Out of 300 g substance [decomposes as per 1st order], how much will remains after 18 hr? (`t_(0.5)=3`hr)

Options

A. 4.6 gm
B. 5.6 gm
C. 9.2 gm
D. 6.4 gm

Answer

To solve the problem of how much of a 300 g substance remains after 18 hours when it decomposes according to first-order kinetics with a half-life of 3 hours, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Rate Constant (k)**:
The half-life (\(t_{0.5}\)) of a first-order reaction is related to the rate constant (k) by the formula:
\[ k = \frac{0.693}{t_{0.5}} \]
Given \(t_{0.5} = 3\) hours, we can calculate k:
\[ k = \frac{0.693}{3} = 0.231 \, \text{hr}^{-1} \]

2. **Use the First-Order Kinetics Formula**:
The amount of substance remaining after time t can be calculated using the first-order kinetics equation:
\[ \ln\left(\frac{[A_0]}{[A]}\right) = kt \]
Where:
- \([A_0]\) = initial amount of substance = 300 g
- \([A]\) = amount of substance remaining after time t
- \(t\) = time = 18 hours
- \(k\) = rate constant calculated in step 1

3. **Plug in the Values**:
Rearranging the equation gives:
\[ \ln\left(\frac{300}{[A]}\right) = 0.231 \times 18 \]
Calculate the right-hand side:
\[ 0.231 \times 18 = 4.158 \]
Thus, we have:
\[ \ln\left(\frac{300}{[A]}\right) = 4.158 \]

4. **Exponentiate to Solve for [A]**:
To solve for \([A]\), we exponentiate both sides:
\[ \frac{300}{[A]} = e^{4.158} \]
Calculate \(e^{4.158}\):
\[ e^{4.158} \approx 63.5 \]
Therefore:
\[ 300 = 63.5 \times [A] \]
Rearranging gives:
\[ [A] = \frac{300}{63.5} \approx 4.73 \, \text{g} \]

5. **Final Calculation**:
After rounding, the amount of substance remaining after 18 hours is approximately 4.6 g.

### Final Answer:
The amount of substance remaining after 18 hours is **4.6 g**.

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