In which of the following equilibria, Kp and Ke are NOT equal?
- PCl5(g) ⇌ PCl3(g) + Cl2(g)
- H2(g)+I2(g) ⇌ 2HI(g)
- CO(g) +H2O(g) ⇌ CO2(g) +H2(g)
- 2BrCl(g) ⇌ Br2(g) + Cl2(g)
The Correct Option is A
Solution and Explanation
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 \( K_p \neq K_c \).
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