Question

For the reaction,
\[ 2A(g) + B_2(g) \rightleftharpoons 2AB_2(g) \] the equilibrium constant, \(K_p\) at \(300\,K\) is \(16.0\). The value of \(K_p\) for \(AB_2(g) \rightleftharpoons A(g) + \dfrac{1}{2}B_2(g)\) is

• A. 8
• B. 0.25
• C. 0.125
• D. 32

Hint:

If reaction is reversed, \(K\) becomes \(1/K\). If reaction is multiplied by \(n\), new \(K = K^n\).

Answer 1

The Correct Option is B

Step 1: Given equilibrium.
\[ 2A + B_2 \rightleftharpoons 2AB_2 \quad K_p = 16 \] Step 2: Required reaction is reverse and half.
Target:
\[ AB_2 \rightleftharpoons A + \frac{1}{2}B_2 \] This is exactly \(\frac{1}{2}\) of the reverse of the given reaction.
Step 3: Reverse reaction constant.
Reverse of given reaction:
\[ 2AB_2 \rightleftharpoons 2A + B_2 \Rightarrow K_p' = \frac{1}{16} \] Step 4: Take half reaction.
When coefficients are divided by 2, equilibrium constant becomes square root:
\[ K_p'' = \sqrt{K_p'} = \sqrt{\frac{1}{16}} = \frac{1}{4} = 0.25 \] Final Answer: \[ \boxed{0.25} \]