4 public questions tagged with this topic.
Kp = Kc (RT)Δ n , where Δ n = 2 - 1 = 1 . Thus, Kc = (Kp/RT) = (1.8/0.0831 × 500) = (1.8/41.55) ≈ 0.043 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier
Kp = Kc (RT)Δ n , Δ n = 1 - 2 = -1 , RT = 0.0831 × 300 = 24.93 . Thus, Kc = Kp × RT = 0.25 × 24.93 ≈ 6.23 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant
Kp = Kc (RT)Δ n , Δ n = 2 - 1 = 1 , RT = 0.0831 × 298 ≈ 24.76 . Thus, Kc = (Kp/RT) = (0.98/24.76) ≈ 0.04 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
Kp = Kc (RT)Δ n , Δ n = 1 - 2 = -1 , RT = 0.0831 × 500 = 41.55 . Thus, Kc = Kp × RT = 0.1 × 41.55 = 4.155 .
Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Chemical Equilibrium - Law of Mass Action and Equilibrium Constant
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