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