The equilibrium constant for the reaction `w+x hArr y+z` is `9`. If one mole of each of w and x are mixed and there is no change in volume, the number of moles of y for formed is

Options

A. (a) `0.10`
B. (b) `0.50`
C. (c) `0.75`
D. (d) `0.54`

Explanation

To solve the problem step by step, we will analyze the equilibrium reaction and the information given.

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction is given as:
\[ W + X \rightleftharpoons Y + Z \]

2. **Identify Initial Moles**:
We start with 1 mole of each reactant:
- Moles of W = 1
- Moles of X = 1
- Moles of Y = 0
- Moles of Z = 0

3. **Define Change in Moles**:
Let \( x \) be the number of moles of W and X that react to form Y and Z at equilibrium. Therefore:
- Moles of W at equilibrium = \( 1 - x \)
- Moles of X at equilibrium = \( 1 - x \)
- Moles of Y at equilibrium = \( x \)
- Moles of Z at equilibrium = \( x \)

4. **Write the Equilibrium Constant Expression**:
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[Y][Z]}{[W][X]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{x \cdot x}{(1 - x)(1 - x)} = \frac{x^2}{(1 - x)^2} \]

5. **Substitute the Given Equilibrium Constant**:
We know that \( K_c = 9 \):
\[ 9 = \frac{x^2}{(1 - x)^2} \]

6. **Cross-Multiply and Rearrange**:
Rearranging gives:
\[ 9(1 - x)^2 = x^2 \]
Expanding this:
\[ 9(1 - 2x + x^2) = x^2 \]
\[ 9 - 18x + 9x^2 = x^2 \]
\[ 8x^2 - 18x + 9 = 0 \]

7. **Solve the Quadratic Equation**:
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 8, b = -18, c = 9 \):
\[ x = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 8 \cdot 9}}{2 \cdot 8} \]
\[ x = \frac{18 \pm \sqrt{324 - 288}}{16} \]
\[ x = \frac{18 \pm \sqrt{36}}{16} \]
\[ x = \frac{18 \pm 6}{16} \]
This gives two possible solutions:
\[ x = \frac{24}{16} = 1.5 \quad \text{(not possible since we started with 1 mole each)} \]
\[ x = \frac{12}{16} = 0.75 \]

8. **Determine the Number of Moles of Y Formed**:
Since \( x \) represents the number of moles of Y formed at equilibrium, we find:
\[ \text{Moles of Y} = 0.75 \]

### Final Answer:
The number of moles of Y formed is **0.75 moles**.

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