In which one of the following equilibria, Kp ≠ Kc?

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JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)
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  1. 2C (s) + O2 (g) ⇌ 2CO (g) 
  2. 2HI (g) ⇌ H2 (g) + I2 (g)
  3. NO (g) + SO2 (g) ⇌ NO (g) + SO3 (g)
  4. 2NO (g) ⇌ N2 (g) + O2 (g)

Answer (Detailed Solution Below)

Option 1 : 2C (s) + O2 (g) ⇌ 2CO (g) 
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JEE Main 04 April 2024 Shift 1
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Detailed Solution

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Calculation:

We know that, KP = Kc(RT)∆ng

KP and Kc are equilibrium constant.

R is gas constant

T is temperature

∆ng is differences of moles.

∴ If ∆ng ≠ 0 then Kp ≠ Kc

Now, 2C (s) + O2 (g) ⇌ 2CO (g)

∆ng = +1

∴ KP = Kc (RT)1

Hence, Kp ≠ Kc
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