Question:

For the given hypothetical reactions, the equilibrium constants are as follows:
\[ X \rightleftharpoons Y; \, K_1 = 1.0 \]
\[ Y \rightleftharpoons Z; \, K_2 = 2.0 \]
\[ Z \rightleftharpoons W; \, K_3 = 4.0 \]
The equilibrium constant for the reaction
\[ X \rightleftharpoons W\]
is:

Updated On: Nov 3, 2025
  • 6.0
  • 12.0
  • 8.0
  • 7.0
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The Correct Option is C

Approach Solution - 1

To determine the equilibrium constant for the overall reaction \( X \rightleftharpoons W \), we need to combine the given reactions and their respective equilibrium constants (\( K_1, K_2, \) and \( K_3 \)) in the following sequence:

  1. The reactions are:
    • \( X \rightleftharpoons Y \) with \( K_1 = 1.0 \)
    • \( Y \rightleftharpoons Z \) with \( K_2 = 2.0 \)
    • \( Z \rightleftharpoons W \) with \( K_3 = 4.0 \)
  2. The overall reaction \( X \rightleftharpoons W \) is the sum of the three individual reactions:
    • \( X \rightleftharpoons Y \), \( Y \rightleftharpoons Z \), and \( Z \rightleftharpoons W \)
  3. To find the equilibrium constant \( K \) for the overall reaction, we multiply the equilibrium constants of the individual reactions:
    • \( K = K_1 \times K_2 \times K_3 \)
  4. Substituting the given values:
    • \( K = 1.0 \times 2.0 \times 4.0 \)
  5. Simplifying the multiplication:
    • \( K = 8.0 \)

Thus, the equilibrium constant for the reaction \( X \rightleftharpoons W \) is \(8.0\).

Let's evaluate the options given:

  • 6.0: This value comes from incorrect multiplication.
  • 12.0: This option is irrelevant to the multiplication results.
  • 8.0: Correct, as derived from \( 1.0 \times 2.0 \times 4.0 \).
  • 7.0: This option is incorrect and irrelevant.

The correct answer is 8.0.

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Approach Solution -2

The equilibrium constant for the net reaction $X \rightleftharpoons W$ is the product of the individual constants:
\[K = K_1 \cdot K_2 \cdot K_3.\]
Substitute values:
\[K = 1 \cdot 2 \cdot 4 = 8.\]
Final Answer:
8.0

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