A first order reaction is 80% completed in 10 minutes .Find specific reaction rate constant second .

Options

A. `0.510s^(-1)`
B. `0.0027s^(-1)`
C. `0.161s^(-1)`
D. `0.2S^(-1)`

Answer

To solve the problem of finding the specific rate constant for a first-order reaction that is 80% completed in 10 minutes, we can follow these steps:

### Step 1: Understand the Reaction Completion
Since the reaction is 80% complete, it means that 20% of the reactant remains.

### Step 2: Convert Time to Seconds
The time given is 10 minutes. We need to convert this into seconds:
\[ 10 \text{ minutes} = 10 \times 60 = 600 \text{ seconds} \]

### Step 3: Write the First-Order Rate Constant Equation
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A]} \right) \]
Where:
- \( [A_0] \) is the initial concentration,
- \( [A] \) is the concentration at time \( t \),
- \( t \) is the time in seconds.

### Step 4: Determine Initial and Remaining Concentrations
Let’s assume the initial concentration \( [A_0] = 100 \) (for simplicity). Since 80% has reacted, the remaining concentration \( [A] \) is:
\[ [A] = 20\% \text{ of } [A_0] = 0.20 \times 100 = 20 \]

### Step 5: Substitute Values into the Equation
Now, substituting the values into the equation:
\[ k = \frac{2.303}{600} \log \left( \frac{100}{20} \right) \]
\[ = \frac{2.303}{600} \log(5) \]

### Step 6: Calculate the Logarithm
Using a logarithm table or calculator, we find:
\[ \log(5) \approx 0.6989 \]

### Step 7: Substitute and Calculate \( k \)
Now substituting the value of \( \log(5) \):
\[ k = \frac{2.303}{600} \times 0.6989 \]
\[ = \frac{1.6097}{600} \]
\[ = 0.00268 \text{ s}^{-1} \]

### Step 8: Round Off the Answer
Rounding off gives:
\[ k \approx 0.0027 \text{ s}^{-1} \]

### Conclusion
Thus, the specific reaction rate constant \( k \) is approximately:
\[ \boxed{0.0027 \text{ s}^{-1}} \]