Consider a reaction `A(g) rarr 3B(g) + 2C(g)` with rate constant is `1.386 xx 10^(-2) "min"^(-1)` in a non-rigid closed container starting with `2` moles of `A` in `12.5 L` vessel initially, if reaction is allowed to take place at constant pressure and at `298 K` the conc. of after `100 min` is

Options

A. `0.18 M`
B. `0.03 M`
C. `0.09 M`
D. `0.01 M`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Determine the initial concentration of A
We start with 2 moles of A in a 12.5 L container. The initial concentration \([A]_0\) can be calculated using the formula:
\[ [A]_0 = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{2 \text{ moles}}{12.5 \text{ L}} = 0.16 \text{ M} \]

### Step 2: Use the first-order rate equation
The reaction is first-order, and we can use the first-order rate equation:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = k \cdot t \]
Where:
- \(k = 1.386 \times 10^{-2} \text{ min}^{-1}\)
- \(t = 100 \text{ min}\)

### Step 3: Calculate \([A]\) after 100 minutes
Rearranging the equation gives:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = (1.386 \times 10^{-2}) \cdot 100 \]
Calculating the right side:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = 1.386 \]
Now, exponentiating both sides:
\[ \frac{[A]_0}{[A]} = e^{1.386} \approx 4.00 \]
Thus,
\[ [A] = \frac{[A]_0}{4.00} = \frac{0.16 \text{ M}}{4.00} = 0.04 \text{ M} \]

### Step 4: Calculate the moles of A remaining
Now we can find the moles of A remaining after 100 minutes:
\[ \text{Moles of A remaining} = [A] \times \text{Volume} = 0.04 \text{ M} \times 12.5 \text{ L} = 0.5 \text{ moles} \]

### Step 5: Calculate the moles of A decomposed
The initial moles of A were 2 moles, so the moles of A decomposed are:
\[ \text{Moles of A decomposed} = 2 - 0.5 = 1.5 \text{ moles} \]

### Step 6: Calculate the moles of B and C formed
From the stoichiometry of the reaction \(A \rightarrow 3B + 2C\):
- For every 1 mole of A that decomposes, 3 moles of B are formed.
- Therefore, the moles of B formed are:
\[ \text{Moles of B} = 3 \times \text{Moles of A decomposed} = 3 \times 1.5 = 4.5 \text{ moles} \]
- The moles of C formed are:
\[ \text{Moles of C} = 2 \times \text{Moles of A decomposed} = 2 \times 1.5 = 3.0 \text{ moles} \]

### Step 7: Calculate the total moles in the container
Total moles in the container after the reaction:
\[ \text{Total moles} = \text{Moles of A remaining} + \text{Moles of B} + \text{Moles of C} = 0.5 + 4.5 + 3.0 = 8.0 \text{ moles} \]

### Step 8: Calculate the concentration of B
Finally, the concentration of B can be calculated as:
\[ [B] = \frac{\text{Moles of B}}{\text{Volume}} = \frac{4.5 \text{ moles}}{12.5 \text{ L}} = 0.36 \text{ M} \]

### Final Answer
The concentration of B after 100 minutes is approximately \(0.36 \text{ M}\).

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