Question

Consider this initial-rate data at a certain temperature for the reaction described by 2no2

Answer 1

 NO2........O3........rate 
0.65.........0.8.......2.6x10^4 
1.1...........0.8.......4.4x10^4 
1.76........1.4........12.32x10^4 

rate2/rate1 = k[NO2 2]^a[O3 2]^b / k[NO2 1]^a[O3 1]^b 
K and O3 cancel 
4.4 / 2.6 = (1.1 / 0.65)^a 
1.7 = 1.7^a.....a = 1 

rate 3 / rate 2 = k[NO2 3][O3 3]^b / k[NO2 2][O3 2]^b 
k cancels 
since you know the order of the rxn with respect to NO2, plug in the values and solve the equation. the [O3]^b is a separate value so reduce all numbers to a final value and then solve for b 
12.32 / 4.4 = (1.76)(1.4)^b / (1.1)(0.8)^b.....1.76/1.1 = 1.6.....1.4 / 0.8 = 1.75 
2.8 = 1.6 x (1.75)^b 
1.75 = (1.75)^b.....b = 1 

rate = k[NO2][O3] 
2.6x10^4 / (0.65)(0.8) = k 
k = 5x10^4