Equilibrium constant K c for the following reaction at 800K is, 4. NH _ 3 1 2 ~N _ 2 + 3…

Chemistry · JEE Advanced · NTA ExamsEquilibrium

Equilibrium constant Kc for the following reaction at 800K is, 4.
\(\mathrm{NH}_{3} \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2} ;\) 
The value of KP for the following reaction will be :
\(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3}\)
  1. \(\left(\frac{800 \mathrm{R}}{4}\right)^{-2}\)
  2. 16 × (800R)2
  3. \(\left[\frac{1}{4 \times 800 \mathrm{R}}\right]^{2}\)
  4. (800R)1/2 4

Answer

(C) [ 1 4 800 R ]^ 2

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