Question

At T(K), the equilibrium constant for the reaction \( A (g) \rightleftharpoons B (g) \) is \( K_c \). If the reaction takes place in the following form \( 2A(g) \rightleftharpoons 2B(g) \), its equilibrium constant is \( K'_c \). The correct relationship between \( K_c \) and \( K'_c \) is:

• A. \( K'_c = (K_c)^2 \)
• B. \( K'_c = (K_c)^{\frac{1}{2}} \)
• C. \( K'_c = (K_c)^{-1} \)
• D. \( K'_c = K_c \)

Hint:

When the stoichiometry of a reaction is multiplied by a factor, the equilibrium constant is raised to the power of that factor.

Answer 1

The Correct Option is A

The equilibrium constant for a reaction is affected by the stoichiometric coefficients of the balanced equation. If the reaction is modified by changing the coefficients, the equilibrium constant changes as well. For the reaction \( A(g) \rightleftharpoons B(g) \), the equilibrium constant is \( K_c \). When the stoichiometry is doubled, i.e., for the reaction \( 2A(g) \rightleftharpoons 2B(g) \), the new equilibrium constant \( K'_c \) will be related to the original equilibrium constant \( K_c \) by the following rule: \[ K'_c = (K_c)^2 \] This is because the equilibrium constant is raised to the power corresponding to the stoichiometric coefficients in the balanced equation.

Answer 2

Step 1: Understanding the Given Reaction
The original reaction is:
\( A (g) \rightleftharpoons B (g) \) with equilibrium constant \( K_c \).

Step 2: Reaction Change and Its Effect on Equilibrium Constant
The new reaction is:
\( 2A (g) \rightleftharpoons 2B (g) \).
This is basically the original reaction multiplied by 2.

Step 3: Relationship Between Equilibrium Constants
When the coefficients of a balanced chemical equation are multiplied by a factor n, the equilibrium constant for the new reaction becomes the original equilibrium constant raised to the power n.
Mathematically,
If reaction is multiplied by n, then
\( K'_c = (K_c)^n \).

Step 4: Applying to the Given Problem
Here, n = 2, so
\( K'_c = (K_c)^2 \).

Step 5: Conclusion
Therefore, the equilibrium constant \( K'_c \) for the reaction \( 2A \rightleftharpoons 2B \) is the square of the original equilibrium constant \( K_c \).