Lab Section – Experiment 3
April 1, 2013
1) X and Y must be integers, how close were my values?
2) Error sources and impact on my values?
3) Explain why rate = [KMnO ] 4 is used when calculating order of
H C O
reaction with respect to 2 2 4 .
4) Answer postlab question 6.5 from the lab site
5) Explain why H 2 is added.
The experimental values found for X and Y were not integers. X was found to be
1.255, and Y was found to be 0.756. These values are not satisfyingly accurate,
but one can see that each was meant to be equal to 1.0.
Significant sources of error can be attributed to inconsistent timing of the
reaction taking place, not recognizing when the reaction had been completed
(color change disparities), and communication breakdowns within a group due to
the abnormally large number of people participating in each group’s experiment.
[KMnO ] 4
Rate = is used when calculating order of reaction with
respect to H 2 O2 4 because the concentrations of Oxalic Acid were not
changed from titration 1 to titration 2. Since H 2 O2 4 (1) = H 2 O2 4 (2),
these values cancel out.
(6.5) It is given that we have 30mL of 0.5M H C O , 15mL of
2 2 x4 y
0.021M KMnO 4 , and 5mL of water. Rate = K[ KMnO ¿ 4 [ H 2 O2¿ 4 .
From earlier questions, it is known that K=2.215E2, x=1.255, and y=0.756.
[ H 2 O2¿ 4 = 50mL =0.30M [ H C2O 2 4 ].
0.021M∗15mL KMnO 4
[KMnO ] 4 = 50mL =0.0063M ].