Half life of a first order reaction is 4s and the initial concentration of the reactant is 0.12 M. The concentration of the reactant left after 16 s is

Options

A. 0.0075 M
B. 0.06M
C. 0.03 M
D. 0.015M

Explanation

To find the concentration of the reactant left after 16 seconds for a first-order reaction, we can follow these steps:

### Step 1: Determine the rate constant (k)
The half-life (T_half) of a first-order reaction is given by the formula:
\[ T_{1/2} = \frac{0.693}{k} \]
Given that \( T_{1/2} = 4 \, \text{s} \), we can rearrange the formula to find k:
\[ k = \frac{0.693}{T_{1/2}} = \frac{0.693}{4} \]

### Step 2: Calculate the value of k
Calculating k:
\[ k = \frac{0.693}{4} = 0.17325 \, \text{s}^{-1} \]

### Step 3: Use the first-order reaction formula
The concentration of the reactant at time t (A_t) can be calculated using the formula:
\[ A_t = A_0 \cdot e^{-kt} \]
Where:
- \( A_0 \) is the initial concentration (0.12 M)
- \( t \) is the time (16 s)
- \( k \) is the rate constant we just calculated

### Step 4: Substitute the values into the formula
Substituting the values:
\[ A_t = 0.12 \cdot e^{-0.17325 \cdot 16} \]

### Step 5: Calculate the exponent
Calculating the exponent:
\[ -kt = -0.17325 \cdot 16 = -2.772 \]

### Step 6: Calculate \( e^{-2.772} \)
Now we calculate:
\[ e^{-2.772} \approx 0.062 \]

### Step 7: Calculate the concentration after 16 seconds
Finally, substituting back into the equation:
\[ A_t = 0.12 \cdot 0.062 \]
\[ A_t \approx 0.00744 \, \text{M} \]

### Conclusion
The concentration of the reactant left after 16 seconds is approximately:
\[ A_t \approx 0.00744 \, \text{M} \]

---