J.R. S. answered 12/16/20
Ph.D. in Biochemistry--University Professor--Chemistry Tutor
Comparing lines 1 & 2:
[A] doubles, [B] and [C] remain constant; rate doubles; ---> 1st order in A
Compare lines 1 & 3:
[B] doubles, [A] and [C] remain constant; rate stays the same ---> zero order in B
Compare lines 2 & 4:
[C] doubles, [A] and [B] remain constant; rate increases 4x ---> 2nd order in C
Rate Law: rate = k[A][C]2
We can use any line to solve for k; using line 1 we have....
1.25 M/s = k[0.1 M][0.25 M)2 = 0.00625 M3 k
k = 200 M-2s-1
Now that we have the rate constant, k, we can find the rate given the concentrations of the reactants:
rate = k[A][C]2 = 200 M-2s-1 (0.573 M)(0.195 M)2
rate = 4.36 Ms-1 = 4.36 M L-1 s-1


J.R. S.
12/16/20
Anthony T.
Rate only increases by 4 between lines 2 and 4.12/16/20