For the reversible reaction, N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) at 500ºC, the value of K p is…
Chemistry · JEE Advanced · NTA Exams — Equilibrium
For the reversible reaction, N2(g) + 3H2(g) \(\text { ⇌ }\)2NH3(g) at 500ºC, the value of Kp is 1.44 × 10–5 when partial pressure is measured in atmospheres. The corresponding value of Kc, with concentration in mole litre–1 is (2000)
- \[\frac{1.44 \times 10^{-5}}{(0.082 \times 500)^{-2}}\]
- \(\frac{1.44 \times 10^{-5}}{(8.314 \times 773)^{-2}}\)
- \(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{2}}\)
- \(\frac{1.44 \times 10^{-5}}{(0.082 \times 773)^{-2}}\)
Answer
(D) 1.44 10^ -5 (0.082 773)^ -2
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