The reaction `A (g) rarr B(g) + 2C (g)` is a first order reaction with rate constant `3465 xx 10^(–6) s^(–1)`. Starting with 0.1 mole of A in 2 litre vessel, find the concentration of A after 200 sec., when the reaction is allowed to take place at constant pressure and temperature.

Options

A. `0.05 M`
B. `0.025 M`
C. `0.0125 M`
D. None of these

Answer

To solve the problem step by step, we will follow the principles of first-order kinetics.

### Step 1: Determine the initial concentration of A
We start with 0.1 moles of A in a 2-liter vessel. The concentration (C₀) is given by the formula:

\[ C_0 = \frac{\text{moles}}{\text{volume in liters}} = \frac{0.1 \, \text{moles}}{2 \, \text{liters}} = 0.05 \, \text{M} \]

### Step 2: Write the first-order rate equation
For a first-order reaction, the rate constant (k) is related to the concentrations by the equation:

\[ k = \frac{1}{t} \ln \left( \frac{C_0}{C_t} \right) \]

Where:
- \( k = 3465 \times 10^{-6} \, \text{s}^{-1} \)
- \( t = 200 \, \text{s} \)
- \( C_0 = 0.05 \, \text{M} \)
- \( C_t \) is the concentration of A at time t.

### Step 3: Substitute the known values into the rate equation
Substituting the known values into the equation:

\[ 3465 \times 10^{-6} = \frac{1}{200} \ln \left( \frac{0.05}{C_t} \right) \]

### Step 4: Rearrange the equation to solve for \( C_t \)
First, multiply both sides by 200 to isolate the logarithm:

\[ 200 \times 3465 \times 10^{-6} = \ln \left( \frac{0.05}{C_t} \right) \]

Calculating the left side:

\[ 0.693 = \ln \left( \frac{0.05}{C_t} \right) \]

### Step 5: Exponentiate both sides to eliminate the natural logarithm
To solve for \( C_t \), we exponentiate both sides:

\[ e^{0.693} = \frac{0.05}{C_t} \]

Calculating \( e^{0.693} \):

\[ e^{0.693} \approx 2 \]

Thus, we have:

\[ 2 = \frac{0.05}{C_t} \]

### Step 6: Solve for \( C_t \)
Rearranging gives:

\[ C_t = \frac{0.05}{2} = 0.025 \, \text{M} \]

### Final Answer
The concentration of A after 200 seconds is \( 0.025 \, \text{M} \).

---