Chem-rate laws and rate constants?

Question

hi i figured out the rate laws however i dont know how to get the rate constant. a little help plz? thanks!(thats a chart down there :P)

--
2 ClO2(aq) + 2 OH -(aq) ClO3-(aq) + ClO2-(aq) + H2O(l)

[ClO2]0 (mol/L)---[OH -]0 (mol/L) --Initial Rate (mol/Ls)

0.0500 -------------- 0.100 ------------ 5.75e10-2
0.100 ------------ 0.100 -------------- 2.30e10-1
0.100 ------------- 0.0500 --------------- 1.15e10-1

(a) Determine the value of the rate constant.
rate law

Answer 1

Comparing experiments 1 and 2 shows that the experiment is second-order with respect to the concentration of ClO2.

Comparing experiments 2 and 3 shows that the experiment is first-order with respect to the concentration of OH-.

Therefore, the rate law can be written as:
rate = k [ClO2]^2 [OH-].

Plug the numbers of any experiment into your rate law. Let's use the first experiment:

rate = k [ClO2]^2 [OH-]
5.75 x 10^-2 mol/L/s = k (0.0500 mol/L)^2 * (0.100 mol/L)
k = 230 (L/mol)^2 * (1/s)

Answer 2

Hold OH- constant, and what happens to the rate when you double ClO2? It quadruples (2nd order).

Hold ClO2 constant, and what happens when you double OH-? It also doubles (1st order).

3rd order overall. Now you can do the rest!

Answer 3

Just put in all the reagents which are in the mechanism up to and including the slowest step. So 2 molecules of NO and 1 molecule of H2 are needed, hence the respective indices.

Answer 4

complicated task. research at a search engine. it will help!