The reaction `A(g)toB(g)+2C(g)` is a first order reaction with rate constant `2.772xx10^(-3)sec^(-1)` reaction is started with only 0.1 mol of A in a container with volume 2 litre and is allowed to take place at constant volume and at constant temperature `300K[R=0.082` litre atm `mol^(-1)K^(-1)]`
`(log2=0.30)`
Q. Concentration of `A` after 250 sec will be:

Options

A. 0.125 M
B. 0.0125 M
C. 0.05 M
D. 0.025M

Answer

To solve the problem, we need to find the concentration of A after 250 seconds for the reaction \( A(g) \to B(g) + 2C(g) \), which is a first-order reaction. Here’s a step-by-step solution:

### Step 1: Calculate Initial Concentration of A
The initial number of moles of A is given as 0.1 mol, and the volume of the container is 2 liters. The concentration of A can be calculated using the formula:

\[ \text{Concentration of A} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Substituting the values:

\[ [A_0] = \frac{0.1 \, \text{mol}}{2 \, \text{L}} = 0.05 \, \text{M} \]

### Step 2: Use the First-Order Reaction Formula
For a first-order reaction, the relationship between the concentration of the reactant at time \( t \) and the initial concentration is given by:

\[ \ln \left( \frac{[A_0]}{[A]} \right) = kt \]

Where:
- \( k = 2.772 \times 10^{-3} \, \text{s}^{-1} \)
- \( t = 250 \, \text{s} \)
- \( [A_0] = 0.05 \, \text{M} \)

### Step 3: Calculate \( kt \)
First, we calculate \( kt \):

\[ kt = (2.772 \times 10^{-3} \, \text{s}^{-1}) \times (250 \, \text{s}) = 0.693 \]

### Step 4: Substitute into the First-Order Equation
Now we can substitute \( kt \) back into the equation:

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

### Step 5: Solve for \( [A] \)
To solve for \( [A] \), we exponentiate both sides:

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

Since \( e^{0.693} \) is approximately equal to 2 (because \( \ln(2) \approx 0.693 \)), we have:

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

Now, rearranging gives:

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

### Final Answer
The concentration of A after 250 seconds is:

\[ [A] = 0.025 \, \text{M} \]