To solve the question regarding the reaction \( A \rightarrow B + C \) with a rate constant of \( 0.001 \, \text{M s}^{-1} \) and an initial concentration of \( A \) as \( 1 \, \text{M} \), we can follow these steps:
### Step 1: Identify the order of the reaction
Since the rate constant is given in units of \( \text{M s}^{-1} \), we can infer that the reaction is first-order. In a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant.
### Step 2: Write the first-order rate equation
For a first-order reaction, the rate equation can be expressed as:
\[
\frac{d[A]}{dt} = -k[A]
\]
Where:
- \( [A] \) is the concentration of reactant \( A \)
- \( k \) is the rate constant
### Step 3: Integrate the rate equation
To find the concentration of \( A \) at any time \( t \), we integrate the rate equation:
\[
\ln[A] = \ln[A_0] - kt
\]
Where:
- \( [A_0] \) is the initial concentration of \( A \)
- \( t \) is the time in seconds
### Step 4: Substitute the values
Given:
- \( [A_0] = 1 \, \text{M} \)
- \( k = 0.001 \, \text{M s}^{-1} \)
- \( t = 10 \, \text{minutes} = 600 \, \text{seconds} \)
Substituting these values into the integrated rate equation:
\[
\ln[A] = \ln(1) - (0.001)(600)
\]
\[
\ln[A] = 0 - 0.6
\]
\[
\ln[A] = -0.6
\]
### Step 5: Calculate the concentration of \( A \)
To find \( [A] \), we exponentiate both sides:
\[
[A] = e^{-0.6} \approx 0.5488 \, \text{M}
\]
### Step 6: Calculate the concentration of \( B \)
Since the reaction produces \( B \) and \( C \) in a 1:1 ratio, the change in concentration of \( A \) is equal to the concentration of \( B \) formed:
\[
\Delta [A] = [A_0] - [A] = 1 - 0.5488 \approx 0.4512 \, \text{M}
\]
Thus, the concentration of \( B \) after 10 minutes is approximately \( 0.4512 \, \text{M} \).
### Final Answer
The concentrations of \( A \) and \( B \) after 10 minutes are approximately:
- \( [A] \approx 0.5488 \, \text{M} \)
- \( [B] \approx 0.4512 \, \text{M} \)
