In the system `X+2YhArrZ`, the equilibrium concentration are,
`[X]=0.06" mol L"^(-1),[Y]=0.12" mol L"^(-1)`,
`[Z]=0.216" mol L"^(-1)`. Find the equilibrium constant of the reaction.

Options

A. 250
B. 500
C. 125
D. 273

Explanation

To find the equilibrium constant \( K_{eq} \) for the reaction \( X + 2Y \rightleftharpoons Z \), we can follow these steps:

### Step 1: Write the expression for the equilibrium constant
The equilibrium constant expression for the reaction is given by:

\[ K_{eq} = \frac{[Z]}{[X][Y]^2} \]

### Step 2: Substitute the equilibrium concentrations into the expression
We have the following equilibrium concentrations:
- \([X] = 0.06 \, \text{mol L}^{-1}\)
- \([Y] = 0.12 \, \text{mol L}^{-1}\)
- \([Z] = 0.216 \, \text{mol L}^{-1}\)

Now we can substitute these values into the equilibrium constant expression:

\[ K_{eq} = \frac{0.216}{(0.06)(0.12)^2} \]

### Step 3: Calculate \([Y]^2\)
First, calculate \((0.12)^2\):

\[ (0.12)^2 = 0.0144 \]

### Step 4: Substitute and calculate the denominator
Now substitute \((0.12)^2\) back into the equation:

\[ K_{eq} = \frac{0.216}{(0.06)(0.0144)} \]

### Step 5: Calculate the denominator
Now calculate the denominator:

\[ (0.06)(0.0144) = 0.000864 \]

### Step 6: Calculate \( K_{eq} \)
Now substitute this value back into the equation for \( K_{eq} \):

\[ K_{eq} = \frac{0.216}{0.000864} \]

### Step 7: Perform the final calculation
Now perform the division:

\[ K_{eq} = 250 \]

### Conclusion
Thus, the equilibrium constant \( K_{eq} \) for the reaction is:

\[ K_{eq} = 250 \]

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