Question

Which of the following relation is correct for gaseous and reversible reactions?

• A. \( \frac{K_C}{K_P} = (RT)^{\Delta n_g} \)
• B. \( K_P = (P)^{\Delta n_g} \)
• C. \( \frac{K_C}{K_X} = (P)^{-\Delta n_g} \)
• D. \( \frac{K_C}{K} = \left( \frac{P}{RT} \right)^{\Delta n_g} \)

Hint:

In reactions involving gases, the equilibrium constant in terms of pressure \( K_P \) is related to the equilibrium constant in terms of concentration \( K_C \) by \( K_P = K_C \times (RT)^{\Delta n_g} \), where \( \Delta n_g \) is the difference in the number of moles of gaseous products and reactants.

Answer 1

The Correct Option is B

For reversible reactions involving gases, the relationship between the equilibrium constants in terms of concentration \( K_C \) and pressure \( K_P \) is given by: \[ K_P = K_C \times (RT)^{\Delta n_g} \] where \( \Delta n_g \) is the change in the number of moles of gas between products and reactants.
This is derived from the ideal gas law.
In the given options, \( K_P \) is correctly related to \( P^{\Delta n_g} \) in option (b), which is a simplified form of the correct relationship.
Thus, the correct answer is (b).