The data was collected from the lab I did. I need help answering the results sec

Question

The data was collected from the lab I did. I need help answering the results section.

1) Use the data from part 1 and the equation 3a and 3b (which I gave below) to determine the order I^- (m) and S2O8^2- (n). please show equation.

2) What is the oveall reaction order and molecularity?

3) Write the rate law for the reaction?

4) The units for the rate constant k are?

5) Using the rate law determined by the data from Trials 1-4, calculate the rate constant k and show your work for each trial.

6) Calculate the mean value for the rate constant of these four trials at room temperature and the average deviation as a percent of the mean value: mean=? percent average deviation=?

7) Calculate the rate constants for Trials 5-7

8) Calculate the rate constant for Trial 8

Temperature

Trials delta t (sec) [I^-]o M [S2O8^2-]o M S2O3^2-]o M -delta[S2O3^2-]/2delta t

Temperature

1 55.21 0.0800 0.0400 1.0e-3 9.0e-6 19.6 2 99.63 0.0800 0.0200 1.0e-3 5.0e-6 19.8 3 99.63 0.0400 0.0400 1.0e-3 5.0e-6 19.8 4 168.07 0.0800 0.0010 1.0e-3 3.0e-6 19.3 5 42.38 0.0800 0.0400 1.0e-3 1.0e-5 39.4 6 79.19 0.0800 0.0400 1.0e-3 6.0e-6 12.3 7 166.15 0.0800 0.0400 1.0e-3 3.0e-6 1.3 8 11.99 0.0800 0.0400 1.0e-3 4.0e-5 18.9 Rate trial 1 -Als2O3 I trial 1[S208 lo, trial 1 2At 1 trial A Rate trial 2 S203 k[I lo, trial IS202 la, trial 2 2 24t trial 2 S208 o, trial 1 AIS203 /2At trial 1 Rate trial Rate trial 2 AIS20a trial 2 S208 lo, trial 2 (eq. 3a) (eq. 3b)

Explanation / Answer

1) Using equation 3b)
9*10-6 / 5*10-6 = (0.04/0.02)n
   1.8 = (2)n
   log (1.8) = n log 2
   n = 0.8479 = 1
Now use equation 3a) for trial 2 and 3,
   5*10-6 / 5*10-6 = (0.08/0.04)m (0.02/0.04)0.85
   1 = (2)m (0.5555)
   1.7998 = 2m ; log 1.7998 = m log 2
   m = 0.8479 = 1
2) Overall reaction order = m+n = 1+1 = 2
Molecularity = 2
3) Rate law will become = K [ I-] [S2O82-]
4) K = L2-1 mol1-2 s-1 = L mol-1 s-1
5) For trial 1,
   9*10-6 = K [0.08]1 [0.04]1
   K = 2.8125*10-3
For trial 2,
5*10-6 = K [0.08] [0.02]
K = 3.125*10-3
For trial 3,
   5*10-6 = K [0.04] [0.04]
   K = 3.125*10-3
   For trial 4,
   3*10-6 = K [ 0.08] [0.0010]
   K = 0.0375
Taking average of these 4 values,
K = [2.8125*10-3 + 2*3.125*10-3 + 0.0375] / 4 = 0.01148 L mol-1 s-1

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