Q2 / 87 · NTA Abhyas (JEE Main) 2020 · Chemistry · Chemical Equilibrium

Equilibrium constant KC for the following reaction at 800 K is, 4
NH3(g)⇌12N2(g)+32H2(g)
The value of KP for the following reaction will be
N2(g)+3H2(g)⇌2NH3(g)

  • A
    800R42-
  • B
    16×(800R)2
  • Correct answer
    C
    14×800R2
  • D
    800R1/24

Solution

NH 3 (g)⇌ 1 2 N 2 (g) +  3 2 H 2 (g);   K C  = 4
N2(g)+3H2(g)⇌2NH3(g) Δng=2-4=-2
So KC for the following reaction will be equal to 142
KP=KC(RT)Δng
=  1 4 2  ×  (800 × R) −2
= 1 4×800 R 2

Hence, the correct answer is option: 'C'.

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