At 550 K, the Kc for the following reaction is `10^(4) mol^(-1)L . X(g) + Y(g) rarr Z(g)` At equilibrium, it was observed that `[X] = 1/2[Y] = 1/2[Z]` . What is the value of `[Z]("in "mol^(-1)L)`at equilibrium ?

Options

A. `2 xx 10^(-4)`
B. `10^(-4)`
C. `2 xx 10^(4)`
D. `10^(4)`

Explanation

To solve the problem, we need to determine the equilibrium concentration of Z given the equilibrium constant \( K_c \) and the relationships between the concentrations of X, Y, and Z.

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
\[ X(g) + Y(g) \rightleftharpoons Z(g) \]

2. **Define the equilibrium concentrations:**
Let the concentration of Z at equilibrium be denoted as \([Z] = z\). According to the problem, we have:
\[ [X] = \frac{1}{2}[Y] = \frac{1}{2}[Z] \]
This implies:
\[ [Y] = 2[X] \quad \text{and} \quad [X] = \frac{1}{2}[Z] \]

3. **Express \([X]\) and \([Y]\) in terms of \([Z]\):**
From \([Z] = z\), we can express:
\[ [X] = \frac{1}{2}z \quad \text{and} \quad [Y] = 2[X] = 2 \left(\frac{1}{2}z\right) = z \]

4. **Substitute into the equilibrium constant expression:**
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[Z]}{[X][Y]} \]
Substituting the expressions for \([X]\), \([Y]\), and \([Z]\):
\[ K_c = \frac{z}{\left(\frac{1}{2}z\right)(z)} = \frac{z}{\frac{1}{2}z^2} = \frac{2}{z} \]

5. **Set \( K_c \) equal to the given value:**
We know that \( K_c = 10^4 \):
\[ 10^4 = \frac{2}{z} \]

6. **Solve for \( z \):**
Rearranging gives:
\[ z = \frac{2}{10^4} = 2 \times 10^{-4} \, \text{mol/L} \]

7. **Final result:**
The equilibrium concentration of Z is:
\[ [Z] = 2 \times 10^{-4} \, \text{mol/L} \]

### Summary:
The value of \([Z]\) at equilibrium is \( 2 \times 10^{-4} \, \text{mol/L} \).