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Step-by-Step Solution
Step 1: Identify the Given Reactions and Their Equilibrium Constants
1) The first reaction:
$N_{2}(g) + O_{2}(g) \rightleftharpoons 2\,NO(g)$
The equilibrium constant for this reaction is $K_{1}$.
2) The second reaction:
$2\,NO(g) + O_{2}(g) \rightleftharpoons 2\,NO_{2}(g)$
The equilibrium constant for this reaction is $K_{2}$.
Step 2: State the Target Reaction and Its Desired Equilibrium Constant
We are asked to find the equilibrium constant, $K$, for the reaction:
$NO_{2}(g) \rightleftharpoons \frac{1}{2}N_{2}(g) + O_{2}(g)$
Step 3: Express the Intermediate Reactions in Terms of Square Root of $K_{1}$ and $K_{2}$
First, rewrite each reaction to form NO and NO2 from N2 and O2 in a single-molecule change format (i.e., per 1 mole of NO or NO2 instead of 2):
- For the reaction forming 1 mole of NO:
$N_{2}(g) + O_{2}(g) \rightleftharpoons 2\,NO(g)$
If we want to form just 1 NO, divide everything by 2:
$\frac{1}{2}N_{2}(g) + \frac{1}{2}O_{2}(g) \rightleftharpoons NO(g)$
Now the equilibrium constant for this modified reaction is $\sqrt{K_{1}}$ (because we halved the stoichiometric coefficients).
- For the reaction forming 1 mole of NO2:
$2\,NO(g) + O_{2}(g) \rightleftharpoons 2\,NO_{2}(g)$
Divide everything by 2:
$NO(g) + \frac{1}{2}O_{2}(g) \rightleftharpoons NO_{2}(g)$
The equilibrium constant for this modified reaction is $\sqrt{K_{2}}$.
Step 4: Combine the Two Modified Reactions to Form NO2 Directly from N2 and O2
Adding the modified reactions:
1) $\frac{1}{2}N_{2}(g) + \frac{1}{2}O_{2}(g) \rightleftharpoons NO(g)$ with equilibrium constant $\sqrt{K_{1}}$.
2) $NO(g) + \frac{1}{2}O_{2}(g) \rightleftharpoons NO_{2}(g)$ with equilibrium constant $\sqrt{K_{2}}$.
Upon addition, we get:
$\frac{1}{2}N_{2}(g) + O_{2}(g) \rightleftharpoons NO_{2}(g)$
The equilibrium constant for the combined reaction is the product of the individual equilibrium constants:
$K_{\text{combined}} = \sqrt{K_{1}} \times \sqrt{K_{2}} = \sqrt{K_{1} \, K_{2}}$
Step 5: Find the Desired Equilibrium Constant by Taking the Reverse of the Combined Reaction
The reaction we need is exactly the reverse of:
$\frac{1}{2}N_{2}(g) + O_{2}(g) \rightleftharpoons NO_{2}(g)$
That reverse reaction is:
$NO_{2}(g) \rightleftharpoons \frac{1}{2}N_{2}(g) + O_{2}(g)$
When a reaction is reversed, its equilibrium constant becomes the reciprocal. Hence, the required equilibrium constant is:
$K = \frac{1}{\sqrt{K_{1} \, K_{2}}} = \left(\frac{1}{K_{1} \, K_{2}}\right)^{\tfrac{1}{2}}$
Step 6: State the Final Answer
$\displaystyle K = \left[ \frac{1}{K_{1}\,K_{2}} \right]^{\frac{1}{2}}$
