For an equilibrium `[2A(g) implies B(g) + C(g)]` equilibrium constant is 0.09. If 0.2 mol of A(g) is initially taken in a vessel of volume 10 L then concentration of B[g) at equilibrium is

Options

A. `3.75` x `10^-3` M
B. `2.45` x `10^-3` M
C. `1.5` x `10^-3` M
D. `7.5` x `10^-3` M

Answer

To solve the problem, we need to find the concentration of B at equilibrium for the reaction:

\[ 2A(g) \rightleftharpoons B(g) + C(g) \]

Given:
- The equilibrium constant \( K_c = 0.09 \)
- Initial moles of A = 0.2 mol
- Volume of the vessel = 10 L

### Step 1: Calculate the initial concentration of A

The concentration of A can be calculated using the formula:

\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Substituting the values:

\[ \text{Concentration of A} = \frac{0.2 \text{ mol}}{10 \text{ L}} = 0.02 \text{ M} \]

### Step 2: Set up the equilibrium expression

The equilibrium expression for the reaction is given by:

\[ K_c = \frac{[B][C]}{[A]^2} \]

### Step 3: Define changes in concentration at equilibrium

Let \( x \) be the change in concentration of B and C at equilibrium. Since 2 moles of A produce 1 mole of B and 1 mole of C, we can express the changes as follows:

- Change in concentration of A = \(-2x\)
- Change in concentration of B = \(+x\)
- Change in concentration of C = \(+x\)

At equilibrium, the concentrations will be:

- \([A] = 0.02 - 2x\)
- \([B] = x\)
- \([C] = x\)

### Step 4: Substitute into the equilibrium expression

Substituting the equilibrium concentrations into the equilibrium expression:

\[ 0.09 = \frac{x \cdot x}{(0.02 - 2x)^2} \]

This simplifies to:

\[ 0.09 = \frac{x^2}{(0.02 - 2x)^2} \]

### Step 5: Cross-multiply and simplify

Cross-multiplying gives:

\[ 0.09(0.02 - 2x)^2 = x^2 \]

Expanding the left side:

\[ 0.09(0.0004 - 0.08x + 4x^2) = x^2 \]

This simplifies to:

\[ 0.000036 - 0.0072x + 0.36x^2 = x^2 \]

Rearranging the equation:

\[ 0.36x^2 - x^2 - 0.0072x + 0.000036 = 0 \]

This simplifies to:

\[ -0.64x^2 - 0.0072x + 0.000036 = 0 \]

### Step 6: Solve the quadratic equation

Multiplying through by -1 to make calculations easier:

\[ 0.64x^2 + 0.0072x - 0.000036 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Where \( a = 0.64, b = 0.0072, c = -0.000036 \):

Calculating the discriminant:

\[ b^2 - 4ac = (0.0072)^2 - 4(0.64)(-0.000036) \]

Calculating the values:

\[ = 0.00005184 + 0.00009216 = 0.000144 \]

Now substituting into the quadratic formula:

\[ x = \frac{-0.0072 \pm \sqrt{0.000144}}{2 \cdot 0.64} \]

Calculating \( \sqrt{0.000144} = 0.012 \):

\[ x = \frac{-0.0072 \pm 0.012}{1.28} \]

Calculating the two possible values for \( x \):

1. \( x = \frac{0.0048}{1.28} \approx 0.00375 \)
2. \( x = \frac{-0.0192}{1.28} \) (not valid since concentration cannot be negative)

### Step 7: Find the concentration of B at equilibrium

Thus, the concentration of B at equilibrium is:

\[ [B] = x \approx 0.00375 \text{ M} \]

### Final Answer

The concentration of B at equilibrium is approximately **0.00375 M**.

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