InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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.A. `1.25 xx 10^(-3) mol L^(-1)s^(-)`B. `1.0 xx 10^(-2) mol L^(-1)s^(-)`C. `2.50 xx 10^(-3) mol L^(-1)s^(-)`D. `2.0 xx 10^(-2) mol L^(-1)s^(-)` |
|
Answer» Correct Answer - A `A + B rarr` product `r = K[A]^(1) [B]^(2)` `r_(1) = K[1]^(1) [1]^(2) = 1 xx 10^(-2) (K = 1 xx 10^(-2))` `r_(1) = K[(1)/(2)][(1)/(2)]^(2) = 1 xx 10^(-2) xx (1)/(8)` = `1.25 xx 10^(-3) "mol L"^(-1) S^(-1)` |
|