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 \( K_c \) for the reaction \( X + Y \rightleftharpoons Z \), we can follow these steps:

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

### Step 2: Set up the initial moles
Initially, we have:
- Moles of \( X = 5 \)
- Moles of \( Y = 3 \)
- Moles of \( Z = 0 \)

### Step 3: Determine the change in moles at equilibrium
At equilibrium, it is given that 2 moles of \( Z \) are formed. This means:
- Moles of \( Z \) formed = 2

Since the formation of 2 moles of \( Z \) corresponds to the consumption of 2 moles of \( X \) and 2 moles of \( Y \), we have:
- Moles of \( X \) consumed = 2
- Moles of \( Y \) consumed = 2

### Step 4: Calculate the moles at equilibrium
Now, we can calculate the moles of each substance at equilibrium:
- Moles of \( X \) at equilibrium = Initial moles of \( X \) - Moles of \( X \) consumed
\[ = 5 - 2 = 3 \]

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

- Moles of \( Z \) at equilibrium = Moles of \( Z \) formed
\[ = 2 \]

### Step 5: Write the expression for the equilibrium constant \( K_c \)
The expression for the equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[Z]}{[X][Y]} \]
where \([Z]\), \([X]\), and \([Y]\) are the concentrations of \( Z \), \( X \), and \( Y \) at equilibrium, respectively.

### Step 6: Calculate the concentrations
Assuming the volume of the reaction mixture is 1 liter, the concentrations will be equal to the number of moles:
- \([Z] = 2 \, \text{mol/L}\)
- \([X] = 3 \, \text{mol/L}\)
- \([Y] = 1 \, \text{mol/L}\)

### Step 7: Substitute the concentrations into the \( K_c \) expression
Now, substituting the concentrations into the \( K_c \) expression:
\[ K_c = \frac{[Z]}{[X][Y]} = \frac{2}{3 \times 1} = \frac{2}{3} \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is:
\[ K_c = \frac{2}{3} \]

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