1. Title: Lesson 6 Activation Energy Learning Objectives: – Understand the term activation energy – Calculate activation energy from experimental data

2. Main Menu Recap 1.Two species, P and Q, react together according to the following equation. P + Q → R The accepted mechanism for this reaction is P + P → 2Pfast 2P + Q → R + Pslow What is the order with respect to P and Q? PQ A.11 B.12 C.21 D.22

3. Main Menu Activation Energy  Activation energy is the minimum energy to colliding particles need in order to react  You can think of it as:  The energy required to begin breaking bonds  The energy that particles need to overcome the mutual repulsion of their electron shells.  Can you think of an analogy?

4. Main Menu The rate constant k is temperature dependent  10 o C increase generally leads to doubling of rate.  From rate equation, we can see temperature has no effect on concentration of reactants so it must affect k.  From the Maxwell-Boltzmann distribution curve, the value of the activation energy will dictate the extent of change in number of particles that can react at a higher temperature. Large E a  temp rise causes significant increase in number of particles reacting Small E a  temp rise causes a less significant increase in number of particles reacting. Lower E a

5. Main Menu The Arrhenius Equation  We met the rate constant, k, a couple of lessons ago  The Arrhenius Equation tells us how k is related to a variety of factors: Where: k is the rate constant E a is the activation energy T is the temperature measured in Kelvin R is the gas constant, 8.314 J mol -1 K -1. e is Euler’s number A is the ‘frequency factor’ or Arrhenius constant or pre-exponential factor This equation can be found in section 1 of the data booklet! ‘A’ takes into account the frequency with which successful collisions will occur. Like ‘k’ it has the same units that vary with order of reaction.

6. Main Menu What happens if you increase the temperature by 10°C from, say, 20°C to 30°C (293 K to 303 K)?  The frequency factor, A, in the equation is approximately constant for such a small temperature change. We need to look at how e -(E A / RT) changes the fraction of molecules with energies equal to or in excess of the activation energy.  Let's assume an activation energy of 50 kJ mol -1. In the equation, we have to write that as 50000 J mol -1. The value of the gas constant, R, is 8.31 J K -1 mol -1.  The fraction of the molecules able to react has almost doubled by increasing the temperature by 10°C. That causes the rate of reaction to almost double.

7. Main Menu Rearranging Arrhenius  If we take logs of both sides, we can re-express the Arrhenius equation as follows:  This may not look like it, but is actually an equation in the form y = mx + c Where: ‘y’ is ln k ‘m’ is -E a /R ‘x’ is 1/T ‘c’ is ln A

8. Main Menu To determine E a Experimentally: (Assuming we know the rate equation)  Measure the rate of reaction at various different temperatures.  Keeping all concentrations the same  Calculate the rate constant, k, at each temperature.  Plot a graph of ln k (y-axis) vs 1/T (x-axis)  The gradient of this graph is equal to ‘-E a /R’, this can be rearranged to calculate E a.

9. Use the following data to find the activation energy value for the reaction H 2 + I 2  2HI which is second order overall. Temp / o C200300500600 Rate constant k / mol -1 dm 3 s -1 3.07 x 10 -9 2.76 x 10 -6 3.02 x 10 -2 6.07 x 10 -1 Temp/K473573773873 ln k5.2983175.7037826.2146086.39693 1/T / K -1 0.0021140.0017450.0012940.001145 - 15.4 0.00077 Gradient = -15.4 / 0.00077 = - 20000 K  E a = - Grad x R = -(-20000) x 8.314 = +166280 J mol -1 = +166.28 kJ mol -1 DO NOT include origin for x-axis

10. Use the following data to find (a) the activation energy value for the reaction 2N 2 O(g)  2N 2 (g) + O 2 (g) and (b) the rate constant at 900K. - 7.00 0.00023 (a) Gradient = - 7.00 / 0.00023 = - 30400 K  E a = - Grad x R = -(-30400) x 8.314 = +252746 J mol -1 = +252.8 kJ mol -1 Rate constant k /mol -1 dm 3 s -1 0.00110.38001.670011.5900 Temp /K 838100110531125 ln k -6.812-0.9680.5132.450 1/T /K -1 0.0011930.0009990.0009500.000889 (b) At T = 900K  1/T = 0.00111 K -1  from graph, ln k = -4.3  k = e -4.3  0.0136 mol -1 dm 3 s -1 0.00111 4.3

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12. Solving simultaneous equations  Activation energy can also be calculated from values of the rate constant, k, at only two temperatures.  At T 1, k 1 :  At T 2, k 2 :  By subtracting the second equation from the first, the following equation can be derived:  This equation can be found in section 1 of the data booklet.

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16. Main Menu In practice…  You will be using the method described previously to determine the activation energy for: S 2 O 3 2- (aq) + 2H + (aq)  SO 2 (aq) + S(s) + H 2 O(l)  Follow the instructions hereinstructions here  You may wish to use the spreadsheet template here for your calculations: http://mrjdfield.edublogs.org/2014/02/14/topic-6-kinetics/ http://mrjdfield.edublogs.org/2014/02/14/topic-6-kinetics/

17. Main Menu Recap  Activation energy can be determined by the gradient of a graph of ln k vs 1/T