1. 1 i)
No; the reaction is termolecular, it’s unlikely that three molecules will simultaneously collide with enough energy and proper orientation.
ii)
When comparing experiments 1 and 3, [B] is tripled and the rate stays constant ∴[B] is zeroth order
When comparing experiments 3 and 2, [A] is tripled and the rate is tripled ∴[A] is first order
rate = k[A]
iii)
No; if it occurred in one step, the rate law would be rate = k[A][B]2

2. iv) v) rate = k[A] k = 3.0 x 10-3 min-1 3.0 x 10-4 = k[0.10]
Rate of disappearance of A is 9.0 x M/min
Rate of disappearance of B is 1.8 x 10-3 M/min
Rate of appearance of D is 1.8 x M/min

3. 2 a)
When comparing experiments 1 and 2, [NO] is doubled and the rate quadruples ∴[NO] is second order
When comparing experiments 1 and 3, [NO] is halved (which would one-fourth the rate) and [H2] is doubled. The rate halves.
Since [NO] would have decreased it by a factor of four, but the rate decreased by a factor of two, [H2] is responsible for increasing it by 2 times ∴[H2] is first order b)
rate = k[NO]2[H2]
k = 120 M-2 s-1
0.6 x 10-5 = k[5.0 x 10-3]2[2.0 x 10-3]

4. c) rate = k[NO]2[H2] rate = 120[3.0 x 10-2]2[1.2x 10-2]
rate = 1.3 x M/s
Rate = k[NO]2[H2]
i) 2 NO + H2 → N2O + H2O (slow)
ii) 2 NO → N2O2 (fast)
N2O2+ H2 → N2O + H2O (slow)
Rate = k[N2O2][H2]
N2O2 cannot be in the rate expression
Write an equilibrium for the previous step
[NO]2 =[N2O2]
Rate = k[NO]2[H2]

5. iii) NO + H2 → H2O + N (fast)
N + NO → N2O (slow)
Rate = k[N][NO]
N cannot be in the rate expression
Write an equilibrium for the previous step
[NO][H2] =[N]
Rate = k[NO]2[H2]
All three are consistent, but mechanism 1 is termolecular; it’s unlikely that three molecules will simultaneously collide with enough energy and proper orientation so it’s not 1.