1. Integrated Rate Equation C. Y. Yeung (CHW, 2009) p.01

2. To study Integrated Equation, e.g. : During the reaction, both [A] and [B] decrease! p.02 large excess of [B] should be used. [A] = -k’t + [A] 0 (zeroth order) In order to ensure that the decreasing rate is due to decreasing [A], not [B] … i.e. keep [B] as “effectively constant”.

3. p.03 rate = = k’ [A] 1 - d[A] dt = k’[A] - d[A] dt First Order Rxn (m = 1) = k’t + C - ln [A] when t = 0, [A] = [A] 0  C = - ln [A] 0  - ln [A] = k’t - ln [A] 0 integrated rate eqn. (first order) = k’dt - d[A] [A] = k’ dt  d[A] 1[A] - ln [A] = - k’t + ln [A] 0

4. p.04 Thus, Timet0t0 t1t1 t2t2 t3t3 …. [A][A] 0 [A] 1 [A] 2 [A] 3 …. ln[A]ln [A] 0 ln [A] 1 ln [A] 2 ln [A] 3 …. ln [A] = - k’t + ln [A] 0 ln [A] t ln [A] 0 slope = - k’ (first order)

5. p.05 rate = = k’ [A] 2 - d[A] dt = k’[A] 2 - d[A] dt Second Order Rxn (m = 2) = k’t + C [A] -1 [A] -1 when t = 0, [A] = [A] 0  C = [A] 0 -1  [A] -1 = k’t + [A] 0 -1 integrated rate eqn. (second order) = k’dt - d[A] [A] 2 = k’ dt  d[A] 1 [A] 2 -

6. p.06 Thus, Timet0t0 t1t1 t2t2 t3t3 …. [A][A] 0 [A] 1 [A] 2 [A] 3 …. [A] -1 [A] 0 -1 [A] 1 -1 [A] 2 -1 [A] 3 -1 …. [A] -1 t [A] 0 -1 slope = k’ (second order)  [A] -1 = k’t + [A] 0 -1

7. p.07 Summary … 3 Integrated Rate Eqns ln [A] = - k’t + ln [A] 0 m = 0 m = 1 m = 2 [A] -1 = k’t + [A] 0 -1 [A] = - k’t + [A] 0

8. p.08 p. 76 Q.4 Decomposition of H 2 O 2

9. p.09 (a)To show 1 st order : ln [H 2 O 2 ] = - k t + ln [H 2 O 2 ] 0

10. p.10 Plot ln [H 2 O 2 ] against time The graph gives a straight line, therefore the reaction is 1 st order w.r.t. [H 2 O 2 ].

11. p.11 (b)Expression for the Rate Equation : rate = k[H 2 O 2 ] Calculate “k” : slope = -k = -0.0495,  k = 0.0495 min -1 Calculate half life (time at which [A] = ½[A] 0 ) : ln (1/2[H 2 O 2 ]) = - (0.0495) t + ln [H 2 O 2 ] 0 ln (1/2) = - (0.0495) t t = 14.0 mins

12. p.12 At the beginning, [H 2 O 2 ] = 3.0 mol dm -3 When expt. started, [H 2 O 2 ] = 0.750 mol dm -3 (c)How long the [H 2 O 2 ] in the contaminated bottle? ln (0.750) = - (0.0495) t + ln (3.0) t = 28.0 mins

13. p.13 Expt. 8 Decomposition of H 2 O 2 Flask A (150cm 3 water) 10cm 3 1.00 mol dm -3 H 2 O 2 50cm 3 borate buffer 10cm 3 diluted KMnO 4 start stop watch! 10cm 3 sample (around 5 mins) 10cm 3 1.0M H 2 SO 4 Flask B Titrate against dilute KMnO 4 (4×10 -3 M)

14. What happens in Flasks A and B …? p.14 Flask A H 2 O 2 + 2OH -  O 2 + 2H 2 O + 2e -` MnO 4 - + 2H 2 O + 3e -  MnO 2 + 4OH - × 3 × 2 3H 2 O 2 + 2MnO 4 -  3O 2 + 2H 2 O + 2OH - + 2MnO 2 2H 2 O 2 O 2 + 2H 2 O MnO 2 Flask B MnO 2 is killed by H 2 SO 4. 2MnO 4 - + 5H 2 O 2 + 6H +  2Mn 2+ + 8H 2 O + 5O 2

15. p.15 Date Treatment … For 1 st order rxn, ln [A] = - k t + ln [A] 0  ln ([A] 0 /[A]) = k t As vol. of MnO 4 - used  [A],  ln (V 0 /V) = k t If a straight line is plotted [ln(V 0 /V) vs t]  1st order, and slope = k!

16. Next …. p.16 Activation Energy and Arrhenius Equation (p.??? ) Assignment p.48 Q.7-11 [due date: 16/2(Mon)] Lab report [due date: 23/2(Mon)] p.75 Q.2-3 [due date: 16/2(Mon)]