Given that the equilibrium constant K_p for the reaction A(g) ⇌ 2 B(g) is 0.0450, what is the value of K_p for the reaction 2 A(g) ⇌ 4 B(g)?

• A. 0.2025
• B. 0.0900
• C. 0.0450
• D. 0.00203

Answer: (D)

Verified step by step guidance

1. Identify the given reaction and its equilibrium constant: For the reaction \(A(g) \rightleftharpoons 2 B(g)\), the equilibrium constant is \(K_p = 0.0450\).
2. Write the new reaction and relate it to the original: The new reaction is \$2 A(g) \rightleftharpoons 4 B(g)$, which is exactly the original reaction multiplied by 2.
3. Recall the rule for equilibrium constants when a reaction is multiplied by a factor \(n\): The new equilibrium constant \(K_p^{\prime}\) is related to the original by \(K_p^{\prime} = (K_p)^n\).
4. Apply this rule with \(n=2\): Calculate \(K_p^{\prime} = (0.0450)^2\) to find the equilibrium constant for the new reaction.
5. Express the final answer as \(K_p^{\prime}\) without calculating the numerical value, since the problem asks for the method rather than the final number.