Arrhenius equation LO- Carry out calculations involving the Arrhenius equation. So far we have looked quantitatively at how to show the effect of a concentration change on rate of reaction. For example, The rate eqn for a rxn at 298K is: Rate = k [A] [B]. Some results are shown for the rxn. [A] mol dm-3 [B] mol dm-3 Initial rate mol dm-3 s-1 0.50 x 10-3 1.00 x 10-3 0.15 x 10-3 Questions 1) Calculate a value for K and its units. 2) When the temperature was increased to 310K, the rate of reaction increased to 0.30 x 10-3 mol dm-3 s-1. The concentration of the reactants were kept constant. Calculate a value for k. What can you deduce from your answer? The value of k increases when T increases. This shows that the effect of T on ror is taken into account by k.

Graph to represent relationship between temperature and k

Using the Arrhenius equation The Arrhenius equation shows how changes in temperature and activation energy affect the value of k. Activation energy- J mol-1 Gas constant- 8.31 J K-1 mol-1 Euler’s number- a constant (like pi) with a value 2.7182…… Rate constant- number of successful collisions resulting in a rxn per second Temperature- K k = A x e–(Ea / RT) Frequency factor (or attempt frequency)- this is a constant that represents the number of collisions per second (leading to a rxn or not) with the correct orientation to react. Fraction of molecules with enough energy to react

Using the Arrhenius equation k = A x e–(Ea / RT) 293K 50000 J mol -1 Example 1) A chemical reaction is carried out at 200C and has an Ea of 50kJ mol-1. Calculate the fraction of molecules that are able to react at this temperature. (Gas constant = 8.31 J K-1 mol-1) The value of A is 4.79x109 s-1. Calculate the value of k. Working A) Work out the value of –(Ea/RT) -(50000 / (8.31 x 293) = -20.5353 Work out the fraction of molecules able to react e-20.5353 = 1.21x10-9 B) Calculate K k = 4.79x109 x 1.21x10-9 = 5.78 e

What effect has the 100C (10K) increase in T had on the value of K? 2) The same chemical reaction from 1) is carried out at 300C. Calculate the fraction of molecules that are able to react at this temperature. (Gas constant = 8.31 J K-1 mol-1) The value of A is 4.79x109 s-1. Calculate the value of k. A) B) k = 4.79x109 x 2.38x10-9 = 11.4 What effect has the 100C (10K) increase in T had on the value of K? How will this affect the rate of reaction when the rate equation is applied (assuming all concentrations in the rate eqn remain constant)? The value of k has doubled so the ror will double.

Question By working out values for k, calculate how much faster a reaction with an activation energy of 40 kJ mol-1 would be at 40°C than at 20°C, assuming the concentrations of everything are the same. R = 8.31 J K-1 mol-1. The value of A for these reactions is 6.14x106. Does a 100C increase in T always lead to a doubling of the rate of reaction? Explain your answer.

Answer At 20°C (= 293 K), the fraction of molecules having energies equal to or greater than EA is e-(40000/(8.31 x 293)) = 7.33 x 10-8 The value of k at 200C is k = 7.33x10-8 x 6.14x106 = 0.45 At 40°C (= 313 K), the fraction of molecules having energies equal to or greater than EA is e-(40000/(8.31 x 313)) = 2.10 x 10 -7 The value of k at 400C is k = 2.10x10-7 x 6.14x106 = 1.29 The value of the rate constant, k, has increased by a factor of 1.29 / 0.45 = 2.86 If the “rates double for an increase of 10°C” rule applied, this would give an increase in rate of 4 times. This rule only works for reactions with an activation energy of about 50 kJ mol-1 at temperatures around room temperature.

ln k = ln A – Ea / RT Other forms of the Arrhenius equation This lesson and next we will look at other forms of the Arrhenius equation. You are not expected to understand the derivation of the eqn and it will be given to you in the exam. Another form of the Arrhenius eqn is ln k = ln A – Ea / RT ln is a form of logarithm. You just need to be able to use the button on your calculator. Question A slow reaction has a rate constant k = 6.51 × 10−3 mol−1 dm3 at 300 K. Use the equation ln k = ln A – Ea / RT to calculate a value, in kJ mol−1, for the activation energy of this reaction. The constant A = 2.57 × 1010 mol−1 dm3. The gas constant R = 8.31 J K−1 mol−1.  (2)

Answer Ea = RT(lnA – lnk) or -Ea = RT(lnk – lnA) 1 Ea = 8.31 × 300 (23.97 – (–5.03)) = 72300 J mol–1 / 1000 = 72.3 kJmol-1