The equilibrium constant for the reaction N _ 2 ( ~g )+ O _ 2 ( ~g ) 2 NO ( ~g ) at…
Chemistry · JEE Main · NTA Exams — Equilibrium
The equilibrium constant for the reaction
\(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \div 2 \mathrm{NO}(\mathrm{~g})\)
at temperature T is 4 × 10-4. The value of Kc for the reaction
\(\mathrm{NO}(\mathrm{~g})=\frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) at the same temperature:
- 2.5 × 102
- 50
- 4 × 10-4
- 0.02
Answer
(D) 0.02
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