Step 1: Concept Overview:
This problem uses rules for manipulating equilibrium constants (K) based on changes to the corresponding chemical equations. This is similar to Hess's Law for enthalpy, but with different math.
Step 2: Key Formulas:
The rules are:
1. Reverse a reaction: \(K_{new} = 1/K_{old}\).
2. Multiply a reaction by 'n': \(K_{new} = (K_{old})^n\).
3. Add reactions: \(K_{new} = K_a \times K_b\).
Step 3: Detailed Solution:
Target reaction: \(\frac{1}{2}\)P\(_2\) + \(\frac{1}{2}\)Q\(_2\) + \(\frac{1}{2}\)R\(_2\) \(\rightleftharpoons\) PQR.
Manipulate given reactions to match the target.
Reaction 1: 2PQ \(\rightleftharpoons\) P\(_2\) + Q\(_2\), K\(_1\) = 4 \(\times\) 10\(^4\).
Reverse Reaction 1 because P\(_2\) and Q\(_2\) are on the left in the target reaction.
Reversed Reaction 1: P\(_2\) + Q\(_2\) \(\rightleftharpoons\) 2PQ.
\(K'_{1} = \frac{1}{K_1} = \frac{1}{4 \times 10^4} = 0.25 \times 10^{-4}\).
Multiply the reversed reaction by \(\frac{1}{2}\) to match coefficients in the target.
Modified Reaction 1: \(\frac{1}{2}\)P\(_2\) + \(\frac{1}{2}\)Q\(_2\) \(\rightleftharpoons\) PQ.
\(K''_{1} = (K'_{1})^{1/2} = (0.25 \times 10^{-4})^{1/2} = \sqrt{0.25} \times \sqrt{10^{-4}} = 0.5 \times 10^{-2}\).
Reaction 2: PQ + \(\frac{1}{2}\)R\(_2\) \(\rightleftharpoons\) PQR, K\(_2\) = 5 \(\times\) 10\(^{-3}\).
Reaction 2 already has PQR on the right and \(\frac{1}{2}\)R\(_2\) on the left.
Add the modified reactions:
(\(\frac{1}{2}\)P\(_2\) + \(\frac{1}{2}\)Q\(_2\)) + (PQ + \(\frac{1}{2}\)R\(_2\)) \(\rightleftharpoons\) (PQ) + (PQR)
PQ cancels out, leaving:
\(\frac{1}{2}\)P\(_2\) + \(\frac{1}{2}\)Q\(_2\) + \(\frac{1}{2}\)R\(_2\) \(\rightleftharpoons\) PQR
Step 4: Final Answer:
Multiply the equilibrium constants.
\[ K_3 = K''_{1} \times K_2 = (0.5 \times 10^{-2}) \times (5 \times 10^{-3}) \]\[ K_3 = 2.5 \times 10^{-5} \]This corresponds to option (2).