[Kinetics] Using Rate as a Conversion Factor

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Discussion Overview

The discussion revolves around a homework problem related to the decomposition of ozone into oxygen, specifically calculating the number of molecules of O2 produced per liter of atmosphere per day based on a given rate of reaction. The conversation includes stoichiometric calculations and interpretations of textbook answers.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for the number of O2 molecules formed, indicating a rate of 6.5 x 10-4 M O3/s and using stoichiometry to derive an answer of approximately 5.071248 x 1025 molecules/L.
  • Another participant suggests that the discrepancy with the textbook answer may be due to a misunderstanding of the units, proposing that the calculation is correct but the textbook answer lacks clarity regarding the time component.
  • A later reply questions the accuracy of the textbook answer, suggesting it may be a typographical error, and proposes that the correct interpretation should be 5.1 x 1025 instead of 5.1 x 1025 without the exponent notation.
  • One participant expresses frustration with the reliability of textbook answers, noting that errors are common in scientific literature.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the textbook answer, with some believing it to be a typo while others focus on the clarity of the unit representation. There is no consensus on the definitive correctness of the textbook answer.

Contextual Notes

Participants note potential limitations in the textbook's presentation of the answer, including the lack of clarity regarding the time unit and possible typographical errors. The discussion does not resolve these issues.

brbrett
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Homework Statement


Certain pollutants encourage the following decomposition of ozone: 2O 3(g) -> 3O 2(g), at a rate of 6.5 x 10-4 M O 3/s. How many molecules of O2 gas are formed in each litre of atmosphere every day by this process? (As this problem provides a rate in units of mol/L/s and requires molecules/L as an answer, we can simply leave the unit "L" in the denominator the entire time.)

Homework Equations


Stoichiometry, mole conversions, Mole concentration (molarity)

The Attempt at a Solution


2O 3(g) -> 3O 2(g)
6.5 E 23 * 3/2 = 0.000975molO2
0.000975moleO2/s * 6.02*1023/1mol * 86400/day = 5.071248 E 25
This answer is not correct according to my book. The book says the answer is 5.1 * 1025 molecules/L. Not sure why they didn't mention days in the answer, considering how that is part of the question.

I appreciate any help! I only just started learning this, and may have forgotten something regarding stoichiometry since I haven't done it for a couple of months. Please avoid telling the steps, since I want to understand it~
 
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brbrett said:

Homework Statement


How many molecules of O2 gas are formed in each litre of atmosphere every day by this process? (As this problem provides a rate in units of mol/L/s and requires molecules/L as an answer, we can simply leave the unit "L" in the denominator the entire time.)

The Attempt at a Solution


2O 3(g) -> 3O 2(g)
6.5 E 23 * 3/2 = 0.000975molO2
0.000975moleO2/s * 6.02*1023/1mol * 86400/day = 5.071248 E 25
This answer is not correct according to my book. The book says the answer is 5.1 * 1025 molecules/L. Not sure why they didn't mention days in the answer, considering how that is part of the question.
~
so I think the problem is with the units only? numerically your answer is same as the answer given in the book .Your attempt is also right ..
First you calculated the moles of O2 released per L in one second.Then you converted it for per day.You can drop the unit of time ..because they are asking in one day how many molecules of gas is released per L ...in the book answer provided is in molecules /L in one day.
 
I did think I was right. Just looked, and it seems there is a typo in the book. It says 5.1 x 1025, but if you change it to 5.1 x 10^25, it matches my answer (or close enough anyways). Gotta love it when textbook answers aren't proofread... >.>
 
some how I thought it as a typo error on this forum ...I thought you are asking why they did not use per day in the unit.
 
brbrett said:
I did think I was right. Just looked, and it seems there is a typo in the book. It says 5.1 x 1025, but if you change it to 5.1 x 10^25, it matches my answer (or close enough anyways). Gotta love it when textbook answers aren't proofread... >.>
I'm almost certain that the 5.1 x 1025 in your book is a typo. Their answer is a bit over 5000, and I doubt anyone would write that number as 5.1 x 1025. 5.1 x 1025 makes sense to me.

Lots of science and math books have incorrect answers, especially the more advanced ones. I've seen books that had the text formatted in one edition, but in a later edition, someone dropped the ball on formatting.
 

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