The rate constant of a first order reaction of `0.0693"min"^(-1)`. What is the time (in min) required for reducing an initial concentration of 20 mole `"lit"^(-1)` to 2.5 mole `"lit"^(-1)`?

Options

A. 40
B. 41
C. 20
D. 30

Answer

To solve the problem, we will use the first-order reaction rate equation. The formula for the first-order reaction is given by:

\[ k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A]} \right) \]

Where:
- \( k \) = rate constant (0.0693 min\(^{-1}\))
- \( t \) = time (in minutes)
- \( [A_0] \) = initial concentration (20 mol/L)
- \( [A] \) = final concentration (2.5 mol/L)

### Step 1: Substitute the values into the equation

We start by substituting the known values into the equation:

\[ 0.0693 = \frac{2.303}{t} \log \left( \frac{20}{2.5} \right) \]

### Step 2: Calculate the logarithm

Next, we calculate the logarithm:

\[ \frac{20}{2.5} = 8 \]
Thus,
\[ \log(8) \approx 0.903 \]

### Step 3: Substitute the logarithm back into the equation

Now we substitute \(\log(8)\) back into the equation:

\[ 0.0693 = \frac{2.303}{t} \times 0.903 \]

### Step 4: Rearrange the equation to solve for \(t\)

Rearranging the equation to isolate \(t\):

\[ t = \frac{2.303 \times 0.903}{0.0693} \]

### Step 5: Calculate the time \(t\)

Now we perform the multiplication and division:

\[ t = \frac{2.303 \times 0.903}{0.0693} \approx \frac{2.080}{0.0693} \approx 30.0 \text{ min} \]

### Final Answer

Thus, the time required to reduce the concentration from 20 mol/L to 2.5 mol/L is approximately **30 minutes**.

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