Equilibrium concentrations of A and B involved in following equilibrium are 0.01 M and 0.02 M respectively: `[A(g)hArr B(g)]` If 0.01 M of A is added at equilibrium, then at new equilibrium concentration of B will be

Options

A. `6.67 xx10^-2` M
B. `2.67 xx 10^-2` M
C. `1.33 xx 10^-2` M
D. `1.33 xx 10^-1` M

Answer

To solve the problem, we need to find the new equilibrium concentration of B after adding 0.01 M of A to the system at equilibrium.

### Step-by-Step Solution:

1. **Identify Initial Equilibrium Concentrations**:
- At equilibrium, the concentration of A, [A], is 0.01 M.
- The concentration of B, [B], is 0.02 M.

2. **Write the Equilibrium Expression**:
- The equilibrium constant (K) for the reaction \( A(g) \rightleftharpoons B(g) \) is given by:
\[ K = \frac{[B]}{[A]} \]
- Substituting the initial concentrations:
\[ K = \frac{0.02}{0.01} = 2 \]

3. **Add 0.01 M of A**:
- When 0.01 M of A is added, the new concentration of A becomes:
\[ [A] = 0.01 + 0.01 = 0.02 \, \text{M} \]
- The concentration of B remains at 0.02 M initially.

4. **Apply Le Chatelier's Principle**:
- According to Le Chatelier's principle, adding more A will shift the equilibrium to the right (towards B).
- Let \( x \) be the change in concentration of A that reacts to form B.

5. **Set Up New Equilibrium Concentrations**:
- New concentration of A at equilibrium:
\[ [A] = 0.02 - x \]
- New concentration of B at equilibrium:
\[ [B] = 0.02 + x \]

6. **Set Up the Equilibrium Constant Equation**:
- The equilibrium constant remains the same, so:
\[ K = \frac{[B]}{[A]} = 2 \]
- Substituting the new concentrations:
\[ 2 = \frac{0.02 + x}{0.02 - x} \]

7. **Solve for x**:
- Cross-multiply to solve for \( x \):
\[ 2(0.02 - x) = 0.02 + x \]
\[ 0.04 - 2x = 0.02 + x \]
\[ 0.04 - 0.02 = 2x + x \]
\[ 0.02 = 3x \]
\[ x = \frac{0.02}{3} \approx 0.00667 \, \text{M} \]

8. **Calculate New Concentration of B**:
- Substitute \( x \) back to find the new concentration of B:
\[ [B] = 0.02 + x = 0.02 + 0.00667 \approx 0.02667 \, \text{M} \]

9. **Final Answer**:
- The new equilibrium concentration of B is approximately:
\[ [B] \approx 2.67 \times 10^{-2} \, \text{M} \]

### Conclusion:
The new equilibrium concentration of B after adding 0.01 M of A is **2.67 x 10^-2 M**.