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: Write the equilibrium constants for the given reactions.
For the first reaction: \[ X_2 + Y_2 \rightleftharpoons 2XY, \quad K' = \frac{1}{K_1} = \frac{1}{2.5 \times 10^{-5}} = 4 \times 10^4 \] For the second reaction: \[ \frac{1}{2} Z_2 + XY \rightleftharpoons XYZ, \quad K_2 = 5 \times 10^{-3} \]
Step 2: Combine the two reactions to get the desired equilibrium.
By adding equation (A) and equation (B): \[ \frac{1}{2} X_2 + \frac{1}{2} Y_2 + \frac{1}{2} Z_2 + XY \rightleftharpoons XYZ \]
Step 3: Calculate the overall equilibrium constant.
The overall equilibrium constant \(K_{\text{eq}}\) is the product of the individual equilibrium constants: \[ K_{\text{eq}} = K' \times K_2 = \left( \frac{1}{25 \times 10^{-6}} \right)^{1/2} \times 5 \times 10^{-3} \] \[ K_{\text{eq}} = 5 \times 10^{-3} \times \left( \frac{1}{25 \times 10^{-6}} \right)^{1/2} = 1 \]