In a reaction `A+2B hArr 2C, ` 2.0 moles of `'A'` 3 moles of 'B' and 2.0 moles of 'C' are placed in a 2.0 L flask and the equilibrium concentration of 'C' is 0.5 mol`//` L . The equilibrium constant (K) for the reaction is

Options

A. 0.073
B. 0.147
C. 0.05
D. 0.026

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine Initial Concentrations
Given:
- Moles of A = 2.0 moles
- Moles of B = 3.0 moles
- Moles of C = 2.0 moles
- Volume of the flask = 2.0 L

We can calculate the initial concentrations of A, B, and C using the formula:
\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume}} \]

Calculating:
- Initial concentration of A:
\[ [C_A]_{initial} = \frac{2.0 \, \text{moles}}{2.0 \, \text{L}} = 1.0 \, \text{mol/L} \]
- Initial concentration of B:
\[ [C_B]_{initial} = \frac{3.0 \, \text{moles}}{2.0 \, \text{L}} = 1.5 \, \text{mol/L} \]
- Initial concentration of C:
\[ [C_C]_{initial} = \frac{2.0 \, \text{moles}}{2.0 \, \text{L}} = 1.0 \, \text{mol/L} \]

### Step 2: Write the Equilibrium Expression
The equilibrium reaction is:
\[ A + 2B \rightleftharpoons 2C \]

The equilibrium constant \( K_c \) is given by:
\[ K_c = \frac{[C]^2}{[A][B]^2} \]

### Step 3: Determine Changes at Equilibrium
We know that at equilibrium, the concentration of C is given as:
\[ [C]_{equilibrium} = 0.5 \, \text{mol/L} \]

Since the reaction produces 2 moles of C for every 1 mole of A and 2 moles of B consumed, we can determine the changes in concentrations:
- Let \( x \) be the change in concentration of C from initial to equilibrium. Since the concentration of C decreases from 1.0 mol/L to 0.5 mol/L, we have:
\[ x = 1.0 - 0.5 = 0.5 \, \text{mol/L} \]

Thus, the changes for A and B will be:
- Change in A: \( -\frac{x}{2} = -0.25 \, \text{mol/L} \)
- Change in B: \( -x = -0.5 \, \text{mol/L} \)

### Step 4: Calculate Equilibrium Concentrations
Now we can find the equilibrium concentrations:
- Equilibrium concentration of A:
\[ [C_A]_{equilibrium} = [C_A]_{initial} - 0.25 = 1.0 - 0.25 = 0.75 \, \text{mol/L} \]
- Equilibrium concentration of B:
\[ [C_B]_{equilibrium} = [C_B]_{initial} - 0.5 = 1.5 - 0.5 = 1.0 \, \text{mol/L} \]

### Step 5: Substitute into the Equilibrium Expression
Now we substitute the equilibrium concentrations into the \( K_c \) expression:
\[ K_c = \frac{(0.5)^2}{(0.75)(1.0)^2} \]
Calculating:
\[ K_c = \frac{0.25}{0.75} = \frac{1}{3} \approx 0.333 \]

### Step 6: Final Result
Thus, the equilibrium constant \( K_c \) for the reaction is approximately:
\[ K_c \approx 0.333 \]