What time does it take for reactants to reduce to 3/4 of initial concentration if the rate constant is 7.5 x `10^-3 S^-1`?

Options

A. 38.4s
B. 40.2s
C. 39.3s
D. 36.8s

Answer

To solve the problem of determining the time it takes for the reactants to reduce to 3/4 of their initial concentration, we can follow these steps:

### Step 1: Understand the given information
- The rate constant (k) is given as \( 7.5 \times 10^{-3} \, \text{s}^{-1} \).
- We need to find the time (T) when the concentration of the reactants reduces to \( \frac{3}{4} \) of the initial concentration.

### Step 2: Define initial concentration
Let the initial concentration \( [R_0] = 100 \) (this is an arbitrary value for calculation purposes). Therefore, \( \frac{3}{4} \) of the initial concentration is:
\[ [R] = \frac{3}{4} \times [R_0] = \frac{3}{4} \times 100 = 75 \]

### Step 3: Use the first-order reaction formula
For a first-order reaction, the relationship between concentration and time can be expressed as:
\[ k = \frac{2.303}{T} \log \left( \frac{[R_0]}{[R]} \right) \]
Rearranging this gives:
\[ T = \frac{2.303}{k} \log \left( \frac{[R_0]}{[R]} \right) \]

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
- \( k = 7.5 \times 10^{-3} \, \text{s}^{-1} \)
- \( [R_0] = 100 \)
- \( [R] = 75 \)

We have:
\[ T = \frac{2.303}{7.5 \times 10^{-3}} \log \left( \frac{100}{75} \right) \]

### Step 5: Calculate the logarithm
Calculating the ratio:
\[ \frac{100}{75} = \frac{4}{3} \]
Now, find the logarithm:
\[ \log \left( \frac{4}{3} \right) \approx 0.1249 \]

### Step 6: Substitute the logarithm back into the equation
Now substituting back into the time equation:
\[ T = \frac{2.303}{7.5 \times 10^{-3}} \times 0.1249 \]

### Step 7: Calculate T
Calculating \( T \):
\[ T \approx \frac{2.303 \times 0.1249}{7.5 \times 10^{-3}} \approx \frac{0.2884}{0.0075} \approx 38.4 \, \text{s} \]

### Final Answer
The time it takes for the reactants to reduce to \( \frac{3}{4} \) of their initial concentration is approximately **38.4 seconds**.

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