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Derive the relation between Kp and Kc for the following reactions.
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Detailed Solution
In chemical equilibrium involving gases, the equilibrium constants Kp and Kc are related by the equation:
Kp = Kc (RT)Δn
Where:
- Kp = equilibrium constant in terms of partial pressures
- Kc = equilibrium constant in terms of concentrations
- R = universal gas constant
- T = temperature in Kelvin
- Δn = change in the number of moles of gas, calculated as nproducts − nreactants
i) Reaction: 2 HI (g) ⇌ H2 (g) + I2 (g)
For this reaction, the change in the number of moles (Δn) is calculated as:
Δn = nproducts − nreactants = (1 + 1) − 2 = 0
Substituting into the Kp and Kc relation formula:
Kp = Kc (RT)0
Since any number raised to the power of 0 is 1:
Kp = Kc
ii) Reaction: N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g)
For this reaction, the change in the number of moles (Δn) is:
Δn = nproducts − nreactants = 2 − (1 + 3) = −2
Substituting into the Kp and Kc relation formula:
Kp = Kc (RT)−2
Since (RT)−2 is less than 1:
Kp < Kc
iii) Reaction: 2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g)
For this reaction, the change in the number of moles (Δn) is:
Δn = nproducts − nreactants = 2 − (2 + 1) = −1
Substituting into the Kp and Kc relation formula:
Kp = Kc (RT)−1
Since (RT)−1 is less than 1:
Kp < Kc
iv) Reaction: CaCO3 (s) ⇌ CaO (s) + CO2 (g)
For this reaction, the change in the number of moles (Δn) is:
Δn = nproducts − nreactants = 1 − 0 = 1
Substituting into the Kp and Kc relation formula:
Kp = Kc (RT)1
Since (RT)1 is greater than 1:
Kp > Kc