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The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(g) at temperature T is
4×10-4. The value of Kc for the reaction,
NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:
NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:
[AIEEE 2012]
- a)4x10-4
- b)50.0
- c)0.02
- d)2.5x102
Correct answer is option 'B'. Can you explain this answer?
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The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(...
Most Upvoted Answer
The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(...
B is correct BECAUSE (´ﻌ`)
Kc for ...
NO(g)→ ½ N2(g) + ½O2(g)
:::is equal to 1/√Kc of this equation N2(g)+O2(g) → 2NO(g)
thus
Kc of
NO(g)→ ½ N2(g) + ½O2(g)
is
1/√ 4 x 10^ -4
= 100/2
=50
Kc for ...
NO(g)→ ½ N2(g) + ½O2(g)
:::is equal to 1/√Kc of this equation N2(g)+O2(g) → 2NO(g)
thus
Kc of
NO(g)→ ½ N2(g) + ½O2(g)
is
1/√ 4 x 10^ -4
= 100/2
=50
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The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(...
Given Information:
The equilibrium constant (Kc) for the reaction N2(g) + O2(g) → 2NO(g) at temperature T is 4×10^-4.
Explanation:
Calculating Kc for the given reaction:
The reaction given is:
NO(g) → 1/2 N2(g) + 1/2 O2(g)
The equilibrium constant (Kc) for this reaction can be calculated by taking the reciprocal of the equilibrium constant for the reverse reaction and raising it to the power of the stoichiometric coefficients:
Kc(reverse) = 1/Kc(forward) = 1/(4×10^-4) = 2500
Answer:
Therefore, the value of Kc for the reaction NO(g) → 1/2 N2(g) + 1/2 O2(g) at the same temperature is 2500, which is equivalent to 50.0 (option b).
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The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(g) at temperature T is4×10-4. The value of Kc for the reaction,NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:[AIEEE 2012]a)4x10-4b)50.0c)0.02d)2.5x102Correct answer is option 'B'. Can you explain this answer?
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The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(g) at temperature T is4×10-4. The value of Kc for the reaction,NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:[AIEEE 2012]a)4x10-4b)50.0c)0.02d)2.5x102Correct answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(g) at temperature T is4×10-4. The value of Kc for the reaction,NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:[AIEEE 2012]a)4x10-4b)50.0c)0.02d)2.5x102Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(g) at temperature T is4×10-4. The value of Kc for the reaction,NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:[AIEEE 2012]a)4x10-4b)50.0c)0.02d)2.5x102Correct answer is option 'B'. Can you explain this answer?.
The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(g) at temperature T is4×10-4. The value of Kc for the reaction,NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:[AIEEE 2012]a)4x10-4b)50.0c)0.02d)2.5x102Correct answer is option 'B'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(g) at temperature T is4×10-4. The value of Kc for the reaction,NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:[AIEEE 2012]a)4x10-4b)50.0c)0.02d)2.5x102Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(g) at temperature T is4×10-4. The value of Kc for the reaction,NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:[AIEEE 2012]a)4x10-4b)50.0c)0.02d)2.5x102Correct answer is option 'B'. Can you explain this answer?.
Solutions for The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(g) at temperature T is4×10-4. The value of Kc for the reaction,NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:[AIEEE 2012]a)4x10-4b)50.0c)0.02d)2.5x102Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(g) at temperature T is4×10-4. The value of Kc for the reaction,NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:[AIEEE 2012]a)4x10-4b)50.0c)0.02d)2.5x102Correct answer is option 'B'. Can you explain this answer?, a detailed solution for The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(g) at temperature T is4×10-4. The value of Kc for the reaction,NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:[AIEEE 2012]a)4x10-4b)50.0c)0.02d)2.5x102Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(g) at temperature T is4×10-4. The value of Kc for the reaction,NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:[AIEEE 2012]a)4x10-4b)50.0c)0.02d)2.5x102Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The equilibrium constant (Kc) for the reaction N2(g)+O2(g) → 2NO(g) at temperature T is4×10-4. The value of Kc for the reaction,NO(g)→ ½ N2(g) + ½O2(g) at the same temperature is:[AIEEE 2012]a)4x10-4b)50.0c)0.02d)2.5x102Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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