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

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 10-4 M/min Rate of disappearance of B is 1.8 x 10-3 M/min Rate of appearance of D is 1.8 x 10-3 M/min

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]

c) rate = k[NO]2[H2] rate = 120[3.0 x 10-2]2[1.2x 10-2] rate = 1.3 x 10-3 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]

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.