2) 1) 3) a c b 4) Forward reaction has a greater Ea since it has the higher crest from start; both graphs start and end at the same points ∴ same magnitude ΔH. (negative for the forward / exothermic reaction and positive for the reverse / endothermic reaction) 5) A catalyst is not a part of the overall reaction, but it is a part of the elementary steps. A catalyst lowers the activation energy

6)

∴ reaction occurs more rapidly [B] tripled tripled first order with respect to [A] b) [A] doubled octupled third order with respect to [B] c) rate = k[A][B]3 d) e) 640 M-3s-1 1.024 Ms-1 8. a) Under high pressure, more collisions occur. ∴ reaction occurs more rapidly b) Reactions generally proceed quicker @ high temperature, high pressure, high concentration, and larger surface area

b) d) e) 1a) c) 2) 3) b) d c d c b products 5kJ/mol True 60 kJ/mol Activated complex 3) Ea Ea’ reactants ΔH products b) 60 kJ/mol

4 a) b) c) 1) 2) 3) O (g) 2 O3 (g) → 3 O2 (g) O2 (g) double When comparing experiments 1 and 2, [B] is 1.5 times larger and the rate is 1.5 times larger ∴[B] is first order 2) 80 M/hr 3) So C disappears at three times the rate Z appears 45 M/hr

4) rate = k[A]2[B][C] When comparing experiments 1 and 2, [B] is 1.5 times larger and the rate is 1.5 times larger ∴[B] is first order When comparing experiments 3 and 4, [C] is double and the rate is double ∴[C] is first order When comparing experiments 1 and 3, [C] is double and [A] is double. The rate is 8 times larger . . . . . . . since [C] is first order, [C] is responsible for 2 times, thus [A] is responsible for 4 times ∴ [A] is second order

¼ (divide rate of experiment 1 by four) 5) 2.5 M/hr When comparing experiments 1 and 5 [B] is half as big – rate would half [C] is double – rate would double [A] is half as big – rate would be one-fourth ¼ (divide rate of experiment 1 by four) ½ x 2 x ¼ = 6) 1.25 x 106 M-3hr-1 (do this for each trial, Σ, and divide by five) 10 = k[0.1]2[0.02][0.04]

(do this for each trial, Σ, and divide by six) 7) rate = k[A]2[B] 1.04 M-2s-1 When comparing experiments 1 and 2, [B] is double and the rate is double ∴[B] is first order When comparing experiments 3 and 4, [A] is double and the rate is quadruple ∴[A] is second order 0.26 x 10-9 = k[1.0 x 10-3]2[0.25 x 10-3] (do this for each trial, Σ, and divide by six)

HNO2 (aq) + NH4+ (aq) → 2 H2O (l) + H+ (aq) + N2 (g) 8) i. HNO2(aq) + H+(aq) → H2O(l) + NO+(aq) (fast) ii. NH4+(aq) → NH3(aq) + H+(aq) (fast) iii. NO+(aq) + NH3(aq) → NH3NO+(aq) (slow) iv. NH3NO+(aq) → H2O(l) + H+(aq) + N2(aq) (fast) HNO2 (aq) + NH4+ (aq) → 2 H2O (l) + H+ (aq) + N2 (g) i. [HNO2][H+] = [NO+] ii. [NH4+] = [NH3][H+] i. HNO2(aq) + H+(aq) → H2O(l) + NO+(aq) (fast) ii. NH4+(aq) → NH3(aq) + H+(aq) (fast) iii. NO+(aq) + NH3(aq) → NH3NO+(aq) (slow) iv. NH3NO+(aq) → H2O(l) + H+(aq) + N2(aq) (fast) Rate = k[NO+][NH3] Neither of these species can be in the rate expression Write equilibriums for the previous steps

8) rate = k[HNO2][NH4+] i. HNO2(aq) + H+(aq) → H2O(l) + NO+(aq) (fast) ii. NH4+(aq) → NH3(aq) + H+(aq) (fast) iii. NO+(aq) + NH3(aq) → NH3NO+(aq) (slow) iv. NH3NO+(aq) → H2O(l) + H+(aq) + N2(aq) (fast) Rate = k[NO+][NH3] [HNO2][H+] = [NO+] i. HNO2(aq) + H+(aq) → H2O(l) + NO+(aq) ii. NH4+(aq) → NH3(aq) + H+(aq) [NH4+] = [NH3][H+] [NH4+] = [NH3] [H+] [NH4+] Rate = k[HNO2][H+] [H+] rate = k[HNO2][NH4+]

8) 9) HNO2 (aq) + NH4+ (aq) → 2 H2O (l) + H+ (aq) + N2 (g) NO+ , NH3 i. HNO2(aq) + H+(aq) → H2O(l) + NO+(aq) (fast) ii. NH4+(aq) → NH3(aq) + H+(aq) (fast) iii. NO+(aq) + NH3(aq) → NH3NO+(aq) (slow) iv. NH3NO+(aq) → H2O(l) + H+(aq) + N2(aq) (fast) HNO2 (aq) + NH4+ (aq) → 2 H2O (l) + H+ (aq) + N2 (g) NO+ , NH3 , NH3NO+ 9) 51 sec Step 1: 2NO → N2O2 0.85 min Step 2: N2O2 + H2 → N2O + H2O 0.015 hr Step 3: N2O + H2 → N2 + H2O 55 sec 54 sec 2NO + 2H2 → N2+ 2H2O Which step is the slowest step?

9) rate = k[NO]2[H2]2 Rate = k[N2O][H2] Step 1: 2NO → N2O2 Step 2: N2O2 + H2 → N2O + H2O Step 3: N2O + H2 → N2 + H2O Rate = k[N2O][H2] N2O cannot be in the rate expression Write equilibriums for the previous steps [NO]2 =[N2O2] Step 1: 2NO → N2O2 Step 2: N2O2 + H2 → N2O + H2O [N2O2][H2] = [N2O] [NO]2[H2] = [N2O] rate = k[NO]2[H2]2