Question

Kc for the reaction A(g) + 2B(g) \(\rightleftharpoons\) C(g) + 2D(g) is 4.0 at 300 K. What is the value of K'c for the reaction, 2C(g) + 4D(g) \(\rightleftharpoons\) 2A(g) + 4B(g) at 300 K?

• A. 8.0
• B. 1/8
• C. 1/2
• D. 16
• E. 1/16

Hint:

Always apply operations in order: Reverse the reaction first (\(1/K\)), then raise to the power of the coefficient multiplier.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The equilibrium constant (\(K_c\)) changes predictably when the chemical equation is manipulated:
1. If the reaction is reversed, the new constant is \(1/K_c\).
2. If the coefficients are multiplied by a factor \(n\), the new constant is \(K_c^n\).

Step 3: Detailed Explanation:
Original Reaction (1): \(A + 2B \rightleftharpoons C + 2D\) with \(K_c = 4.0\).
Target Reaction (2): \(2C + 4D \rightleftharpoons 2A + 4B\).

Compare Reaction (2) to Reaction (1):
1. Reaction (1) has been reversed. New constant = \(1/4\).
2. The entire equation has then been multiplied by a factor of 2.
Therefore:
\[ K'_c = \left( \frac{1}{K_c} \right)^2 \]
\[ K'_c = \left( \frac{1}{4} \right)^2 = \frac{1}{16} \]

Step 4: Final Answer:
The value of \(K'_c\) is 1/16.