For the reaction in equilibrium `A harr B`
`([B])/([A])=4.0 xx 10^(8)" "(-d[A])/(dt)=2.3 xx 10^(6)S^(-1)[A]" "(-d)/(dt)[B]=K[B]` Thus, K is

Options

A. `1.1 xx 10^(-15) S^(-1)`
B. `5.8 xx 10^(-3) S^(-1)`
C. `1.7 xx 10^(2) S^(-1)`
D. `9.2 xx 10^(14) S^(-1)`

Answer

To solve the problem, we need to find the equilibrium constant \( K \) for the reaction \( A \rightleftharpoons B \). We are given the following information:

1. The equilibrium expression \( \frac{[B]}{[A]} = 4.0 \times 10^8 \)
2. The rate of change of concentration of \( A \): \( -\frac{d[A]}{dt} = 2.3 \times 10^6 \, \text{s}^{-1} \)
3. The rate of change of concentration of \( B \): \( -\frac{d[B]}{dt} = K[B] \)

### Step-by-Step Solution:

**Step 1: Write the equilibrium expression for the reaction.**

The equilibrium constant \( K \) for the reaction \( A \rightleftharpoons B \) is given by:
\[ K = \frac{[B]}{[A]} \]
From the problem, we know that:
\[ \frac{[B]}{[A]} = 4.0 \times 10^8 \]

**Step 2: Relate the rates of change of concentrations.**

At equilibrium, the rate of formation of \( B \) is equal to the rate of consumption of \( A \):
\[ -\frac{d[A]}{dt} = \frac{d[B]}{dt} \]
We can express this as:
\[ -\frac{d[A]}{dt} = K[B] \]

**Step 3: Substitute the known values into the equation.**

We know:
\[ -\frac{d[A]}{dt} = 2.3 \times 10^6 \, \text{s}^{-1} \]
Substituting this into the equation gives:
\[ 2.3 \times 10^6 = K[B] \]

**Step 4: Express \( [B] \) in terms of \( [A] \) using the equilibrium expression.**

From the equilibrium expression, we can express \( [B] \) as:
\[ [B] = 4.0 \times 10^8 [A] \]

**Step 5: Substitute \( [B] \) back into the rate equation.**

Now substituting \( [B] \) into the rate equation:
\[ 2.3 \times 10^6 = K(4.0 \times 10^8 [A]) \]

**Step 6: Solve for \( K \).**

Rearranging gives:
\[ K = \frac{2.3 \times 10^6}{4.0 \times 10^8 [A]} \]

**Step 7: Since we need \( K \) in terms of the equilibrium ratio, we can simplify further.**

We can simplify this to:
\[ K = \frac{4.0 \times 10^8}{2.3 \times 10^6} \]

**Step 8: Calculate \( K \).**

Calculating gives:
\[ K \approx 5.8 \times 10^{-3} \, \text{s}^{-1} \]

### Final Answer:
Thus, the value of \( K \) is:
\[ K \approx 5.8 \times 10^{-3} \, \text{s}^{-1} \]