A first order reaction is 20% complete in 10 minutes. What is the specific rate constant for the reaction?

Options

A. 0.0970 `"min^(-1)`
B. 0.009 `"min"^(-1)`
C. 0.0223 `"min"^(-1)`
D. `2.223 "min"^(-1)`

Explanation

To find the specific rate constant for the first-order reaction that is 20% complete in 10 minutes, we can follow these steps:

### Step 1: Understand the problem
We know that the reaction is first-order, and it is 20% complete in 10 minutes. This means that 20% of the initial reactant has been converted to products.

### Step 2: Define the initial concentration
Let's assume the initial concentration of the reactant (A) is 100 units. Since the reaction is 20% complete, the amount of reactant that has reacted (X) is:
\[ X = 20\% \text{ of } A = 0.20 \times 100 = 20 \]
Thus, the concentration of the reactant remaining at time \( t \) is:
\[ A - X = 100 - 20 = 80 \]

### Step 3: Write the first-order rate equation
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{A}{A - X} \right) \]
Where:
- \( t \) is the time (10 minutes in this case),
- \( A \) is the initial concentration (100),
- \( A - X \) is the concentration remaining (80).

### Step 4: Substitute the values into the equation
Now, we can substitute the values into the equation:
\[ k = \frac{2.303}{10} \log \left( \frac{100}{80} \right) \]

### Step 5: Calculate the logarithm
First, calculate the logarithm:
\[ \log \left( \frac{100}{80} \right) = \log (1.25) \]
Using a calculator, we find:
\[ \log (1.25) \approx 0.09691 \]

### Step 6: Substitute the logarithm back into the equation
Now substitute this value back into the equation for \( k \):
\[ k = \frac{2.303}{10} \times 0.09691 \]

### Step 7: Calculate \( k \)
Now perform the multiplication:
\[ k = 0.2303 \times 0.09691 \approx 0.0223 \text{ min}^{-1} \]

### Final Answer
The specific rate constant \( k \) for the reaction is approximately:
\[ k \approx 0.0223 \text{ min}^{-1} \]

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