Question

At T(K), the equilibrium constant of \(A_2(g) + B_2(g) \rightleftharpoons C(g)\) is \(2.7 \times 10^{-5}\). What is the equilibrium constant for \(\frac{1}{3} A_2(g) + \frac{1}{3} B_2(g) \rightleftharpoons \frac{1}{3} C(g)\) at the same temperature?

• A. \((2.7 \times 10^{-5})^3\)
• B. \(6 \times 10^{-2}\)
• C. \(\sqrt{2.7 \times 10^{-5}}\)
• D. \(3 \times 10^{-2}\)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When a balanced chemical equation is multiplied by a stoichiometric factor \(n\), the new equilibrium constant \(K'\) is the original equilibrium constant \(K\) raised to the power of \(n\).
Step 2: Key Formula or Approach:
If Reaction 1 is: \(\text{A}_2 + \text{B}_2 \rightleftharpoons \text{C}\) with equilibrium constant \(K_1\).
And Reaction 2 is: \(\frac{1}{3}\text{A}_2 + \frac{1}{3}\text{B}_2 \rightleftharpoons \frac{1}{3}\text{C}\).
Reaction 2 is obtained by multiplying Reaction 1 by \(n = \frac{1}{3}\).
Therefore, \(K_2 = (K_1)^n = (K_1)^{1/3}\).
Step 3: Detailed Explanation:
Given \(K_1 = 2.7 \times 10^{-5}\).
We need to calculate \((2.7 \times 10^{-5})^{1/3}\).
To make the cube root calculation simpler, rewrite the scientific notation so the exponent is a multiple of 3:
\(2.7 \times 10^{-5} = 27 \times 10^{-6}\).
Now apply the cube root:
\(K_2 = (27 \times 10^{-6})^{1/3}\)
\(K_2 = (27)^{1/3} \times (10^{-6})^{1/3}\)
Since \(3^3 = 27\), \((27)^{1/3} = 3\).
\((10^{-6})^{1/3} = 10^{-2}\).
\(K_2 = 3 \times 10^{-2}\).
Step 4: Final Answer:
The new equilibrium constant is \(3 \times 10^{-2}\).