Reaction `A+BrarrC+D` follows rate law .`r=k[A]^(1//2)[B]^(1//2)` Starting with 1 M of A and B each. What is the time taken fro concetration of A become `0.1M`?
(Given`k=2.303xx10^(-2)sec^(-1))`

Options

A. 10 sec
B. 100 sec
C. 1000 sec
D. 434 sec

Answer

To solve the problem, we need to determine the time taken for the concentration of A to decrease from 1 M to 0.1 M, given the rate law and the rate constant.

### Step-by-Step Solution:

1. **Identify the Rate Law**:
The rate law for the reaction is given as:
\[ r = k[A]^{1/2}[B]^{1/2} \]
Here, \( k = 2.303 \times 10^{-2} \, \text{sec}^{-1} \).

2. **Initial Concentrations**:
We start with initial concentrations:
\[ [A]_0 = 1 \, \text{M}, \quad [B]_0 = 1 \, \text{M} \]
We want to find the time taken for the concentration of A to become:
\[ [A] = 0.1 \, \text{M} \]

3. **Determine the Change in Concentration**:
The change in concentration of A is:
\[ \Delta[A] = [A]_0 - [A] = 1 \, \text{M} - 0.1 \, \text{M} = 0.9 \, \text{M} \]

4. **Use Integrated Rate Law**:
Since the reaction follows a rate law that can be treated as pseudo-first order (both A and B have the same initial concentration), we can use the integrated rate equation for first-order reactions:
\[ \ln\left(\frac{[A]_0}{[A]}\right) = kt \]
Substituting the values:
\[ \ln\left(\frac{1}{0.1}\right) = 2.303 \times 10^{-2} \times t \]

5. **Calculate the Natural Logarithm**:
Calculate the natural logarithm:
\[ \ln(10) \approx 2.303 \]
Therefore:
\[ 2.303 = 2.303 \times 10^{-2} \times t \]

6. **Solve for Time (t)**:
Rearranging the equation to solve for \( t \):
\[ t = \frac{2.303}{2.303 \times 10^{-2}} = \frac{2.303}{0.02303} \approx 1000 \, \text{seconds} \]

### Final Answer:
The time taken for the concentration of A to become 0.1 M is approximately **1000 seconds**.