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NUMBER OF MOLECUES WITH A PARTICULAR ENERGY

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 ).

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NUMBER OF MOLECUES WITH A PARTICULAR ENERGY

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  1. 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

  2. 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) NUMBER OF MOLECUES WITH A PARTICULARENERGY Ea MOLECULAR ENERGY

  3. INCREASING TEMPERATURE TEMPERATURE T2 > T1 MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY T1 T2 NUMBER OF MOLECUES WITH A PARTICULARENERGY EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER Ea MOLECULAR ENERGY 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

  4. INCREASING TEMPERATURE T3 MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY T1 NUMBER OF MOLECUES WITH A PARTICULARENERGY T2 TEMPERATURE T2 > T1 > T3 MOLECULAR ENERGY REVIEW no particles have zero energy/velocity some particles have very low and some have very high energies/velocities most have intermediate velocities as the temperature increases the curves flatten, broaden and shift to higher energies

  5. Heterogenous vs homogenous

  6. ADDING A CATALYST • 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. WITHOUT A CATALYST WITH A CATALYST

  7. ADDING A CATALYST MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULARENERGY EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER Ea MOLECULAR ENERGY 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 curveafterEashowing that more molecules have energies in excess of the activation energy

  8. 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)

  9. 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.

  10. 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

  11. Arhhenius equation R = 8.314Jk-1mol-1 T is absolute temperature Fraction of molecules possessing sufficient energy for the reaction, is given by e -Ea/RT

  12. Arhhenius equation • Take natural log of both sides: ln k = (-Ea/R)x (1/T) + ln A Y= mx + c

  13. 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.

  14. What did we measure in our practical?

  15. Gradient = -Ea/R R= 8.314J/K/mol Answer: Activation energy = 48.1KJmol-1

  16. 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)

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