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Step-by-Step Solution
Step 1: Identify the Given Equilibrium and Its Constant
The original reaction is:
$ \text{N}_2(g) + \text{O}_2(g) \leftrightharpoons 2\,\text{NO}(g) $
The equilibrium constant for this reaction is:
$ K_c = 4 \times 10^{-4} $
Step 2: Write the Target Reaction and Relate It to the Given Reaction
We want the equilibrium constant for the reaction:
$ \text{NO}(g) \leftrightharpoons \tfrac{1}{2}\,\text{N}_2(g) + \tfrac{1}{2}\,\text{O}_2(g). $
Notice that this reaction is effectively the reverse of half the given reaction. Specifically, the given reaction produces $2\,\text{NO}$ from $\text{N}_2$ and $\text{O}_2$, while our target reaction consumes $1\,\text{NO}$ to produce half a molecule of each $\text{N}_2$ and $\text{O}_2$.
Step 3: Use the Rules for Manipulating Equilibrium Constants
1. When we reverse a reaction, the new equilibrium constant is the reciprocal of the original.
2. When we multiply or divide the coefficients of a balanced equation by a factor, we raise the equilibrium constant to that same factor’s power.
Here, we have the reaction reversed and also multiplied by $\tfrac{1}{2}$ compared to the original reaction. If the original reaction is:
$ \text{N}_2 + \text{O}_2 \leftrightharpoons 2\,\text{NO}, $
and $ K_c(\text{original}) = 4 \times 10^{-4}, $
then for the reversed reaction $ 2\,\text{NO} \leftrightharpoons \text{N}_2 + \text{O}_2, $ the new equilibrium constant would be $ \frac{1}{4 \times 10^{-4}}. $
Now, since we need only half of that reversed reaction:
$ \text{NO} \leftrightharpoons \tfrac{1}{2}\,\text{N}_2 + \tfrac{1}{2}\,\text{O}_2, $
we take the square root of the reciprocal of the original $K_c$. Hence
$ K'_c = \frac{1}{\sqrt{4 \times 10^{-4}}} .$
Step 4: Calculate the New Equilibrium Constant
We have:
$ \sqrt{4 \times 10^{-4}} = \sqrt{4} \times \sqrt{10^{-4}} = 2 \times 10^{-2}. $
Therefore:
$ K'_c = \frac{1}{2 \times 10^{-2}} = \frac{1}{0.02} = 50. $
Final Answer
The equilibrium constant for the reaction
$ \text{NO}(g) \leftrightharpoons \tfrac{1}{2}\,\text{N}_2(g) + \tfrac{1}{2}\,\text{O}_2(g) $
is
50.
