J.R. S. answered 03/31/21
Ph.D. University Professor with 10+ years Tutoring Experience
(a). Comparing experiments 1 and 2, we see that [NO] is constant and [O2] doubles. We see that the rate doubles also. This tells us that the reaction is FIRST order in O2
Next, we compare experiments 5 to 2 and we see that [O2] is constant and [NO] doubles. Here we see that the rate increases by 4 times (4x), going from 0.014 to 0.057 mol/L-s. This tells us the reaction is SECOND order in NO.
The rate law is:
Rate = k[NO]2[O2]
(b) If [NO] is tripled and [O2] is quadrupled, the initial rate would change as follows:
Rate = k (3)2(4)1 = 36k so the rate would increase by 36 times.