5 mole of X are mixed with 3 moles of Y. At equilibrium for the reaction, `X + Y rarr Z` 2 moles of Z are formed. The equilibrium constant for the reaction will be

Options

A. `2//3`
B. `1//2`
C. `3//2`
D. `1//4`

Explanation

To find the equilibrium constant for the reaction \( X + Y \rightleftharpoons Z \), we will follow these steps:

### Step 1: Write the balanced equation
The balanced equation for the reaction is:
\[ X + Y \rightleftharpoons Z \]

### Step 2: Identify initial moles of reactants and products
Initially, we have:
- Moles of \( X \) = 5
- Moles of \( Y \) = 3
- Moles of \( Z \) = 0 (since no product is formed initially)

### Step 3: Set up the change in moles at equilibrium
Let \( x \) be the amount of \( Z \) formed at equilibrium. According to the problem, 2 moles of \( Z \) are formed at equilibrium:
\[ x = 2 \]

### Step 4: Calculate moles of reactants at equilibrium
Using the stoichiometry of the reaction:
- Moles of \( X \) at equilibrium = Initial moles of \( X \) - moles of \( Z \) formed
\[ = 5 - 2 = 3 \]

- Moles of \( Y \) at equilibrium = Initial moles of \( Y \) - moles of \( Z \) formed
\[ = 3 - 2 = 1 \]

### Step 5: Write the equilibrium concentrations
Assuming the reaction occurs in a volume of 1 liter, the concentrations at equilibrium are:
- Concentration of \( X \) = \( \frac{3 \text{ moles}}{1 \text{ L}} = 3 \, \text{M} \)
- Concentration of \( Y \) = \( \frac{1 \text{ mole}}{1 \text{ L}} = 1 \, \text{M} \)
- Concentration of \( Z \) = \( \frac{2 \text{ moles}}{1 \text{ L}} = 2 \, \text{M} \)

### Step 6: Write the expression for the equilibrium constant \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[\text{products}]}{[\text{reactants}]} \]

For our reaction:
\[ K_c = \frac{[Z]}{[X][Y]} \]

### Step 7: Substitute the equilibrium concentrations into the expression
Substituting the concentrations we found:
\[ K_c = \frac{[Z]}{[X][Y]} = \frac{2}{3 \times 1} \]

### Step 8: Calculate \( K_c \)
Calculating the value:
\[ K_c = \frac{2}{3} \]

### Conclusion
The equilibrium constant \( K_c \) for the reaction is:
\[ K_c = \frac{2}{3} \]