Question

In which of the following \( K_c = K_p \)? a) \( \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \) b) \( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \)

• A. Reaction a)
• B. Reaction b)
• C. Both reactions
• D. None of the above

Hint:

For \( K_c = K_p \) to hold, the change in the number of moles of gas \( \Delta n \) must be zero.

Answer 1

The Correct Option is A

The equilibrium constant \( K_p \) and \( K_c \) are related by the following equation: \[ K_p = K_c \left( RT \right)^{\Delta n} \] Where: - \( \Delta n \) is the change in the number of moles of gas (products - reactants), - \( R \) is the gas constant, - \( T \) is the temperature. For the reaction to have \( K_c = K_p \), the change in the number of moles of gas \( \Delta n \) must be zero. Let's examine both reactions: ### Reaction a: \( \text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g) \) In this reaction, the number of moles of gas on both sides is the same: 1 mole of \( \text{PCl}_5 \) on the left and 1 mole of \( \text{PCl}_3 \) and 1 mole of \( \text{Cl}_2 \) on the right. Thus, \( \Delta n = 0 \), and therefore \( K_c = K_p \). ### Reaction b: \( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \) In this reaction, there are 2 moles of gas on the left and 2 moles of gas on the right, so \( \Delta n = 0 \), and \( K_c = K_p \) holds true here as well. Thus, the correct answer is: \[ \boxed{(C) \, \text{Both reactions}} \]