For a non-equilibrium process `A + B to` Produts the rate is first order with respect to A and second order with respect to B. If 1.0 Mole each of A and B are introduced into a one liter vessel and the intial rate is `1.0 xx 10^(-2)`mol `L^(-1)s^(-1)`, the rate when half of the reaction have been eonsumed is:

Options

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

Explanation

`rate = k [A][B]^2`
`:. 10^(-2)= k [1][1]^2`
or `k = 10^(-2) litre^2 mol^(-2)sec^(-1)`
`:. Rate II= 10^(-2)[0.5]xx [0.5]^2`
`= 1.2 xx 10^(-3)` mol/litre- sec