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. 0.21
B. 0.5
C. 0.75
D. 0.05

Explanation

To find the equilibrium constant \( K_c \) for the reaction \( A + 2B \rightleftharpoons 2C \), we will follow these steps:

### Step 1: Write the balanced equation and the expression for \( K_c \)
The balanced equation is:
\[ A + 2B \rightleftharpoons 2C \]

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

### Step 2: Determine initial moles and concentrations
We have:
- Initial moles of \( A = 2.0 \) moles
- Initial moles of \( B = 3.0 \) moles
- Initial moles of \( C = 2.0 \) moles
- Volume of the flask = 2.0 L

Now, we calculate the initial concentrations:
- \([A]_{initial} = \frac{2.0 \, \text{moles}}{2.0 \, \text{L}} = 1.0 \, \text{mol/L}\)
- \([B]_{initial} = \frac{3.0 \, \text{moles}}{2.0 \, \text{L}} = 1.5 \, \text{mol/L}\)
- \([C]_{initial} = \frac{2.0 \, \text{moles}}{2.0 \, \text{L}} = 1.0 \, \text{mol/L}\)

### Step 3: Set up the change in concentration at equilibrium
Let \( x \) be the amount of \( A \) that reacts at equilibrium. The changes in concentrations will be:
- \([A]_{eq} = 1.0 - x\)
- \([B]_{eq} = 1.5 - 2x\)
- \([C]_{eq} = 1.0 + 2x\)

### Step 4: Use the given equilibrium concentration of \( C \)
We know that the equilibrium concentration of \( C \) is \( 0.5 \, \text{mol/L} \):
\[ [C]_{eq} = 1.0 + 2x = 0.5 \]

### Step 5: Solve for \( x \)
Rearranging the equation:
\[ 2x = 0.5 - 1.0 \]
\[ 2x = -0.5 \]
\[ x = -0.25 \]

### Step 6: Calculate equilibrium concentrations
Now substitute \( x \) back into the equations for equilibrium concentrations:
- \([A]_{eq} = 1.0 - (-0.25) = 1.25 \, \text{mol/L}\)
- \([B]_{eq} = 1.5 - 2(-0.25) = 1.5 + 0.5 = 2.0 \, \text{mol/L}\)
- \([C]_{eq} = 0.5 \, \text{mol/L}\) (given)

### Step 7: Substitute equilibrium concentrations into \( K_c \) expression
Now we can substitute these values into the \( K_c \) expression:
\[ K_c = \frac{[C]^2}{[A][B]^2} = \frac{(0.5)^2}{(1.25)(2.0)^2} \]

Calculating:
\[ K_c = \frac{0.25}{(1.25)(4)} = \frac{0.25}{5} = 0.05 \]

### Final Answer
Thus, the equilibrium constant \( K_c \) for the reaction is:
\[ K_c = 0.05 \]

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