Decomposition of a gas is of first order. It takes 80 minutes for 80% of the gas to be decomposed when its initial concentration is `8 xx 10^(-3)` mole/litre. Calculate the specific reaction rate.

Explanation

To solve the problem of calculating the specific reaction rate for the first-order decomposition of a gas, we can follow these steps:

### Step 1: Understand the First-Order Reaction Equation
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{1}{T} \ln \left( \frac{A_0}{A_t} \right) \]
where:
- \( T \) is the time taken for the reaction (in minutes),
- \( A_0 \) is the initial concentration of the reactant,
- \( A_t \) is the concentration of the reactant at time \( T \).

### Step 2: Identify Given Values
From the problem, we know:
- Time \( T = 80 \) minutes,
- Initial concentration \( A_0 = 8 \times 10^{-3} \) moles/litre,
- 80% of the gas is decomposed, which means 20% remains. Therefore, \( A_t = 20\% \) of \( A_0 \).

### Step 3: Calculate \( A_t \)
Calculate \( A_t \):
\[ A_t = 0.20 \times A_0 = 0.20 \times (8 \times 10^{-3}) = 1.6 \times 10^{-3} \text{ moles/litre} \]

### Step 4: Substitute Values into the Rate Constant Formula
Now, substitute \( A_0 \) and \( A_t \) into the formula for \( k \):
\[ k = \frac{1}{80} \ln \left( \frac{8 \times 10^{-3}}{1.6 \times 10^{-3}} \right) \]

### Step 5: Simplify the Fraction Inside the Logarithm
Calculate the fraction:
\[ \frac{8 \times 10^{-3}}{1.6 \times 10^{-3}} = 5 \]
Thus, we have:
\[ k = \frac{1}{80} \ln(5) \]

### Step 6: Calculate \( \ln(5) \)
Using a calculator or logarithm table, we find:
\[ \ln(5) \approx 1.609 \]

### Step 7: Substitute \( \ln(5) \) Back into the Equation
Now substitute \( \ln(5) \) into the equation for \( k \):
\[ k = \frac{1.609}{80} \approx 0.02011 \text{ min}^{-1} \]

### Step 8: Final Result
Thus, the specific reaction rate \( k \) is approximately:
\[ k \approx 0.0201 \text{ min}^{-1} \]