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 and initial concentration of A.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Rate constant \( k = 1.0 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1} \)
- Initial concentration of A, \( [A]_0 = 1.0 \, \text{mol L}^{-1} \)
- Time, \( t = 100 \, \text{minutes} \)

2. **Calculate the Amount of A Reacted:**
The amount of A that reacts can be calculated using the formula:
\[ \text{Reactant used} = k \times t \]
Substituting the values:
\[ \text{Reactant used} = (1.0 \times 10^{-3} \, \text{mol L}^{-1} \text{min}^{-1}) \times (100 \, \text{min}) = 0.1 \, \text{mol L}^{-1} \]

3. **Determine the Remaining Concentration of A:**
The concentration of A after 100 minutes can be calculated as:
\[ [A] = [A]_0 - \text{Reactant used} \]
Substituting the values:
\[ [A] = 1.0 \, \text{mol L}^{-1} - 0.1 \, \text{mol L}^{-1} = 0.9 \, \text{mol L}^{-1} \]

4. **Calculate the Concentration of B Formed:**
From the stoichiometry of the reaction \( A \rightarrow 2B \), for every 1 mole of A that reacts, 2 moles of B are produced. Therefore, the concentration of B formed can be calculated as:
\[ [B] = 2 \times \text{Reactant used} \]
Substituting the values:
\[ [B] = 2 \times 0.1 \, \text{mol L}^{-1} = 0.2 \, \text{mol L}^{-1} \]

5. **Final Answer:**
The concentration of B after 100 minutes is:
\[ [B] = 0.2 \, \text{mol L}^{-1} \]