Lecture Note

IB Chemistry: Rate Expression Rate Expression Distinguish between the terms rate constant, overall order of reaction and order of reaction with respect to a particular reactant. ● Rate Constant: This is a constant value that determines the rate of a chemical reaction and is frequently represented by the symbol, k. Units for rate constant: Order Units 1 s -1 2 dm 3 mol -1 s -1 3 dm 6 mol -2 s -1 Rate is expressed as : Rate = k[A] x [B] y . This is called the rate expression. “k” is known as the “rate constant” and is the constant of proportionality. This value remains constant and can be calculated by rearranging Rate = k[A] x [B] y to : 1) If [A] is doubled, [B] kept constant, and rate stays the same, what is the order of reaction? [2] x = [A] = amount of times number has changed, in this case, it has doubled so A=2) [1] = [B] [2] x [1] 0 = 1 [2] x =1 x= 0 Hence, the order of reaction here is zero. Rate Expression is therefore: k[A] 0 [B] = k[B] 2) If [A] is kept constant and [B] is doubled, then the initial rate also doubles [1] 0 [2] y = 2 Y = 1

Hence, the order of reaction here is one. Rate Expression= k[B] 1 3) If [A] is doubled, [B] is kept constant, the rate increases 4 times. [2] x [1] 0 = 4 X=2 Hence, the order of reaction here is two. Rate Expression = k[A] 2 4) Lastly, if [A] is doubled, [B] is kept constant and the rate increases 8 times. [2] y [B] 2 = 8 y= 3 Hence, the order of reaction here is three. Rate Expression = k[A] 3 [B] 2 Overall Order of Reaction: “This is the sum of the individual order of reactions” So for the last one, the individual orders of reactions as seen through the exponents are 3 and 2, so the overall order of reaction will be: 3+2 = 5 Overall order of reaction = 5. Deduce the rate expression for a reaction from experimental data. Let’s try to answer these questions from the experimental data below: 1) What is the rate expression of the reaction? 2) Overall Rate of Reaction 3) Rate constant and its units at 298K. Note: The data below is contrived and is used solely for exam purposes. Experiment number Initial concentration of [A] (g) / moldm -3 Initial Concentration of [B] (g)/ moldm -3 Initial rate of formation of [C]/moldm -3 s -1 1 3.0×10 -3 5.0×10 -3 4.0×10 -3 2 3.0×10 —3 1.0×10 -2 8.0×10 -3 3 2.0×10 -3 3.0×10 -3 1.0×10 -3 4 8.0×10 -3 3.0×10 -3 1.6 x 10 -2

1) Let’s closely examine the experimental data. If we look at experiment 1 and 2, if we keep [A] constant at 3.0×10 -3 , and we double, the concentration for [B], the rate of formation [C] also doubles. Let’s now map this out mathematically. [A] = 1 (as the concentration remains constant) [B] = 2 ( as the concentration doubles) [C] = 2 (as the rate doubles) [1][2] x = 2 Using some simple maths, we can deduce x= 1, so the reaction for [B] is first order. If we look at experiment 3 and 4, [B] is kept constant whilst [A] is quadrupled (x4). As a result, we can the rate of reaction is (1.6×10 -2 /1.0×10 -3 ), which equals 16. Hence, from experiment 3 and 4 [A] = 4 (as the concentration increases by 4 times) [B]= 1 (as it is kept constant) [C] = 16 (rate increases 16 times) [4] y [1] = 16 Y =2, hence [A] is a 2 nd order reaction, Hence, the rate expression is: Rate = k [A] 2 [B] 1 , where k is the rate constant. Solve problems involving the rate expression. Let’s continue on from 16.1.2 2) Rate= k [A] 2 [B] 1 The overall order of reaction is simply “x+y” or the sum of the individual orders of reaction. This can easily be found by simply adding the “exponent” signs which in this case are 1 and 2 So Overall order of reaction = 2+1 =3 3) Rate= k [A] 2 [B] 1 K = Rate/ [A] 2 [B] 1 Now, simply take in any of the experimental values in the table and plug in the values for rate, [A] 2 , and [B] to arrive at the rate constant. I’ll get the results from Experiment 1. Since the overall reaction is 3 rd order, the units we will use are: dm 6 mol -2 s -1 K= (4.0×10 -3 ) / (3.0×10 -3 ) 2 x (5.0×10 -3 )

K= 8.888 x 10 4 K= 8.888 x 10 4 dm 6 mol -2 s -1 Sketch, identify and analyse graphical representations for zero-, first- and second-order reactions. 1) Zeroth Order: Rate stays constant, regardless of concentration. 2) First Order: Rate is proportional to the Concentration. 3). Second Order: The graph is a curve, as there is a quadratic relation between rate and concentration.

- Traits Periodic Element
- Summary of Table Periodic
- Periodic Element Assingment
- Development System Periodic Element
- Arrangement System Periodic Element Modern
- Particles or Mole Conversions Chemistry
- Molecular Compounds
- Mixture Ecamples Measures and Units Chemistry Notes
- Introduction of Acids
- General Chemistry Introduction
- Conversion Problems
- Comparison of Temperature Scales
- Atomtic Structure
- Unit Conversions
- The Mole Avogadro's Number

IB Chemistry: Rate Expression

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