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? P Q A. 1 B. 2 C. D.

Arrhenius Equation

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?

Activation Energy

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 Ea is the activation energy T is the temperature measured in Kelvins R is the gas constant, 8.314 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 -Ea/R ‘x’ is 1/T ‘c’ is ln A

To determine Ea 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 ‘-Ea/R’, this can be rearranged to calculate Ea.

Arrhenius Equation A plot of ln k vs 1/T will give a straight line The slope of line will equal Ea/R The activation energy may be found by multiplying the slope by “R”

Graphing with Arrhenius Rate constants for the reaction CO(g) + NO2(g)  CO2(g) + NO(g) Were measured at four different temperatures. Plot the data to determine activation energy in kJ/mol.

Graphing 318 0.332 308 0.184 298 0.101 288 0.0521 T (Kelvin) k (M1 s1)‏

Graphing Steps: Make a column of ln k data Make a column of inverse temp (1/T) Plot ln k vs 1/T Calculate the slope Multiply slope by R

Arrhenius Another Way! Another useful arrangement of the Arrhenius equation enables calculation of: 1) Ea (with k at two different temps) 2) the rate constant at a different temperature (with Ea, k and temps)

Arrhenius Another Way! Use the data to calculate activation energy of the reaction mathematically 7.0 x 101 500 6.1 x 102 450 2.9 x 103 400 k (s1) T (Kelvin)

Arrhenius

Problems Find the activation energy (in kJ/mol) of the reaction if the rate constant at 600K is 3.4 M-1s-1 and 31.0 at 750K. Find the rate constant if the temperature is 289K, Activation Energy is 200kJ/mol and pre-exponential factor is 9 M-1s-1 Find the new rate constant at 310K if the rate constant is 7 M-1s-1 at 370K, Activation Energy is 900kJ/mol  Calculate the activation energy if the pre-exponential factor is 15 M-1s-1, rate constant is 12M-1s-1and it is at 22K Find the new temperature if the rate constant at that temperature is 15M-1s-1 while at temperature 389K the rate constant is 7M-1s1,  the Activation Energy is 600kJ/mol