Question

Consider the following reactions in which all the reactants and products are in gaseous state. 
2PQ \(\rightleftharpoons\) P\(_2\) + Q\(_2\)              K\(_1\) = 4 \(\times\) 10\(^4\) 
PQ + \(\frac{1}{2}\)R\(_2\) \(\rightleftharpoons\) PQR             K\(_2\) = 5 \(\times\) 10\(^{-3}\) 
The value of K\(_3\) for the equilibrium \(\frac{1}{2}\)P\(_2\) + \(\frac{1}{2}\)Q\(_2\) + \(\frac{1}{2}\)R\(_2\) \(\rightleftharpoons\) PQR is 
 

• A. 2.5 x 10\(^{-1}\)
• B. 2.5 x 10\(^{-5}\)
• C. 1.25 x 10\(^{-5}\)
• D. 1.25 x 10\(^{-2}\)

Hint:

Remember the key difference in operations: for enthalpy (\(\Delta H\)), you add/subtract. For equilibrium constants (K), you multiply/divide. When changing coefficients by a factor 'n', \(\Delta H\) is multiplied by 'n', but K is raised to the power of 'n'.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem involves applying the rules for manipulating equilibrium constants (K) based on manipulations of the corresponding chemical equations. This is analogous to Hess's Law for enthalpy changes, but the mathematical operations are different.
Step 2: Key Formula or Approach:
The rules are as follows:
1. If a reaction is reversed, the new equilibrium constant is the reciprocal of the original: \(K_{new} = 1/K_{old}\).
2. If a reaction is multiplied by a factor 'n', the new equilibrium constant is the original raised to the power of 'n': \(K_{new} = (K_{old})^n\).
3. If two reactions are added together, the new equilibrium constant is the product of the original constants: \(K_{new} = K_a \times K_b\).
Step 3: Detailed Explanation:
Our target reaction is: \(\frac{1}{2}\)P\(_2\) + \(\frac{1}{2}\)Q\(_2\) + \(\frac{1}{2}\)R\(_2\) \(\rightleftharpoons\) PQR.
Let's manipulate the given reactions to obtain the target reaction.
Reaction 1: 2PQ \(\rightleftharpoons\) P\(_2\) + Q\(_2\), K\(_1\) = 4 \(\times\) 10\(^4\).
Our target has P\(_2\) and Q\(_2\) on the left side. So, we must reverse Reaction 1.
Reversed Reaction 1: P\(_2\) + Q\(_2\) \(\rightleftharpoons\) 2PQ.
The new constant is \(K'_{1} = \frac{1}{K_1} = \frac{1}{4 \times 10^4} = 0.25 \times 10^{-4}\).
Our target has coefficients of \(\frac{1}{2}\). So, we must multiply the reversed reaction by \(\frac{1}{2}\).
Modified Reaction 1: \(\frac{1}{2}\)P\(_2\) + \(\frac{1}{2}\)Q\(_2\) \(\rightleftharpoons\) PQ.
The new constant is \(K''_{1} = (K'_{1})^{1/2} = (0.25 \times 10^{-4})^{1/2} = \sqrt{0.25} \times \sqrt{10^{-4}} = 0.5 \times 10^{-2}\).
Reaction 2: PQ + \(\frac{1}{2}\)R\(_2\) \(\rightleftharpoons\) PQR, K\(_2\) = 5 \(\times\) 10\(^{-3}\).
This reaction already has PQR on the right and \(\frac{1}{2}\)R\(_2\) on the left, which matches our target.
Add the modified reactions:
(\(\frac{1}{2}\)P\(_2\) + \(\frac{1}{2}\)Q\(_2\)) + (PQ + \(\frac{1}{2}\)R\(_2\)) \(\rightleftharpoons\) (PQ) + (PQR)
The PQ intermediate cancels out, leaving the target reaction:
\(\frac{1}{2}\)P\(_2\) + \(\frac{1}{2}\)Q\(_2\) + \(\frac{1}{2}\)R\(_2\) \(\rightleftharpoons\) PQR
Step 4: Final Answer:
Since we added the two modified reactions, we must multiply their equilibrium constants to get the final K\(_3\).
\[ K_3 = K''_{1} \times K_2 = (0.5 \times 10^{-2}) \times (5 \times 10^{-3}) \] \[ K_3 = 2.5 \times 10^{-5} \] This corresponds to option (2).