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When N2O5 is heated at certain temperature, it dissociates as N2O5 (g) ⇌ N2O3 (g) + O2 (g); Kc = 2.5. At the same time N2O3 also decomposes as:
N2O3 (g) ⇌ N2O (g) + O2 (g).
If initially 4.0 moles of N2O5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O2 at equilibrium is 2.5 M. Equilibrium concentration of N2O5 is :
N2O3 (g) ⇌ N2O (g) + O2 (g).
If initially 4.0 moles of N2O5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O2 at equilibrium is 2.5 M. Equilibrium concentration of N2O5 is :
- a)1.0 M
- b)1.5 M
- c)2.166 M
- d)1.846 M
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
When N2O5 is heated at certain temperature, it dissociates as N2O5 (g)...
Q [O2] = x + y = 2.5
for N2O5, Kc = [N2O5] [O2]/ [N2O5]
and 2.5 = (x +y (x -y)/4-x
∴ x = 2.166
[N2O5] = 4 – x = 1.846
Most Upvoted Answer
When N2O5 is heated at certain temperature, it dissociates as N2O5 (g)...
Solution:
Given reaction: N2O5 (g) N2O3 (g) O2 (g) ; Kc = 2.5
1. Write the equilibrium expression for the given reaction.
Kc = [N2O3] [O2] / [N2O5]
2. Write the equilibrium expression for the second reaction.
Kc' = [N2O] [O2] / [N2O3]
3. Write the balanced equation for the overall reaction.
N2O5 (g) N2O3 (g) O2 (g)
4. Calculate the initial concentration of N2O5.
Initial concentration of N2O5 = 4.0 mol / 1.0 L = 4.0 M
5. Let x be the amount of N2O5 that dissociates to form N2O3 and O2.
N2O5 (g) N2O3 (g) O2 (g)
Initial: 4.0 M 0 M 0 M
Change: -x +x +x
Equilibrium: 4.0-x M x M x M
6. Use the equilibrium concentrations to calculate the equilibrium constant.
Kc = [N2O3] [O2] / [N2O5]
2.5 = x * x / (4.0 - x)
7. Solve for x using the quadratic formula.
x = (4.0 - √(16 - 10x)) / 5
8. Use the equilibrium concentrations of N2O3 and O2 to calculate the equilibrium concentration of N2O5.
[N2O3] = x = 1.846 M
[O2] = x = 2.5 M
Kc' = [N2O] [O2] / [N2O3]
2.5 = [N2O] * 2.5 / 1.846
[N2O] = 3.389 M
[N2O5] = (4.0 - x) = 2.154 M
Therefore, the equilibrium concentration of N2O5 is 1.846 M.
Given reaction: N2O5 (g) N2O3 (g) O2 (g) ; Kc = 2.5
1. Write the equilibrium expression for the given reaction.
Kc = [N2O3] [O2] / [N2O5]
2. Write the equilibrium expression for the second reaction.
Kc' = [N2O] [O2] / [N2O3]
3. Write the balanced equation for the overall reaction.
N2O5 (g) N2O3 (g) O2 (g)
4. Calculate the initial concentration of N2O5.
Initial concentration of N2O5 = 4.0 mol / 1.0 L = 4.0 M
5. Let x be the amount of N2O5 that dissociates to form N2O3 and O2.
N2O5 (g) N2O3 (g) O2 (g)
Initial: 4.0 M 0 M 0 M
Change: -x +x +x
Equilibrium: 4.0-x M x M x M
6. Use the equilibrium concentrations to calculate the equilibrium constant.
Kc = [N2O3] [O2] / [N2O5]
2.5 = x * x / (4.0 - x)
7. Solve for x using the quadratic formula.
x = (4.0 - √(16 - 10x)) / 5
8. Use the equilibrium concentrations of N2O3 and O2 to calculate the equilibrium concentration of N2O5.
[N2O3] = x = 1.846 M
[O2] = x = 2.5 M
Kc' = [N2O] [O2] / [N2O3]
2.5 = [N2O] * 2.5 / 1.846
[N2O] = 3.389 M
[N2O5] = (4.0 - x) = 2.154 M
Therefore, the equilibrium concentration of N2O5 is 1.846 M.
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When N2O5 is heated at certain temperature, it dissociates as N2O5 (g) N2O3 (g) + O2 (g); Kc = 2.5. At the same time N2O3 also decomposes as:N2O3 (g) N2O (g) + O2 (g).If initially 4.0 moles of N2O5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O2 at equilibrium is 2.5 M. Equilibrium concentration of N2O5 is :a)1.0 Mb)1.5 Mc)2.166 Md)1.846 MCorrect answer is option 'D'. Can you explain this answer?
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When N2O5 is heated at certain temperature, it dissociates as N2O5 (g) N2O3 (g) + O2 (g); Kc = 2.5. At the same time N2O3 also decomposes as:N2O3 (g) N2O (g) + O2 (g).If initially 4.0 moles of N2O5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O2 at equilibrium is 2.5 M. Equilibrium concentration of N2O5 is :a)1.0 Mb)1.5 Mc)2.166 Md)1.846 MCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When N2O5 is heated at certain temperature, it dissociates as N2O5 (g) N2O3 (g) + O2 (g); Kc = 2.5. At the same time N2O3 also decomposes as:N2O3 (g) N2O (g) + O2 (g).If initially 4.0 moles of N2O5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O2 at equilibrium is 2.5 M. Equilibrium concentration of N2O5 is :a)1.0 Mb)1.5 Mc)2.166 Md)1.846 MCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When N2O5 is heated at certain temperature, it dissociates as N2O5 (g) N2O3 (g) + O2 (g); Kc = 2.5. At the same time N2O3 also decomposes as:N2O3 (g) N2O (g) + O2 (g).If initially 4.0 moles of N2O5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O2 at equilibrium is 2.5 M. Equilibrium concentration of N2O5 is :a)1.0 Mb)1.5 Mc)2.166 Md)1.846 MCorrect answer is option 'D'. Can you explain this answer?.
When N2O5 is heated at certain temperature, it dissociates as N2O5 (g) N2O3 (g) + O2 (g); Kc = 2.5. At the same time N2O3 also decomposes as:N2O3 (g) N2O (g) + O2 (g).If initially 4.0 moles of N2O5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O2 at equilibrium is 2.5 M. Equilibrium concentration of N2O5 is :a)1.0 Mb)1.5 Mc)2.166 Md)1.846 MCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When N2O5 is heated at certain temperature, it dissociates as N2O5 (g) N2O3 (g) + O2 (g); Kc = 2.5. At the same time N2O3 also decomposes as:N2O3 (g) N2O (g) + O2 (g).If initially 4.0 moles of N2O5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O2 at equilibrium is 2.5 M. Equilibrium concentration of N2O5 is :a)1.0 Mb)1.5 Mc)2.166 Md)1.846 MCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When N2O5 is heated at certain temperature, it dissociates as N2O5 (g) N2O3 (g) + O2 (g); Kc = 2.5. At the same time N2O3 also decomposes as:N2O3 (g) N2O (g) + O2 (g).If initially 4.0 moles of N2O5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O2 at equilibrium is 2.5 M. Equilibrium concentration of N2O5 is :a)1.0 Mb)1.5 Mc)2.166 Md)1.846 MCorrect answer is option 'D'. Can you explain this answer?.
Solutions for When N2O5 is heated at certain temperature, it dissociates as N2O5 (g) N2O3 (g) + O2 (g); Kc = 2.5. At the same time N2O3 also decomposes as:N2O3 (g) N2O (g) + O2 (g).If initially 4.0 moles of N2O5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O2 at equilibrium is 2.5 M. Equilibrium concentration of N2O5 is :a)1.0 Mb)1.5 Mc)2.166 Md)1.846 MCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice When N2O5 is heated at certain temperature, it dissociates as N2O5 (g) N2O3 (g) + O2 (g); Kc = 2.5. At the same time N2O3 also decomposes as:N2O3 (g) N2O (g) + O2 (g).If initially 4.0 moles of N2O5 are taken in 1.0 litre flask and allowed to dissociate. Concentration of O2 at equilibrium is 2.5 M. Equilibrium concentration of N2O5 is :a)1.0 Mb)1.5 Mc)2.166 Md)1.846 MCorrect answer is option 'D'. Can you explain this answer? tests, examples and also practice JEE tests.
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