In 69.3 min, a first order reaction is 50% incom plete. How much reactants are left after 161 min?

Options

A. 0.8
B. 0.4
C. 0.2
D. 60^

Explanation

To solve the problem step by step, we will follow the principles of first-order kinetics.

### Step 1: Determine the rate constant (k)
We know that in a first-order reaction, the half-life (T_half) is given by the formula:
\[ T_{1/2} = \frac{0.693}{k} \]

From the problem, we know that the reaction is 50% complete in 69.3 minutes. Therefore, the half-life (T_half) is 69.3 minutes.

Using the half-life to find k:
\[ k = \frac{0.693}{T_{1/2}} = \frac{0.693}{69.3} = 0.01 \text{ min}^{-1} \]

### Step 2: Use the first-order kinetics equation
The first-order kinetics equation is given by:
\[ \ln\left(\frac{A_0}{A}\right) = kt \]
Where:
- \( A_0 \) is the initial concentration of the reactant,
- \( A \) is the concentration of the reactant at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time in minutes.

### Step 3: Set up the equation for the time of interest
We are interested in the amount of reactant left after 161 minutes. We can rearrange the equation to find \( A \):
\[ A = A_0 e^{-kt} \]

### Step 4: Substitute the known values
Let’s assume the initial concentration \( A_0 = 100 \) (this is arbitrary for percentage calculations). Now substituting the values:
- \( k = 0.01 \text{ min}^{-1} \)
- \( t = 161 \text{ min} \)

So we have:
\[ A = 100 e^{-0.01 \times 161} \]

Calculating the exponent:
\[ -0.01 \times 161 = -1.61 \]

Now substituting this back into the equation:
\[ A = 100 e^{-1.61} \]

### Step 5: Calculate \( e^{-1.61} \)
Using a calculator:
\[ e^{-1.61} \approx 0.200 \]

Now substituting this value:
\[ A \approx 100 \times 0.200 = 20 \]

### Step 6: Conclusion
Thus, after 161 minutes, the amount of reactant left is approximately **20**.