LESSON 8 Activation Energy
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 → P2 fast P2 + Q → R + P slow What is the order with respect to P and Q? PQ A.11 B.12 C.21 D.22
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LESSON 8: ACTIVATION ENERGY Objectives: Understand the term activation energy
The common relationship between rate and temperature is that for every 10 degrees Celsius you are essentially doubling the rate. This is not a law, it is generally used as a rule of thumb. We have learned that the rate of a reaction depend on two things. k, the rate constant Concentration of reactants Since increasing the temp doesn’t effect the concentration, its effect goes to k. k is a general measure of the rate of a reaction at a given temperature.
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?
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 Kelvins R is the gas constant, J mol -1 K - 1. e is Euler’s number A is the ‘frequency factor’
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
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.
RECAP Activation energy can be determined by the gradient of a graph of ln k vs 1/T