Chem 128 Kinetics Lab

• Some background
• Specific Questions and goals
• Some representative calculations
• Some typical data (from the afternoon lab section)

First, let's look at the Bromine + Acetone reaction

• Goals (Your report should answer these three questions)
  • 1. What is the order of the reaction with respect to H+, Acetone and Bromine
  • 2. what is the rate constant for the reaction at room temperature
  • 3. what is the energy of activation for the reaction

• How to analyze results
  • Remember [Br₂] is much smaller than the concentration of acetone or H+
  • So during a reaction run, both acetone and H+ stay nearly constant

• Run 1-- let's us see the Absorbance and how it changes with time.
  • you probably found a very nice straight line, at least until the absorbance goes to zero
  • when [Br₂] drops to half of the starting value,
    • the slope stays the same
  • so the rate of reaction (slope) is unaffected by [Br2]
    • in other words the reaction is zero order in Br₂-- see Goal #1 above
• Run 4 should confirm that
  • it starts with less Bromine, but other conditions are the same as run 1
    • slope of the graph is the same
    • slope (rate) doesn't change when we change [Br2]

• How does the rate change when we decrease [acetone] in run 2
  • measure new slope
  • compare to run #1
  • compute order n = log(slope1 / slope2) / log ([acetone1]/[acetone2])
    • n = log (S1/S2) / log ( 5/3)
  • n= order with respect to acetone -- see Goal #1 above
    • m is usually rounded off to the nearest integer

• How does the rate change when we decrease [H+] in run 3
  • measure new slope
  • compare to run #1
  • compute order m = log(slope1/slope3) / log ([H+,1]/[H+,2])
    • m = log (S1/S2) / log ( 5/3)
    • m= order with respect to acetone -- see Goal #1 above
      • m is usually rounded off to the nearest integer

• Now we need to find the rate constant
  • First, let's determine the actual rate of reaction
  • Let's chose run 1 (rate is different for runs 2, 3)
  • Slope of the graph is {delta Abs} / {delta- time}
  • We really want {delta [Br₂]} / {delta time}
    • [Br2] = Abs / 160.
    • 160. is the molar extinction coefficient for Br₂ at 400 nm
  • If you used computerized data collection, it allowed you to enter the molar extinction coefficient and it replots [Br₂]. If you did this and asked for the slope, you may have the actual reaction rate.
  • If you determined the slope of the Abs vs time plot...
    • Simply divide that slope by 160

• Now we can compute the rate constant
  • rate(1) = k [H+]^m [acetone]ⁿ where n and m were found above
    • [acetone] = 4.0 M * (10/40)
      • since we added 10 ml in a 40 ml reaction mix
    • [H+] = 1.0 * (10/40)
  • solve for k-- that's Goal #2

• Presumably you did two more runs, identical to run 1 except at different temperature
  • Rate(1, cold) = as above, using slope for the cold reaction
    • find k
  • Rate(1, warm) = as above
  • plot these three rate constants using ln (k) vs /T
    • find slope
    • Ea = -8.314 * slope -- that's Goal #3