Edexcel rate of reaction questions A2

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Edexcel rate of reaction questions A2

  1. 1. Question 1 (a)
  2. 2.  Half life 1 = 38 seconds  Half life 2 = 38 seconds  Half life 3 = 37 seconds  Equal half lives so 1st order reaction with respect to H2O2  Rate = k [H2O2]  k = Rate/ [H2O2] = 1.61 x 10-5s-1 from results at 12 seconds.
  3. 3. Question 1 (c) + (d)
  4. 4. Question 2 (a)
  5. 5. Question 2 (c)
  6. 6.  (a) H2O2 + 2I- + 2H+ = I2 + 2H2O  (b) Rate = k [H2O2] [I- ]  (c) Overall order = 2  (d) First step is the rate determining step – because it involves one molecule of hydrogen peroxide and one ion of iodine reacting and no hydrogen ions to confirm the rate equation.
  7. 7.  (a) Zero order with respect to X – see experiments 1,4 and 5 – an increase in the concentration of X does not affect the reaction rate.  (b) Second order with respect to Y – see experiments 1 and 2 – doubling the concentration of X results in a four times faster reaction.  Rate = k [Y]2
  8. 8.  Rate = k [Y]2  0.0001 = k (0.1)2  So k = 0.001 mol -1dm3s-1  Y + Y = Y2 ‡  Y2 ‡ + X = XY2  Chemists study rate equations in order to determine the mechanisms of reactions and then predict factors that could make reactions more economically viable.
  9. 9.  The sodium thiosulphate reacts with the iodine which is initially produced and remove it from the reaction mixture. This allow a fixed end point of the reaction to be determined once all the sodium thiosulphate is used up.  The starch solution detected the end of the initial reaction and the point at which iodine is no longer reacting with the sodium thiosulpate.
  10. 10. Question 5
  11. 11.  The iron (III) ions are acting as a catalyst and lower the activation energy.

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