The equilibrium constant for the reaction `A_((g))+2B_((g)) rarr C_((g))` is ` 0.25 " dm"^(6) "mole"^(-2)`. In a volume of `5 "dm"^(3)`, what amount of A must be mixed with 4 moles of B to yield 1 mole of C at equilibrium.

Options

A. (a)12.5 moles
B. (b)26 moles
C. (c )25 moles
D. (d)13 moles

Explanation

To solve the problem, we need to determine the amount of A that must be mixed with 4 moles of B to yield 1 mole of C at equilibrium, given the equilibrium constant \( K_c = 0.25 \, \text{dm}^6 \, \text{mole}^{-2} \) and a volume of \( 5 \, \text{dm}^3 \).

### Step-by-Step Solution:

1. **Write the balanced chemical equation:**
\[ A(g) + 2B(g) \rightleftharpoons C(g) \]

2. **Set up the initial conditions:**
- Initial moles of B = 4 moles
- Initial moles of C = 0 moles (since it hasn't been formed yet)
- Let the initial moles of A be \( n \) (which we need to find).

3. **Determine the change in moles at equilibrium:**
- If 1 mole of C is formed, then:
- 1 mole of A is consumed.
- 2 moles of B are consumed.
- Therefore, at equilibrium:
- Moles of A = \( n - 1 \)
- Moles of B = \( 4 - 2 = 2 \)
- Moles of C = 1

4. **Calculate the concentrations at equilibrium:**
- Volume = \( 5 \, \text{dm}^3 \)
- Concentration of A at equilibrium:
\[ [A] = \frac{n - 1}{5} \]
- Concentration of B at equilibrium:
\[ [B] = \frac{2}{5} \]
- Concentration of C at equilibrium:
\[ [C] = \frac{1}{5} \]

5. **Use the equilibrium constant expression:**
\[ K_c = \frac{[C]}{[A][B]^2} \]
Substituting the known values:
\[ 0.25 = \frac{\frac{1}{5}}{\left(\frac{n - 1}{5}\right)\left(\frac{2}{5}\right)^2} \]

6. **Simplify the equation:**
\[ 0.25 = \frac{\frac{1}{5}}{\frac{(n - 1) \cdot 4}{25}} \]
\[ 0.25 = \frac{25}{5 \cdot 4(n - 1)} = \frac{5}{4(n - 1)} \]

7. **Cross-multiply to solve for \( n \):**
\[ 0.25 \cdot 4(n - 1) = 5 \]
\[ n - 1 = \frac{5}{1} = 5 \]
\[ n = 6 \]

8. **Conclusion:**
The amount of A that must be mixed with 4 moles of B to yield 1 mole of C at equilibrium is **6 moles**.