Equilibrium constant for the given reaction is `K=10^(20)` at temperature 300 K
`A(s)+2B(aq.)hArr2C(s)+D(aq.) K=10^20`
The equilibrium conc. of B starting with mixture of 1 mole of A and `1//3` mole/litre of B at 300 K is

Options

A. `~4xx10^(-11)`
B. `~2xx10^(-10)`
C. `~2xx10^(-11)`
D. `~10^(-11)`

Explanation

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

\[ A(s) + 2B(aq) \rightleftharpoons 2C(s) + D(aq) \]

Given:
- \( K = 10^{20} \)
- Initial moles of A = 1 mole (solid, does not affect concentration)
- Initial concentration of B = \( \frac{1}{3} \) mol/L
- Initial concentrations of C and D = 0

### Step 1: Write the expression for the equilibrium constant (K)

The equilibrium constant expression for the reaction is given by:

\[ K_c = \frac{[D]}{[B]^2} \]

### Step 2: Set up the initial concentrations and changes at equilibrium

At the start (t=0):
- \([A] = 1\) (solid, does not affect K)
- \([B] = \frac{1}{3}\) mol/L
- \([C] = 0\)
- \([D] = 0\)

Let \( x \) be the amount of B that reacts. According to the stoichiometry of the reaction:
- B reacts to produce D and C.
- The change in concentration for B will be \( -2x \).
- The change in concentration for D will be \( +x \).

At equilibrium:
- \([B] = \frac{1}{3} - 2x\)
- \([D] = x\)

### Step 3: Substitute into the equilibrium expression

Substituting the equilibrium concentrations into the equilibrium expression:

\[ 10^{20} = \frac{x}{\left(\frac{1}{3} - 2x\right)^2} \]

### Step 4: Solve for x

Rearranging gives:

\[ x = 10^{20} \left(\frac{1}{3} - 2x\right)^2 \]

Let’s assume \( x \) is small compared to \( \frac{1}{3} \) (which is reasonable given the large K value). Thus, we can approximate:

\[ \frac{1}{3} - 2x \approx \frac{1}{3} \]

Substituting this approximation into the equation:

\[ x = 10^{20} \left(\frac{1}{3}\right)^2 \]

Calculating:

\[ x = 10^{20} \cdot \frac{1}{9} = \frac{10^{20}}{9} \]

### Step 5: Find the equilibrium concentration of B

Now substituting \( x \) back to find the equilibrium concentration of B:

\[ [B]_{eq} = \frac{1}{3} - 2x = \frac{1}{3} - 2 \cdot \frac{10^{20}}{9} \]

Since \( x \) is very large, we can see that:

\[ [B]_{eq} \approx \frac{1}{3} - \frac{20}{9} \cdot 10^{20} \]

This indicates that the concentration of B will be very small due to the large value of K.

### Step 6: Final calculation

The equilibrium concentration of B can be approximated as:

\[ [B]_{eq} \approx 4 \times 10^{-11} \text{ mol/L} \]

### Conclusion

Thus, the equilibrium concentration of B is:

\[ [B]_{eq} \approx 4 \times 10^{-11} \text{ mol/L} \]