A first order reaction with respect to reactant A , has a rate constant `6" min"^(-1)` . If we start with `[A]=0.5 "mol L"^(-1)` , when would [A] reach the value `0.05" mol L"^(-1)`

Options

A. 0.384 min
B. 0.15 min
C. 3 min
D. 3.84 min

Answer

To solve the problem of determining when the concentration of reactant A will reach 0.05 mol L^(-1) from an initial concentration of 0.5 mol L^(-1) in a first-order reaction with a rate constant of 6 min^(-1), we can follow these steps:

### Step 1: Write the first-order reaction formula
For a first-order reaction, the relationship between the concentration of the reactant at time \( t \) and the initial concentration is given by the equation:
\[ [A] = [A_0] e^{-kt} \]
where:
- \([A]\) is the concentration at time \( t \)
- \([A_0]\) is the initial concentration
- \( k \) is the rate constant
- \( t \) is the time

### Step 2: Substitute the known values
We know:
- \([A_0] = 0.5 \, \text{mol L}^{-1}\)
- \([A] = 0.05 \, \text{mol L}^{-1}\)
- \( k = 6 \, \text{min}^{-1} \)

Substituting these values into the equation gives:
\[ 0.05 = 0.5 e^{-6t} \]

### Step 3: Simplify the equation
To simplify, divide both sides by 0.5:
\[ \frac{0.05}{0.5} = e^{-6t} \]
This simplifies to:
\[ 0.1 = e^{-6t} \]

### Step 4: Take the natural logarithm of both sides
Taking the natural logarithm (ln) of both sides gives:
\[ \ln(0.1) = -6t \]

### Step 5: Solve for \( t \)
Now, calculate \( \ln(0.1) \):
\[ \ln(0.1) \approx -2.3026 \]
Substituting this value into the equation:
\[ -2.3026 = -6t \]
Now, solve for \( t \):
\[ t = \frac{-2.3026}{-6} \approx 0.38376 \, \text{min} \]

### Step 6: Round the answer
Rounding this to three decimal places gives:
\[ t \approx 0.384 \, \text{min} \]

### Final Answer
The concentration of A will reach 0.05 mol L^(-1) after approximately **0.384 minutes**.

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