` vessel and the initial rate was `1.0 xx 10^(-2) mol L^(-1) s^(-1)`, calculate the rate when half the reactants have been turned into Products.

Options

A. `1.25 xx 10^(-3)`
B. `1.2 xx 10^(-2)`
C. `2.5 xx 10^(-4)`
D. None of these

Answer

A

Explanation

order w.r.t. `A = 1`
order w.r.t. `B = 2`
`:. r = K[A] xx [B]^(2)`
when `A` and `B` are consumed `50%`,
`r' = K xx [(A)/(2)] xx [(B)/(2)]^(2) = r xx (1)/(8)`
`= (1.0 xx 10^(-2)) xx (1)/(8) = 1.25 xx 10^(-3)`