Chemical Kinetics Method of Initial Rates

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Presentation transcript:

Chemical Kinetics Method of Initial Rates

Determine the Rate Law (order of a reaction) experimentally

Method of Initial Rates The rate law for a reaction can be determined by studying what happens to the initial instantaneous rate of reaction when we start with different initial concentrations of the reactants. Strategy!

Same reaction as before. Using the Method of Initial Rates to determine the Rate Law of the decomposition reaction of hydrogen iodide. 2 HI (g)  H2 (g) + I2 (g) Same reaction as before. Determine m.

We will look at the data from two experiments at a time. Using the Method of Initial Rates to determine the Rate Law of the decomposition reaction of hydrogen iodide. 2 HI (g)  H2 (g) + I2 (g) Initial [HI] (M) Initial Instantaneous Rate of Reaction (M s-1) Expt 1: 1.0 x 10-2 4.0 x 10-6 Expt 2: 2.0 x 10-2 1.6 x 10-5 Expt 3: 3.0 x 10-2 3.6 x 10-5 We will look at the data from two experiments at a time.

2 HI (g)  H2 (g) + I2 (g) 2m = 4 m = 2 Compare Expt 1 and 2 Initial [HI] (M) Initial Instantaneous Rate of Reaction (M s-1) Expt 1: 1.0 x 10-2 4.0 x 10-6 Expt 2: 2.0 x 10-2 1.6 x 10-5 Expt 3: 3.0 x 10-2 3.6 x 10-5 Compare Expt 1 and 2 Double the initial Concentration of HI The initial rate of reaction increases 4x 2m = 4 m = 2

2 HI (g)  H2 (g) + I2 (g) 3m = 9 m = 2 Compare Expt 1 and 3 Initial [HI] (M) Initial Instantaneous Rate of Reaction (M s-1) Expt 1: 1.0 x 10-2 4.0 x 10-6 Expt 2: 2.0 x 10-2 1.6 x 10-5 Expt 3: 3.0 x 10-2 3.6 x 10-5 Compare Expt 1 and 3 Triple the initial concentration of HI The initial rate of reaction increases 9x 3m = 9 m = 2

2 HI (g)  H2 (g) + I2 (g) 1.5m = 2.25 m log (1.5) = log (2.25) m = 2 Initial [HI] (M) Initial Instantaneous Rate of Reaction (M s-1) Expt 1: 1.0 x 10-2 4.0 x 10-6 Expt 2: 2.0 x 10-2 1.6 x 10-5 Expt 3: 3.0 x 10-2 3.6 x 10-5 Compare Expt 2 and 3 Increase the initial concentration of HI by 1.5 x The initial rate of reaction increases 2.25x 1.5m = 2.25 m log (1.5) = log (2.25) m = 2

Using the Method of Initial Rates to determine the Rate Law of S2O82-(aq) + 2 I- (aq)  2 SO42- (aq) + I2 (aq) Determine m and n.

Using the Method of Initial Rates to determine the Rate Law of S2O82-(aq) + 2 I- (aq)  2 SO42- (aq) + I2 (aq) Initial Concentration (M) Initial Instantaneous [I- ] [S2O82- ] Rate of Reaction (M s-1) Expt 1: 0.060 0.038 1.4 x 10-5 Expt 2: 0.060 0.076 2.8 x 10-5 Expt 3: 0.030 0.076 1.4 x 10-5 Determine m and n.

S2O82-(aq) + 2 I- (aq)  2 SO42- (aq) + I2 (aq) Initial Concentration (M) Initial Instantaneous [I- ] [S2O82- ] Rate of Reaction (M s-1) Expt 1: 0.060 0.038 1.4 x 10-5 Expt 2: 0.060 0.076 2.8 x 10-5 Expt 3: 0.030 0.076 1.4 x 10-5 n = 1

S2O82-(aq) + 2 I- (aq)  2 SO42- (aq) + I2 (aq) Initial Concentration (M) Initial Instantaneous [I- ] [S2O82- ] Rate of Reaction (M s-1) Expt 1: 0.060 0.038 1.4 x 10-5 Expt 2: 0.060 0.076 2.8 x 10-5 Expt 3: 0.030 0.076 1.4 x 10-5 m = 1

Using the Method of Initial Rates to determine the Rate Law of S2O82-(aq) + 2 I- (aq)  2 SO42- (aq) + I2 (aq) First order with respect to S2O82- . First order with respect to I-. Overall Reaction Order: 2nd Order

Determine the rate constant, k, for S2O82-(aq) + 2 I- (aq)  2 SO42- (aq) + I2 (aq) Initial Concentration (M) Initial Instantaneous [I- ] [S2O82- ] Rate of Reaction (M s-1) Expt 1: 0.060 0.038 1.4 x 10-5 Expt 2: 0.060 0.076 2.8 x 10-5 Expt 3: 0.030 0.076 1.4 x 10-5