The rate equation for the reaction `2A + B rarr` products is rate `= k[A][B]^(2)`. If k at T (K) is `5.0 xx 10^(-6) mol^(-2) L^(2)s^(-1)`, the initial rate of the reaction, when [A] `= 0.05 mol L^(-1)` and `[B] = 0.1 mol L^(-1)` is

Options

A. `1.25 xx 10^(-9) L mol L^(-1) s^(-1)`
B. `1.25 xx 10^(-9) mol L^(-1) s^(-1)`
C. `2.50 xx 10^(-9) mol L^(-1) s^(-1)`
D. `2.50 xx 10^(-9) L mol L^(-1) s^(-1)`

Answer

C

Explanation

For the reaction,
`2A + B rarr` Products
Rate `= k[A][B]^(2)`
Given, Rate constant, k of reaction `= 50 xx 10^(-6) mol^(-2) L^(2)S^(-1)`
[A] `= 0.05 mol L^(-1)`
`[B] = 0.1 mol L^(-1)`
`:.` Rate of reaction `= 50 xx 10^(6)[0.05][0.1]^(2)`
`= 5 xx 10^(-6) xx 5 xx 10^(-4)`
`= 2.5 xx 10^(-9) mol L^(-1) S^(-1)`