Question

In which of the following equilibria, K_p and K_e are NOT equal?

• A. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
• B. H₂(g)+I₂(g) ⇌ 2HI(g)
• C. CO(g) +H₂O(g) ⇌ CO₂(g) +H₂(g)
• D. 2BrCl(g) ⇌ Br₂(g) + Cl₂(g)

Answer 1

The Correct Option is A

Step 1: Relationship Between \( K_p \) and \( K_c \) 

The equilibrium constant \( K_p \) is expressed in terms of partial pressures, and \( K_c \) is expressed in terms of concentrations. They are related by the following equation:

$$ K_p = K_c \left( RT \right)^{\Delta n} $$

• \( R \) is the gas constant.
• \( T \) is the temperature in Kelvin.
• \( \Delta n \) is the change in the number of moles of gas between products and reactants.

Step 2: Condition for \( K_p \) and \( K_c \) to be Equal

For \( K_p \) and \( K_c \) to be equal, \( \Delta n \) must be zero. This means that the number of moles of gas on both sides of the reaction must be the same.

Step 3: Analyze Each Option

Option 1: \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \)

• Here, \( \Delta n = 2 - (1 + 1) = 0 \).
• So, \( K_p = K_c \).

Option 2: \( CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \)

• Here, \( \Delta n = 1 + 1 - 1 - 1 = 0 \).
• So, \( K_p = K_c \).

Option 3: \( 2BrCl(g) \rightleftharpoons Br_2(g) + Cl_2(g) \)

• Here, \( \Delta n = 2 - 2 = 0 \).
• So, \( K_p = K_c \).

Option 4: \( PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \)

• Here, \( \Delta n = 2 - 1 = 1 \).
• So, \( K_p \neq K_c \).

Step 4: Conclusion

The correct answer is: Option (1): \( PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \), as \( K_p \neq K_c \).