Title: Lesson 6 Activation Energy

Learning Objectives:

– Understand the term activation energy

– Calculate activation energy from experimental data

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Recap1.Two species, P and Q, react together according to the

following equation.

P + Q → R

The accepted mechanism for this reaction is

P + P → 2P fast2P + Q → R + P slow

What is the order with respect to P and Q?P Q

A. 1 1

B. 1 2

C. 2 1

D. 2 2

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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?

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The rate constant k is temperature dependent 10oC 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 Ea temp rise causes significant increase in number of particles reacting

• Small Ea temp rise causes a less significant increase in number of particles reacting.

Lower Ea

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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 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.

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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.

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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

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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.

Use the following data to find the activation energy value for the reaction H2 + I2 2HI which is second order overall.

Temp /oC 200 300 500 600

Rate constant k / mol-1dm3s-1 3.07 x 10-9 2.76 x 10-6 3.02 x 10-2 6.07 x 10-1

Temp/K 473 573 773 873

ln k 5.298317 5.703782 6.214608 6.39693

1/T / K-1 0.002114 0.001745 0.001294 0.001145

- 15.4

0.00077

Gradient = -15.4 / 0.00077= - 20000 KEa = - Grad x R= -(-20000) x 8.314= +166280 J mol-1

= +166.28 kJ mol-1

DO NOT includeorigin for x-axis

Use the following data to find (a) the activation energy value for the reaction 2N2O(g) 2N2(g) + O2(g) and (b) the rate constant at 900K.

- 7.00

0.00023

(a) Gradient = - 7.00 / 0.00023= - 30400 KEa = - Grad x R

= -(-30400) x 8.314= +252746 J mol-1

= +252.8 kJ mol-1

Rate constant k /mol-1dm3s-1 0.0011 0.3800 1.6700 11.5900

Temp /K 838 1001 1053 1125

ln k -6.812 -0.968 0.513 2.450

1/T /K-1 0.001193 0.000999 0.000950 0.000889

(b) At T = 900K 1/T = 0.00111 K-1

from graph,ln k = -4.3

k = e-4.3

0.0136 mol-1dm3s-

1

0.00111

4.3

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Solving simultaneous equations

Activation energy can also be calculated from values of the rate constant, k, at only two temperatures.

At T1, k1:

At T2, k2:

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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Solutions

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In practice… You will be using the method described

previously to determine the activation energy for:

S2O32-(aq) + 2H+(aq) SO2(aq) + S(s) + H2O(l)

Follow the instructions here

You may wish to use the spreadsheet template here for your calculations: http://mrjdfield.edublogs.org/2014/02/14/topic-6-kinetics/

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Recap Activation energy can be determined by the

gradient of a graph of ln k vs 1/T