Following equilibrium is established at temperature T.
`A(g)hArrB(g)+C(g`)
at eq. `1M 2 M 2M`.
If volume of the vessel is doubled then find the equilibrium concentration of each species. (Given that : `sqrt40=6.324)`

Answer

To solve the problem, we need to determine the new equilibrium concentrations of species A, B, and C after the volume of the vessel is doubled.

### Step-by-Step Solution:

1. **Write the Equilibrium Reaction**:
The equilibrium reaction is given as:
\[ A(g) \rightleftharpoons B(g) + C(g) \]

2. **Initial Equilibrium Concentrations**:
At equilibrium, the concentrations of the species are:
- \([A] = 1 \, M\)
- \([B] = 2 \, M\)
- \([C] = 2 \, M\)

3. **Effect of Doubling the Volume**:
When the volume of the vessel is doubled, the concentration of each species will change. The concentration is defined as:
\[ \text{Concentration} = \frac{\text{Number of moles}}{\text{Volume}} \]
If the volume is doubled (from \(V\) to \(2V\)), the new concentration (\(C'\)) can be calculated as:
\[ C' = \frac{n}{2V} = \frac{n}{V} \cdot \frac{1}{2} = \frac{C}{2} \]
This means the new concentration is half of the initial concentration.

4. **Calculate New Concentrations**:
Now we can calculate the new concentrations for each species:
- For \(A\):
\[ [A]' = \frac{1 \, M}{2} = 0.5 \, M \]
- For \(B\):
\[ [B]' = \frac{2 \, M}{2} = 1 \, M \]
- For \(C\):
\[ [C]' = \frac{2 \, M}{2} = 1 \, M \]

5. **Final Equilibrium Concentrations**:
After doubling the volume, the new equilibrium concentrations are:
- \([A] = 0.5 \, M\)
- \([B] = 1 \, M\)
- \([C] = 1 \, M\)

### Summary of Results:
- \([A] = 0.5 \, M\)
- \([B] = 1 \, M\)
- \([C] = 1 \, M\)