In a first order reaction the concentration of the reactant is reduced to one-fourth of its initial value in 50 seconds. Cacluate the rate constant the reaction

Answer

To calculate the rate constant (k) for a first-order reaction where the concentration of the reactant is reduced to one-fourth of its initial value in 50 seconds, we can follow these steps:

### Step 1: Understand the relationship for first-order reactions
For a first-order reaction, the rate constant (k) can be calculated using the formula:
\[ k = \frac{2.303}{T} \log_{10} \left(\frac{A}{A-X}\right) \]
where:
- \( A \) = initial concentration of the reactant
- \( A-X \) = concentration of the reactant after time \( T \)
- \( T \) = time in seconds

### Step 2: Identify the values
From the problem statement:
- The initial concentration \( A \) is unknown (let's denote it as \( A \)).
- The concentration is reduced to one-fourth of its initial value, so:
\[ A - X = \frac{A}{4} \]
This implies:
\[ X = A - \frac{A}{4} = \frac{3A}{4} \]
- Therefore, \( A - X = \frac{A}{4} \).
- The time \( T \) is given as 50 seconds.

### Step 3: Substitute values into the formula
Now we can substitute these values into the formula:
\[ k = \frac{2.303}{50} \log_{10} \left(\frac{A}{\frac{A}{4}}\right) \]
This simplifies to:
\[ k = \frac{2.303}{50} \log_{10} (4) \]

### Step 4: Calculate \( \log_{10} (4) \)
Using the property of logarithms:
\[ \log_{10} (4) = \log_{10} (2^2) = 2 \log_{10} (2) \]
We know that \( \log_{10} (2) \approx 0.301 \), so:
\[ \log_{10} (4) \approx 2 \times 0.301 = 0.602 \]

### Step 5: Substitute back to find k
Now substituting \( \log_{10} (4) \) back into the equation for k:
\[ k = \frac{2.303}{50} \times 0.602 \]
Calculating this gives:
\[ k = \frac{2.303 \times 0.602}{50} \]
\[ k \approx \frac{1.384}{50} \]
\[ k \approx 0.02768 \, \text{s}^{-1} \]

### Final Answer
Thus, the rate constant \( k \) for the reaction is approximately:
\[ k \approx 0.0277 \, \text{s}^{-1} \]

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