Question

For the reaction
\[ \text{A} + \text{B} \rightleftharpoons \text{C}, \quad K_p = 2.7 \times 10^{-5} \] Calculate \( K_p \) for the reaction
\[ \frac{1}{3} \text{A} + \frac{1}{3} \text{B} \rightleftharpoons \frac{1}{3} \text{C} \]

• A. \( 3 \times 10^{-3} \)
• B. \( \frac{1}{3} \times 10^{-3} \)
• C. \( 9 \times 10^{-3} \)
• D. \( 3 \times 10^{-2} \)

Answer 1

The Correct Option is D

Step 1: Understand the effect of changing the stoichiometry on the equilibrium constant.
For the given reaction: \[ \text{A} + \text{B} \rightleftharpoons \text{C}, \quad K_p = 2.7 \times 10^{-5} \] We are asked to find \( K_p \) for the reaction: \[ \frac{1}{3} \text{A} + \frac{1}{3} \text{B} \rightleftharpoons \frac{1}{3} \text{C} \] We notice that the coefficients of all species are divided by 3. When we divide the coefficients in the balanced reaction by a constant, the equilibrium constant changes according to the power of the factor by which the coefficients are divided. Specifically, if the reaction is multiplied or divided by a factor \( n \), the new equilibrium constant is: \[ K'_p = (K_p)^n \] Step 2: Apply the rule to the given reaction.
In our case, we have divided all the coefficients by 3. Thus, the new equilibrium constant is: \[ K'_p = (K_p)^{1/3} \] \[ K'_p = (2.7 \times 10^{-5})^{1/3} \] Step 3: Calculate the new equilibrium constant.
Taking the cube root of \( 2.7 \times 10^{-5} \): \[ (2.7 \times 10^{-5})^{1/3} \approx 3 \times 10^{-2} \] \( 3 \times 10^{-2} \).