At 273 degrees Celsius, 2NO +Br2 = 2NOBr. The following data for the initial rat

Question

At 273 degrees Celsius, 2NO +Br2 = 2NOBr. The following data for the initial rate of appearance of NOBr: Experiment Concentration NO (M) Concentration Br2 (M) Initial Rate (M/s) 1 0.10 0.20 24 2 0.25 0.20 150 3 0.50 0.50 60 4 0.35 0.50 735 a) Calculate the average value of the rate constant for the appearance of NOBr from the four data sets. b) How is the rate of appearance of the NOBr related to the rate of disappearance of Br2? c) What is the rate of disappearance of Br2 when the concentration of NO is 0.75 M and Br2 is 0.25 M

Explanation / Answer

please rate :)

Experiment 1 is the "baseline" to compare how the initial rate changes with changing initial reactant concentrations. In Experiment 2, [NO]o is increased from Experiment 1 by (0.25 M) / (0.10) = 2.5x. The response of the reaction is to have the rate increase by a factor of (150 M/s) / (24 M/s) = 6.5x. Now it just so happens that 2.5² = 6.5, so it appears that the rate law is second order in [NO]. In Experiment 3, [Br2] has been increased from Experiment 1 by (0.50 M) / (0.20 M) = 2.5x. The increase in Rate was found to be (60 M/s) / (24 M/s) = 2.5x. So the reaction appears to be first order in [Br2].

One experimental point is not enough, so Experiment 4 was done to increase [NO] by 3.5x and [Br2] by 2.5x. The expected Rate is: (25 M/s) (3.5²) (2.5) = 735 m/s, which is what was found. This is enough proof for the rate law to be:
d[NOBr]/dt = k [NO]² [Br],
where k is the rate constant. k would have to be evaluated in another type of experiment.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.