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Kinetics- Higher Level

Kinetics- Higher Level. Take a more quantative look at the whole topic of rates of reaction More maths involved. The Rate equation. Concentration affects the rate of reaction. Does that mean that the concentration of every reactant affects the rate of reaction? For most reactants:

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Kinetics- Higher Level

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  1. Kinetics- Higher Level • Take a more quantative look at the whole topic of rates of reaction • More maths involved

  2. The Rate equation • Concentration affects the rate of reaction. Does that mean that the concentration of every reactant affects the rate of reaction? • For most reactants: • Rate  [Reactant] • Rate = k[Reactant]

  3. The rate equation-orders of reaction • For a general reaction: • A + B  C + D • Rate  [A]a[B]b • Rate = k[A]a[B]b • The powers to which the reactant concentrations (a&b) are raised are called orders of reaction • a, b are the individual orders of reaction • ab (a+b) are the overall orders of reaction

  4. The rate equation- orders of reaction 2 • For the process: A + B  C + D • If the rate equation is: rate = k[A]1[B]0 This means that the reaction is first order with respect to the concentration of A, zero-order with respect to the concentration of B The reaction is first order overall as: 1 + 0 = 1 When [A] doubles, the rate doubles When [B] doubles, the rate is unchanged as [B]0=1 Hence the reaction is independent of [B]

  5. Calculating orders of reaction • Need to do at least 3 different trials for an experiment to determine orders of reaction for a reaction with 2 reactants • In 2 of the experiments keep [A] constant while varying [B], in another one keep [B] while varying [A] If trials 1 & 2 are compared, [B] stays constant, while [A] doubles, the rate doubles, hence the rate is first order with respect to [A]. If trials 3 & 1 are compared, [A] stays constant while [B] doubles, however the rate stays unchanged, hence the reaction is zero-order with respect to [B]. Hence rate = k[A]

  6. Rate equations- worked example • Consider the following reaction: • 2NO(g) + 2H2(g)  N2(g) + 2H20(g) • The following table shows the initial rate for different [NO] & [H2] • What is the order with respect to [NO] & [H2]? • If the top 2 trials are compared, the [NO] is doubled, while [H2] stays constant, the rate quadruples. Hence the rate is second order with respect to [NO]. If the bottom 2 trials are compared, [NO] stays constant, while [H2] doubles, the rate also doubles. Hence the rate is first order with respect to [H2] • Rate = k[NO]2[H2]

  7. More reaction examples Sometimes not only can a reaction be zero-order with respect to 1 of the reactants, but the catalyst can appear in the rate equation. For example consider the iodination of propanone: CH3COCH3(aq) + I2(aq)  CH2ICOCH3(aq) + HI(aq) The rate equation is: Rate = k[CH3COCH3][H+] [H+] appears in the rate equation as the reaction is catalysed by acid. The reaction is independent of [I2]. The reaction is second-order overall. This means that if the [CH3COCH3] & [H+] are both doubled, the initial rate of reaction will increase by a factor of 4. Similarly if both [CH3COCH3] & [H+] are halved, the initial rate will decrease fourfold. The effect of the order of reaction on the rate can always be deduced by the following method. If the order is 1, then the rate double when concentration doubles as 21=2 If the order is 2 the rate quadruples as 22=4 If the order is 3 the rate increases by a factor of 8 as 23=8

  8. Rate equation- more complex examples Consider the following reaction: BrO3-(aq) + 5Br-(aq) +6H+(aq)  3Br2 + 3H2O The rate equation is found to be: rate = k[BrO3-][Br-][H+] What is the effect on the rate if [BrO3-] is halved, but [H+] is quadrupled (at constant temperature and bromide concentration). The quadrupling of the hydrogen ion concentrations results in a quadrupling of the rate (first-order kinetic behaviour); the halving of the bromate concentration results in the rate being reduced by a factor of 2 (first-order kinetic behaviour). Overall this is equivalent to the rate being doubled.

  9. The rate constant k, is a numerical value, which is characteristic of a particular reaction Each reaction has its own unique rate constant in terms of a value and associated units (at a particular temperature and pressure). It does not depend on the [reactant] but is affected by changes in temperature. For a first order reaction rate = k[A] Hence k = rate/[A] Substituting the units for rate & concentration gives: k = mol dm-3 s-1/mol dm-3 This cancels to units of s-1 For a second order reaction: Rate = k[A][B] k = rate/[A][B] Substituting the units for rate & concentration gives mol dm-3 s-1/(mol dm-3)2 which simplifies to mol dm-3 s-1/mol2 dm-6 before cancelling to mol-1dm3s-1

  10. Worked example Propene reacts with bromine to form 1,2-dibromopropane C3H6(g) + Br2(g)  CH3CHBrCH2Br(l) The rate equation is Rate = k[C3H6][Br2] K = 30.0 dm3 mol-1 s-1 Calculate the initial rate of reaction when the concentrations of propene & bromine are both 0.040 mol dm-3 Rate= 30.0 dm3 mol-1 s-1 x 0.040 mol dm-3 x 0.040 mol dm-3 = 0.048 mol dm-3 s-1

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