Question

Consider the following gaseous equilibrium reactions (I), (II) and (III) with equilibrium constants \( K_1 \), \( K_2 \), and \( K_3 \) respectively:
I) \( \frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3 \)
II) \( 2NO \rightleftharpoons N_2 + O_2 \)
III) \( H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \)
The correct expression for the equilibrium constant for the gaseous equilibrium reaction: \[ 2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O \]

• A. \( \frac{K_1^3}{K_1 \times K_2} \)
• B. \( \frac{K_3^3}{K_1^2 \times K_2} \)
• C. \( \frac{K_3^2}{K_1^2 \times K_2} \)
• D. \( \frac{K_3}{K_1^{1/2} \times K_2} \)

Hint:

For equilibrium constant transformations, use: \[ K_{\text{new}} = K_1^x \times K_2^y \times K_3^z \] where exponents depend on reaction modifications.

Answer 1

The Correct Option is C

Step 1: Understanding Equilibrium Constant Rules When combining chemical reactions: - Reversed reactions: \( K_{\text{new}} = \frac{1}{K_{\text{original}}} \). - Multiplying coefficients: \( K_{\text{new}} = (K_{\text{original}})^n \), where \( n \) is the factor of multiplication. - Summing reactions: \( K_{\text{final}} = K_1 \times K_2 \times K_3 \). Step 2: Expressing the Given Reaction in Terms of \( K_1, K_2, K_3 \) Rewriting the target reaction: \[ 2NH_3 + \frac{5}{2} O_2 \rightleftharpoons 2NO + 3H_2O \] Using reactions: - I) \( NH_3 \) formation: The original equation involves \( \frac{1}{2} N_2 + \frac{3}{2} H_2 \rightleftharpoons NH_3 \), which needs to be reversed and multiplied by 2. - New constant: \( K_1^{-2} \). - II) \( NO \) formation: This reaction already matches, so \( K_2 \) is directly used. - III) \( H_2O \) formation: The original reaction involves \( H_2 + \frac{1}{2} O_2 \rightleftharpoons H_2O \), which needs to be multiplied by 3. - New constant: \( K_3^3 \). Step 3: Final Expression Multiplying all equilibrium constants: \[ K_{\text{final}} = K_3^3 \times K_2 \times K_1^{-2} \] \[ = \frac{K_3^2}{K_1^2 \times K_2} \] Conclusion Thus, the correct answer is: \[ \frac{K_3^2}{K_1^2 \times K_2} \]