Question

Given that the equilibrium constant for the reaction \( 2SO_3(g) + O_2(g) \rightleftharpoons 2SO_2(g) \) has a value of \( 2.78 \times 10^{3} \) at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature? \[ SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2}O_2(g) \]

• A. \( 1.8 \times 10^{-3} \)
• B. \( 3.6 \times 10^{-3} \)
• C. \( 6.0 \times 10^{-2} \)
• D. \( 1.3 \times 10^{-5} \)

Hint:

When the stoichiometry of a reaction is changed, the equilibrium constant is modified by raising the original constant to the power of the change in coefficients.

Answer 1

The Correct Option is A

Step 1: The equilibrium constant for the new reaction can be derived by taking the square root of the equilibrium constant for the original reaction.
Step 2: Since the stoichiometry is halved, the equilibrium constant for the new reaction is: \[ K' = \sqrt{K} = \sqrt{2.78 \times 10^{3}} = 1.8 \times 10^{-3}. \]
Final Answer: \[ \boxed{1.8 \times 10^{-3}} \]