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Step-by-step Solution
Step 1: Identify the Initial Reaction and Its Equilibrium Constant
The given reaction is
$N_2(g) + O_2(g) \rightarrow 2\,NO(g)$
and its equilibrium constant is
$K = 4 \times 10^{-4}$.
Step 2: Compare the New Reaction with the Old Reaction
The new reaction is
$NO(g) \rightarrow \tfrac{1}{2}\,N_2(g) + \tfrac{1}{2}\,O_2(g)$.
Notice that it is the reverse of half the original reaction.
Therefore, the new equilibrium constant ($K'$) can be derived from the old constant ($K$) by the relationship for reversing and changing the stoichiometric coefficients in the reaction.
Step 3: Determine the Relationship Between $K$ and $K'$
The original reaction forms 2 moles of NO. The new reaction essentially decomposes NO into half the amounts of $N_2$ and $O_2$.
When the direction of the reaction is reversed, the new equilibrium constant is the reciprocal of the original. Additionally, reducing the stoichiometric coefficients to half introduces a square-root operation.
Hence,
$K' = \dfrac{1}{\sqrt{K}}$.
Step 4: Substitute the Value of $K$ and Calculate $K'$
Given $K = 4 \times 10^{-4}$, so:
$K' = \dfrac{1}{\sqrt{4 \times 10^{-4}}} = \dfrac{1}{2 \times 10^{-2}} = 50.$
Step 5: State the Final Answer
Therefore, the value of the equilibrium constant for the reaction
$NO(g) \rightarrow \tfrac{1}{2}\,N_2(g) + \tfrac{1}{2}\,O_2(g)$
is $50.0$.
