In the reaction `X(g)+Y(g)hArr2Z(g),2` mole of X,1 mole of Y and 1 mole of Z are placed in a 10 litre vessel and allowed to reach equilibrium .If final concentration of Z is 0.2 M , then `K_(c)`for the given reaction is :

Options

A. `1.60`
B. `(80)/(3)`
C. `(16)/(3)`
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ X(g) + Y(g) \rightleftharpoons 2Z(g) \]

### Step 2: Determine initial moles and concentrations
We start with:
- 2 moles of \( X \)
- 1 mole of \( Y \)
- 1 mole of \( Z \)

Since the volume of the vessel is 10 liters, we can calculate the initial concentrations:
- Concentration of \( X \):
\[ [X] = \frac{2 \text{ moles}}{10 \text{ L}} = 0.2 \, M \]
- Concentration of \( Y \):
\[ [Y] = \frac{1 \text{ mole}}{10 \text{ L}} = 0.1 \, M \]
- Concentration of \( Z \):
\[ [Z] = \frac{1 \text{ mole}}{10 \text{ L}} = 0.1 \, M \]

### Step 3: Define the change in concentration at equilibrium
Let \( a \) be the change in moles of \( X \) and \( Y \) that react to reach equilibrium. According to the stoichiometry of the reaction:
- For \( X \): \( 2 - a \)
- For \( Y \): \( 1 - a \)
- For \( Z \): \( 1 + 2a \)

### Step 4: Use the given final concentration of \( Z \)
We are given that the final concentration of \( Z \) is \( 0.2 \, M \):
\[ [Z] = \frac{1 + 2a}{10} = 0.2 \]
Multiplying both sides by 10 gives:
\[ 1 + 2a = 2 \]
Solving for \( a \):
\[ 2a = 2 - 1 = 1 \quad \Rightarrow \quad a = 0.5 \]

### Step 5: Calculate equilibrium concentrations
Now we can find the equilibrium concentrations:
- Concentration of \( X \):
\[ [X] = \frac{2 - a}{10} = \frac{2 - 0.5}{10} = \frac{1.5}{10} = 0.15 \, M \]
- Concentration of \( Y \):
\[ [Y] = \frac{1 - a}{10} = \frac{1 - 0.5}{10} = \frac{0.5}{10} = 0.05 \, M \]
- Concentration of \( Z \):
\[ [Z] = 0.2 \, M \quad \text{(given)} \]

### Step 6: Calculate \( K_c \)
The equilibrium constant \( K_c \) for the reaction is given by:
\[ K_c = \frac{[Z]^2}{[X][Y]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.2)^2}{(0.15)(0.05)} = \frac{0.04}{0.0075} = \frac{4}{0.75} = \frac{16}{3} \]

### Final Answer
Thus, the value of \( K_c \) for the given reaction is:
\[ K_c = \frac{16}{3} \]

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