Question

Calculate the equilibrium constant for the following equilibrium:
\[ \frac{1}{2} X_2 + \frac{1}{2} Y_2 \rightleftharpoons XYZ \] Given: \[ 2XY \rightleftharpoons X_2 + Y_2, \quad K_1 = 2.5 \times 10^{-5} \] \[ \frac{1}{2} Z_2 + XY \rightleftharpoons XYZ, \quad K_2 = 5 \times 10^{-3} \]

Answer 1

Correct Answer: 1

Step 1: Understanding the Question:
The goal is to find the equilibrium constant ($K_{eq}$) of a target reaction by combining and manipulating given reactions with known equilibrium constants.
Step 2: Key Formula or Approach:
1. If a reaction is reversed, its $K$ becomes $1/K$.
2. If a reaction is multiplied by a factor $n$, its $K$ becomes $K^n$.
3. If two reactions are added, their $K$ values are multiplied.
Step 3: Detailed Explanation:
Given reaction 1: $2XY \rightleftharpoons X_2 + Y_2 \quad K_1 = 2.5 \times 10^{-5}$
Reverse it: $X_2 + Y_2 \rightleftharpoons 2XY \quad K' = \frac{1}{K_1} = \frac{1}{2.5 \times 10^{-5}} = \frac{10^5}{2.5} = 4 \times 10^4$
Multiply by $1/2$: $\frac{1}{2}X_2 + \frac{1}{2}Y_2 \rightleftharpoons XY \quad K'' = (K')^{1/2} = \sqrt{4 \times 10^4} = 200 \quad \dots (1)$

Given reaction 2: $\frac{1}{2}Z_2 + XY \rightleftharpoons XYZ \quad K_2 = 5 \times 10^{-3} \quad \dots (2)$

Adding equations (1) and (2):
$\frac{1}{2}X_2 + \frac{1}{2}Y_2 + \frac{1}{2}Z_2 + XY \rightleftharpoons XY + XYZ$
$\frac{1}{2}X_2 + \frac{1}{2}Y_2 + \frac{1}{2}Z_2 \rightleftharpoons XYZ$

The equilibrium constant for the sum is the product of the individual constants:
$K_{eq} = K'' \times K_2 = 200 \times 5 \times 10^{-3} = 1000 \times 10^{-3} = 1$
Step 4: Final Answer:
The equilibrium constant for the final reaction is 1.