Finding orders from initial rates
Please Help in finding orders with respect to o3 and c2h4
Exp O3. C2H4 Initial rate
1 0.5. 1.0. 1.0
2. 2.0. 1.0. 4.0
3. 4.0. 2.0. 16.0
Since in experiment 1 and 3 the concentration of c2h4 doesn't changed I have found that the order with respect to o3 is 1st order
What is the order with respect to c2h4?
ALSO
Exp. NO. H2. Initial rate
1. 0.10. 0.20. 2.6
2. 0.10. 0.50. 6.5
3. 0.30. 0.50 58.5
The order with respect to H2 is first order but what about NO?
Posted from TSR Mobile
Exp O3. C2H4 Initial rate
1 0.5. 1.0. 1.0
2. 2.0. 1.0. 4.0
3. 4.0. 2.0. 16.0
Since in experiment 1 and 3 the concentration of c2h4 doesn't changed I have found that the order with respect to o3 is 1st order
What is the order with respect to c2h4?
ALSO
Exp. NO. H2. Initial rate
1. 0.10. 0.20. 2.6
2. 0.10. 0.50. 6.5
3. 0.30. 0.50 58.5
The order with respect to H2 is first order but what about NO?
Posted from TSR Mobile
I'm not sure how to actually work it out, I have in my head the order wrt C2H4 being second but I'm not sure how to work it out. If anyone could help?
Posted from TSR Mobile
Posted from TSR Mobile
Original post by Northern-lad
Please Help in finding orders with respect to o3 and c2h4
Exp O3. C2H4 Initial rate
1 0.5. 1.0. 1.0
2. 2.0. 1.0. 4.0
3. 4.0. 2.0. 16.0
Since in experiment 1 and 3 the concentration of c2h4 doesn't changed I have found that the order with respect to o3 is 1st order
What is the order with respect to c2h4?
ALSO
Exp. NO. H2. Initial rate
1. 0.10. 0.20. 2.6
2. 0.10. 0.50. 6.5
3. 0.30. 0.50 58.5
The order with respect to H2 is first order but what about NO?
Posted from TSR Mobile
Exp O3. C2H4 Initial rate
1 0.5. 1.0. 1.0
2. 2.0. 1.0. 4.0
3. 4.0. 2.0. 16.0
Since in experiment 1 and 3 the concentration of c2h4 doesn't changed I have found that the order with respect to o3 is 1st order
What is the order with respect to c2h4?
ALSO
Exp. NO. H2. Initial rate
1. 0.10. 0.20. 2.6
2. 0.10. 0.50. 6.5
3. 0.30. 0.50 58.5
The order with respect to H2 is first order but what about NO?
Posted from TSR Mobile
If the order wrt c2h4 is 1st order (which I think it is), then the order wrt o3 is also 1st order (going from exp. 2 to exp. 3 multiplies conc. of c2h4 by 2, which would put the rate at 8. The conc. of o3 also doubles going from exp. 2 to exp. 3. Doubling the 8 gives the 16, so the order wrt o3 is also 1st order).
Order wrt to H2 is 1st. I've found that the order wrt NO is then 2nd order. This is how I got it:
58.5 divided by 6.5 = 9
conc. of NO is tripled going from exp. 2 to exp. 3 (where the 6.5 and the 58.5 come from, respectively - the conc. of H2 stays the same)
square root of 9 = 3
so conc. of NO is 2nd order
Hope this makes sense. PM me or something if it doesn't.:3
Quick Reply
Related discussions
- chem alevel question
- AQA Chemistry A level Ka Question
- Rate equation question
- AQA A Level Chemistry Paper 3 20th June 2018 Unofficial Markscheme
- Initial rates a level chemistry
- alevel maths differential equations question
- Chem help rates
- Pilot training fees help
- When to start applying for internships.
- Achieving the Highest Growth in the G7 – Is it time to turbocharge our economy?
- Student finance
- 2Al(s) Fe2O3(s) → Al2O3(s) 2Fe(s), struggling with this question
- Diagnostic Radiography Degree Apprenticeship
- Unconscious bias - is it time to rethink how we address VAWG?
- Wjec alevel chemistry analysis
- A Level Biology Question
- Chemistry paper 1 a level
- Ask Us Anything!
- Edexcel A-Level Chem Paper 2 Advanced Organic and Physical Chemistry [Exam Chat]
- Rent
Latest
Trending
Last reply 1 week ago
Im confused about this chemistry question, why does it form these productsTrending
Last reply 1 week ago
Im confused about this chemistry question, why does it form these products
