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

Options

A. `2xx10^(-4)`
B. `10^(-4)`
C. `2xx10^4`
D. `10^4`

Answer

To solve the problem, we need to find the equilibrium concentration of Z in the reaction:

\[ X(g) + Y(g) \rightleftharpoons Z(g) \]

Given that at equilibrium:

\[ [X] = \frac{1}{2}[Y] = \frac{1}{2}[Z] \]

Let’s denote the concentration of Z as \( [Z] = 2x \). Then, we can express the concentrations of X and Y in terms of \( x \):

1. \( [X] = x \)
2. \( [Y] = 2x \)
3. \( [Z] = 2x \)

Now, we can write the expression for the equilibrium constant \( K_c \):

\[ K_c = \frac{[Z]}{[X][Y]} \]

Substituting the expressions for \( [X] \), \( [Y] \), and \( [Z] \):

\[ K_c = \frac{2x}{x \cdot 2x} \]

This simplifies to:

\[ K_c = \frac{2x}{2x^2} = \frac{1}{x} \]

We know from the problem that \( K_c = 10^4 \) mol\(^{-1}\) L. Therefore, we can set up the equation:

\[ \frac{1}{x} = 10^4 \]

From this, we can solve for \( x \):

\[ x = 10^{-4} \text{ mol L}^{-1} \]

Now, since we need to find \( [Z] \):

\[ [Z] = 2x = 2 \times 10^{-4} \text{ mol L}^{-1} = 2 \times 10^{-4} \text{ mol L}^{-1} \]

Thus, the value of \( [Z] \) at equilibrium is:

\[ \boxed{2 \times 10^{-4} \text{ mol L}^{-1}} \]