For a reaction, `AiffB` equilibrium constant is 1.66 and `k_("forward")=0.166 hr^(-1)`.
Calculate the time (in hour) when concentration of B is 80% of its equilibrium concentration. (Given : In 25=3.20)

Answer

To solve the problem step by step, we will follow the outlined approach to calculate the time when the concentration of B is 80% of its equilibrium concentration.

### Step 1: Understand the Reaction and Given Data
The reaction is given as:
\[ A \rightleftharpoons B \]

We are provided with the following data:
- Equilibrium constant \( K = 1.66 \)
- Forward rate constant \( k_{\text{forward}} = 0.166 \, \text{hr}^{-1} \)

### Step 2: Determine the Equilibrium Concentrations
Let the initial concentration of A be 100 units (for simplicity), and the initial concentration of B be 0 units.

At equilibrium, let \( x \) be the change in concentration of A that converts to B. Thus:
- Concentration of A at equilibrium = \( 100 - x \)
- Concentration of B at equilibrium = \( x \)

Using the equilibrium constant expression:
\[ K = \frac{[B]}{[A]} = \frac{x}{100 - x} \]
Substituting the value of \( K \):
\[ 1.66 = \frac{x}{100 - x} \]

### Step 3: Solve for \( x \)
Cross-multiplying gives:
\[ 1.66(100 - x) = x \]
Expanding this:
\[ 166 - 1.66x = x \]
Combining like terms:
\[ 166 = 2.66x \]
Thus:
\[ x = \frac{166}{2.66} \approx 62.4 \]

### Step 4: Calculate 80% of Equilibrium Concentration of B
Now, we need to find 80% of the equilibrium concentration of B:
\[ \text{Concentration of B at equilibrium} = x = 62.4 \]
\[ \text{80% of equilibrium concentration of B} = 0.8 \times 62.4 = 49.92 \]

### Step 5: Find the Concentration of A at 80% of B
The concentration of A at this point will be:
\[ \text{Concentration of A} = 100 - 49.92 = 50.08 \]

### Step 6: Calculate the Rate Constants
We have:
- \( k_{\text{forward}} = 0.166 \, \text{hr}^{-1} \)
- To find \( k_{\text{backward}} \), we can use the relationship:
\[ K = \frac{k_{\text{forward}}}{k_{\text{backward}}} \]
Rearranging gives:
\[ k_{\text{backward}} = \frac{k_{\text{forward}}}{K} = \frac{0.166}{1.66} \approx 0.1 \, \text{hr}^{-1} \]

### Step 7: Calculate the Overall Rate Constant
The overall rate constant \( k \) for the reaction can be expressed as:
\[ k = k_{\text{forward}} + k_{\text{backward}} = 0.166 + 0.1 = 0.266 \, \text{hr}^{-1} \]

### Step 8: Use the Integrated Rate Law to Find Time
Using the integrated rate law for a first-order reaction:
\[ \ln \left( \frac{R_0}{R_t} \right) = kt \]
Where:
- \( R_0 = 100 \) (initial concentration of A)
- \( R_t = 100 - 49.92 = 50.08 \) (concentration of A when B is 80% of equilibrium)

Substituting the values:
\[ \ln \left( \frac{100}{50.08} \right) = 0.266 \cdot t \]
Calculating the left side:
\[ \ln(1.994) \approx 0.693 \]
Thus:
\[ 0.693 = 0.266 \cdot t \]
Solving for \( t \):
\[ t = \frac{0.693}{0.266} \approx 2.60 \, \text{hours} \]

### Final Answer
The time when the concentration of B is 80% of its equilibrium concentration is approximately **2.60 hours**.