Question

Consider the following reactions in which all the reactants and products are present in gaseous state:
\(2xy \rightleftharpoons x_2 + y_2\) \(K_1 = 2.5 \times 10^5\)
\(xy + \frac{1}{2}z_2 \rightleftharpoons xyz\) \(K_2 = 5 \times 10^{-3}\)
The value of \(K_3\) for the equilibrium \(\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz\) is:

• A. \(2.5 \times 10^{-3}\)
• B. \(2.5 \times 10^3\)
• C. \(1.0 \times 10^{-5}\)
• D. \(5 \times 10^{-3}\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When reactions are added, their equilibrium constants are multiplied. When a reaction is reversed, $K$ is inverted ($1/K$). When a reaction is multiplied by a factor $n$, the new constant is $K^n$.
Step 2: Key Formula or Approach:
Identify how to combine the given equations to get the target equation.
Step 3: Detailed Explanation:
Target Eq: $\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz$ 1. Reverse and multiply Eq 1 by $\frac{1}{2}$: $\frac{1}{2}x_2 + \frac{1}{2}y_2 \rightleftharpoons xy$
New constant $K' = (1 / K_1)^{1/2} = (1 / 2.5 \times 10^5)^{1/2} = (4 \times 10^{-6})^{1/2} = 2 \times 10^{-3}$. 2. Keep Eq 2 as is: $xy + \frac{1}{2}z_2 \rightleftharpoons xyz$
Constant $K_2 = 5 \times 10^{-3}$. 3. Add these two modified equations: $(\frac{1}{2}x_2 + \frac{1}{2}y_2) + (xy + \frac{1}{2}z_2) \rightleftharpoons xy + xyz$ Cancel $xy$ from both sides: $\frac{1}{2}x_2 + \frac{1}{2}y_2 + \frac{1}{2}z_2 \rightleftharpoons xyz$ $K_3 = K' \times K_2 = (2 \times 10^{-3}) \times (5 \times 10^{-3}) = 10 \times 10^{-6} = 1.0 \times 10^{-5}$.
Step 4: Final Answer:
The value of $K_3$ is $1.0 \times 10^{-5}$.