The equilibrium constant (Kc) for the reaction N2(g) + O2(g) andrarr; 2NO(g) at temperature T is 4 andtimes; 10andndash;4. Thevalue of Kc for the reaction at the same temperature is:a)0.02b)2.5 andtimes; 102c)4 andtimes; 10andndash;4d)50.0Correct answer is option 'D'. Can you explain this answer?

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The equilibrium constant (K_c) for the reaction N₂(g) + O₂(g) → 2NO(g) at temperature T is 4 × 10^–4. Thevalue of K_c for the reaction

at the same temperature is:

a)
0.02
b)
2.5 × 10²
c)
4 × 10^–4
d)
50.0

Correct answer is option 'D'. Can you explain this answer?

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The equilibrium constant (Kc) for the reaction N2(g) + O2(g) → 2N...

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The equilibrium constant (Kc) for the reaction N2(g) + O2(g) → 2NO(g) at temperature T is 4 × 10–4. Thevalue of Kc for the reaction at the same temperature is:a)0.02b)2.5 × 102c)4 × 10–4d)50.0Correct answer is option 'D'. 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 is 4 × 10–4. Thevalue of Kc for the reaction at the same temperature is:a)0.02b)2.5 × 102c)4 × 10–4d)50.0Correct 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 The equilibrium constant (Kc) for the reaction N2(g) + O2(g) → 2NO(g) at temperature T is 4 × 10–4. Thevalue of Kc for the reaction at the same temperature is:a)0.02b)2.5 × 102c)4 × 10–4d)50.0Correct 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 The equilibrium constant (Kc) for the reaction N2(g) + O2(g) → 2NO(g) at temperature T is 4 × 10–4. Thevalue of Kc for the reaction at the same temperature is:a)0.02b)2.5 × 102c)4 × 10–4d)50.0Correct answer is option 'D'. Can you explain this answer?.

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