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Chemical Kinetics (Past Papers) C. Y. Yeung (CHW, 2009) p.01.

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Presentation on theme: "Chemical Kinetics (Past Papers) C. Y. Yeung (CHW, 2009) p.01."— Presentation transcript:

1 Chemical Kinetics (Past Papers) C. Y. Yeung (CHW, 2009) p.01

2 p.02 p. 77 Q.3 (1995 --- Chemical Method) (a)Withdraw a known volume of the reaction mixture and add to excess dilute H 2 SO 4, titrate against standard MnO 4 -. (b)Plot ln[H 2 O 2 ] against t to obtain a straight line. for first order reaction, slope = -k

3 p.03 p. 77 Q.2 (1995 --- Differential Rate Eqn) (a)I 2 :Use B, C : rate independent of [I 2 ] CH 3 COCH 3 :Use B, D : rate  [CH 3 COCH 3 ] H + : Use B, E : rate  [H + ]  Rate = k [CH 3 COCH 3 ] [H + ] (b)Total volume = 100cm 3 Use B: [CH 3 COCH 3 ] = (10  0.789/58)/(100/1000) = 1.36M = 1.36M  4.96  10 -6 = k (1.36) (0.1)  4.96  10 -6 = k (1.36) (0.1) k = 3.65  10 -5 M -1 s -1 k = 3.65  10 -5 M -1 s -1

4 p.04 p. 77 Q.5 (1996 --- Integrated Rate Eqn)  The reaction is first order. k = -slope = 2  10 -3 min -1

5 p.05 p. 80 Q.13 (2002 --- Integrated Rate Eqn) Note: Vol. of gas at time t = V t  [product] Vol. of gas not formed yet = (V total -V t )  [reactant] A straight line plot indicates that the reaction is first order w.r.t. C 6 H 5 N 2 + Cl -. k = -slope = 0.073 s -1

6 p.06 p. 77 Q.1 (1994 --- Integrated Rate Eqn) k = slope = 0.02 M -1 s -1

7 p.07 Assignment p.77 Q.6(a)-(c), Q.8, Q.9(a)-(b), Q.10, Q.11(a),(b), Q.12, Q.14(b)-(d) [due date: 24/2(Tue)] Next …. Activation Energy and Arrhenius Equation (p. 49-53)


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