i can demonstrate an understanding of the terms ‘rate of reaction’, ‘rate equation’,...
TRANSCRIPT
I can demonstrate an understanding of the terms ‘rate of reaction’, ‘rate equation’, ‘order of reaction’, ‘rate
constant’, ‘half-life’, ‘rate-determining step’, ‘activation energy’, ‘heterogeneous and homogeneous catalyst’
(4.3a)
LO: To recall the effects of temperature and catalysts on activation energies
and draw reaction profiles
The curve shown is T1. Draw curve T2 (higher temperature) and curve T3 (lower temperature).
Shade in the effects on activation energies and explain next to these areas, why this is the case (refer to the molecules taking part)
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MOLECULAR ENERGY
Ea
Explanation
increasing the temperature gives more particles an energy greater than Ea
more reactants are able to overcome the energy barrier and form products
a small rise in temperature can lead to a large increase in rate
T1
T2
TEMPERATURE
T2 > T1
Ea
MAXWELL-BOLTZMANN DISTRIBUTION OF
MOLECULAR ENERGY
MAXWELL-BOLTZMANN DISTRIBUTION OF
MOLECULAR ENERGY
INCREASING TEMPERATUREINCREASING TEMPERATURE
MOLECULAR ENERGY
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EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER
REVIEWno particles have zero energy/velocitysome particles have very low and some have very high energies/velocitiesmost have intermediate velocitiesas the temperature increases the curves flatten, broaden and shift to higher energies
T1
T2
T3
TEMPERATURE
T2 > T1 > T3
MAXWELL-BOLTZMANN DISTRIBUTION OF
MOLECULAR ENERGY
MAXWELL-BOLTZMANN DISTRIBUTION OF
MOLECULAR ENERGY
INCREASING TEMPERATUREINCREASING TEMPERATURE
MOLECULAR ENERGY
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Heterogenous vs homogenous
• Catalysts provide an alternative reaction pathway with a lower activation energy (Ea)
• Decreasing the activation energy means that more particles will have sufficient energy to overcome the energy barrier and react
• Catalysts remain chemically unchanged at the end of the reaction.
ADDING A CATALYSTADDING A CATALYST
WITHOUT A CATALYST WITH A CATALYST
The area under the curve beyond Ea corresponds to the number of molecules with sufficient energy to overcome the energy barrier and react.
Lowering the Activation Energy, Ea, results in a greater area under the curve after Ea
showing that more molecules have energies in excess of the activation energy
Ea
MAXWELL-BOLTZMANN DISTRIBUTION OF
MOLECULAR ENERGY
MAXWELL-BOLTZMANN DISTRIBUTION OF
MOLECULAR ENERGY
ADDING A CATALYSTADDING A CATALYST
EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER
MOLECULAR ENERGY
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• I can deduce from experimental data for reactions with zero, first and second order kinetics:
• i) half-life (the relationship between half-life and rate constant will be given if required)
• ii) order of reaction• iii) rate equation• iv) rate-determining step related to reaction mechanisms• v) activation energy (by graphical methods only; the
Arrhenius equation will be given if needed) (4.3f)
The rate equation
• The rate equation shows the effect of changing the concentrations of the reactants on the rate of the reaction. What about all the other things (like temperature and catalysts, for example) which also change rates of reaction? Where do these fit into this equation?
• These are all included in the so-called rate constant - which is only actually constant if all you are changing is the concentration of the reactants. If you change the temperature or the catalyst, for example, the rate constant changes.
Arhhenius equation
- Can be rearranged to find the activation energy of a reaction
- This would be used in experiments where the temperature has NOT been kept constant.
As a rule of thumb in most biological and chemical reactions, the reaction rate doubles when the temperature increases every 10 kelvin
Arhhenius equation
R = 8.314Jk-1mol-1
T is absolute temperature
Fraction of molecules possessing sufficient energy for the reaction, isgiven by e -Ea/RT
• Take natural log of both sides:
ln k = (-Ea/R)x (1/T) + ln A
Y= mx + c
Arhhenius equation
The determination of activation energy requires kinetic data, i.e., the rate constant, k, of the reaction determined at a variety of temperatures.
We then construct a graph of lnk on the y-axis and 1/T on the x-axis, where T is the temperature in Kelvin.
What did we measure in our practical?
Answer: Activation energy = 48.1KJmol-1
Gradient = -Ea/R
R= 8.314J/K/mol
• I can deduce from experimental data for reactions with zero, first and second order kinetics:
• i) half-life (the relationship between half-life and rate constant will be given if required)
• ii) order of reaction• iii) rate equation• iv) rate-determining step related to reaction mechanisms• v) activation energy (by graphical methods only; the
Arrhenius equation will be given if needed) (4.3f)