Question

For which one of the following equilibria, \( K_P = K_C \)?

• A. \( {N}_2(g) + 3{H}_2(g) \rightleftharpoons 2{NH}_3(g) \)
• B. \( 2{SO}_2(g) + {O}_2(g) \rightleftharpoons 2{SO}_3(g) \)
• C. \( {N}_2{O}_4(g) \rightleftharpoons 2{NO}_2(g) \)
• D. \( {PCl}_5(g) \rightleftharpoons {PCl}_3(g) + {Cl}_2(g) \)
• E. \( {H}_2(g) + {Br}_2(g) \rightleftharpoons 2{HBr}(g) \)

Hint:

For the equilibrium constant expressions \( K_P \) and \( K_C \) to be equal, the number of moles of gaseous reactants must equal the number of moles of gaseous products. When this happens, \( K_P = K_C \).

Answer 1

Answer: (E)

The equilibrium constant expressions \( K_P \) and \( K_C \) relate to pressure and concentration respectively, and they are equal only for reactions where the total number of moles of gas does not change.
- \( K_P \) is used when dealing with partial pressures, and \( K_C \) is used when dealing with concentrations.
- \( K_P \) and \( K_C \) will be equal when the change in the number of moles of gases on both sides of the reaction is zero.
Let’s examine the options:
- (A) The reaction \( {N}_2(g) + 3{H}_2(g) \rightleftharpoons 2{NH}_3(g) \) involves 4 moles of gas on the left and 2 moles of gas on the right, so \( K_P \neq K_C \).
- (B) The reaction \( 2{SO}_2(g) + {O}_2(g) \rightleftharpoons 2{SO}_3(g) \) involves 3 moles of gas on the left and 2 moles of gas on the right, so \( K_P \neq K_C \).
- (C) The reaction \( {N}_2{O}_4(g) \rightleftharpoons 2{NO}_2(g) \) involves 1 mole of gas on the left and 2 moles of gas on the right, so \( K_P \neq K_C \).
- (D) The reaction \( {PCl}_5(g) \rightleftharpoons {PCl}_3(g) + {Cl}_2(g) \) involves 1 mole of gas on the left and 2 moles of gas on the right, so \( K_P \neq K_C \).
- (E) The reaction \( {H}_2(g) + {Br}_2(g) \rightleftharpoons 2{HBr}(g) \) involves 2 moles of gas on the left and 2 moles of gas on the right, so \( K_P = K_C \).
Thus, the correct answer is (E).