Solution:
1. Identify the rate-determining step: The slow step is the rate-determining step, as it dictates the overall rate of the reaction. The slow step is: NOBr2 + NO → 2NOBr
2. Write the rate law based on the slow step: Rate = k[NOBr2][NO]
3. Express the concentration of the intermediate (NOBr2) in terms of reactants: Since the first step is a fast equilibrium, we can write its equilibrium constant (K): K = [NOBr2] / ([NO][Br2]) [NOBr2] = K[NO][Br2]
4. Substitute the expression for [NOBr2] into the rate law: Rate = k(K[NO][Br2])[NO] Rate = kK[NO]2[Br2]
5. Determine the overall order of the reaction: The overall order is the sum of the exponents in the rate law. Order = 2 (for NO) + 1 (for Br2) = 3 Therefore, the overall order of the reaction is 3.
Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: