The rate constant of the reaction `A rarr 2B` is `1.0 xx 10^(-3)` mol `"lit" ^(-1)" min"^(-1)`, if the initial concentration of A is `1.0` mole `lit^(-1)` . What would be the concentration of B after 100 minutes.

Options

A. `0.1" mol lit"^(-1)`
B. `0.2 "mol lit"^(-1)`
C. `0.9" mol lit"^(-1)`
D. `1.8 "mol lit"^(-1)`

Explanation

To solve the problem, we need to determine the concentration of B after 100 minutes for the reaction \( A \rightarrow 2B \) given the rate constant \( k = 1.0 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \) and the initial concentration of A, \( [A]_0 = 1.0 \, \text{mol L}^{-1} \).

### Step-by-Step Solution:

1. **Identify the Reaction Order**:
The reaction \( A \rightarrow 2B \) is a first-order reaction because the rate of reaction depends on the concentration of A.

2. **Write the Rate Equation**:
For a first-order reaction, the rate of reaction can be expressed as:
\[ \text{Rate} = k[A] \]

3. **Determine the Change in Concentration**:
Let \( x \) be the amount of A that reacts after time \( t \). The change in concentration of A is given by:
\[ [A] = [A]_0 - x \]
Since 2 moles of B are produced for every mole of A that reacts, the concentration of B formed will be:
\[ [B] = 2x \]

4. **Use the Integrated Rate Law**:
For a first-order reaction, the integrated rate law is:
\[ \ln\left(\frac{[A]_0}{[A]}\right) = kt \]
Rearranging gives:
\[ [A] = [A]_0 e^{-kt} \]

5. **Substitute Known Values**:
Substitute \( [A]_0 = 1.0 \, \text{mol L}^{-1} \), \( k = 1.0 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \), and \( t = 100 \, \text{min} \):
\[ [A] = 1.0 \, e^{- (1.0 \times 10^{-3} \times 100)} \]
\[ [A] = 1.0 \, e^{-0.1} \]

6. **Calculate \( e^{-0.1} \)**:
Using a calculator or a table, we find:
\[ e^{-0.1} \approx 0.9048 \]
Thus,
\[ [A] \approx 1.0 \times 0.9048 = 0.9048 \, \text{mol L}^{-1} \]

7. **Calculate the Change in Concentration of A**:
The amount of A that has reacted is:
\[ x = [A]_0 - [A] = 1.0 - 0.9048 = 0.0952 \, \text{mol L}^{-1} \]

8. **Calculate the Concentration of B**:
Since \( [B] = 2x \):
\[ [B] = 2 \times 0.0952 = 0.1904 \, \text{mol L}^{-1} \]

### Final Answer:
The concentration of B after 100 minutes is approximately \( 0.1904 \, \text{mol L}^{-1} \).

---