For a reaction 2A + B `rarr` product, rate law is `-(d[A])/(dt)=k[A]`. At a time when `t=(1)/(k)`, concentration of the reactant is " `(C_(0)="initial concentration")`

Options

A. `(C_(0))/(e)`
B. `C_(0)e`
C. `(C_(0))/(e^(2))`
D. `(1)/(C_(0))`

Answer

To solve the problem, we need to determine the concentration of reactant A at time \( t = \frac{1}{k} \) for the reaction \( 2A + B \rightarrow \text{products} \) given the rate law \( -\frac{d[A]}{dt} = k[A] \).

### Step-by-Step Solution:

1. **Understanding the Rate Law**:
The rate law for the reaction is given as:
\[ -\frac{d[A]}{dt} = k[A] \]
This indicates that the reaction is first-order with respect to A.

2. **Rearranging the Rate Law**:
We can rearrange the rate law to express it in a more integrable form:
\[ \frac{d[A]}{[A]} = -k \, dt \]

3. **Integrating the Rate Law**:
We will integrate both sides. The left side will be integrated from the initial concentration \( [A]_0 \) to \( [A]_t \) (the concentration at time \( t \)), and the right side will be integrated from \( 0 \) to \( t \):
\[ \int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -k \int_{0}^{t} dt \]
This gives:
\[ \ln[A]_t - \ln[A]_0 = -kt \]

4. **Applying Limits**:
Simplifying the left side using logarithmic properties:
\[ \ln\left(\frac{[A]_t}{[A]_0}\right) = -kt \]

5. **Substituting Time**:
We need to find the concentration at \( t = \frac{1}{k} \):
\[ \ln\left(\frac{[A]_t}{[A]_0}\right) = -k\left(\frac{1}{k}\right) = -1 \]

6. **Exponentiating Both Sides**:
To eliminate the logarithm, we exponentiate both sides:
\[ \frac{[A]_t}{[A]_0} = e^{-1} \]
Therefore, we have:
\[ [A]_t = [A]_0 \cdot e^{-1} \]

7. **Final Concentration**:
Thus, the concentration of A at \( t = \frac{1}{k} \) is:
\[ [A]_t = \frac{[A]_0}{e} \]

### Conclusion:
The final concentration of the reactant A at time \( t = \frac{1}{k} \) is:
\[ [A]_t = \frac{C_0}{e} \]