Question

In which of the following equilibrium $K_{P}=K_{C}$? ________.

• A. $CaCO_{3(s)} \rightleftharpoons CaO_{(s)}+CO_{2(g)}$
• B. $2SO_{2(g)}+O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
• C. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)}+Cl_{2(g)}$
• D. $H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)}$
• E. $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$

Hint:

Check if the number of gas molecules is the same on both sides.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The equilibrium constant can be expressed in terms of concentrations (\(K_c\)) or partial pressures (\(K_p\)). The relationship between them depends on the change in the number of moles of gaseous species in the reaction. We need to find the reaction where this change is zero.
Step 2: Key Formula or Approach
The relationship between \(K_p\) and \(K_c\) is given by: \[ K_p = K_c (RT)^{\Delta n_g} \] where \(\Delta n_g\) is the change in the number of moles of gas, calculated as: \[ \Delta n_g = (\text{total moles of gaseous products}) - (\text{total moles of gaseous reactants}) \] For \(K_p\) to be equal to \(K_c\), the term \((RT)^{\Delta n_g}\) must equal 1. Since R and T are not zero, this condition is only met when the exponent \(\Delta n_g = 0\).
Step 3: Detailed Explanation
We will calculate \(\Delta n_g\) for each given reaction.
(A) \(CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)\):
- Moles of gaseous products = 1 (for CO\(_2\)). - Moles of gaseous reactants = 0 (CaCO\(_3\) is solid). - \(\Delta n_g = 1 - 0 = 1\). Thus, \(K_p \neq K_c\). (B) \(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\):
- Moles of gaseous products = 2. - Moles of gaseous reactants = \(2 + 1 = 3\). - \(\Delta n_g = 2 - 3 = -1\). Thus, \(K_p \neq K_c\). (C) \(PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)\):
- Moles of gaseous products = \(1 + 1 = 2\). - Moles of gaseous reactants = 1. - \(\Delta n_g = 2 - 1 = 1\). Thus, \(K_p \neq K_c\). (D) \(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\):
- Moles of gaseous products = 2. - Moles of gaseous reactants = \(1 + 1 = 2\). - \(\Delta n_g = 2 - 2 = 0\). Thus, \(K_p = K_c(RT)^0 = K_c\). (E) \(N_2O_4(g) \rightleftharpoons 2NO_2(g)\):
- Moles of gaseous products = 2. - Moles of gaseous reactants = 1. - \(\Delta n_g = 2 - 1 = 1\). Thus, \(K_p \neq K_c\). Step 4: Final Answer
The condition \(K_p = K_c\) is met for the reaction \(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\).