In a first order reaction fraction of the total concentration of the reactant varies with time 't' is equal to

Options

A. `e^(+kt)`
B. `10^(+0.434kt)`
C. `1/2^(-n)`
D. `e^(-kt)`

Answer

To solve the problem of how the fraction of the total concentration of the reactant varies with time in a first-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding First-Order Reactions**:
A first-order reaction is one where the rate of reaction is directly proportional to the concentration of one reactant. The general form of a first-order reaction can be represented as:
\[ A \rightarrow \text{Products} \]

2. **Rate Law Expression**:
The rate of a first-order reaction can be expressed as:
\[ \text{Rate} = -\frac{d[A]}{dt} = k[A] \]
where \( [A] \) is the concentration of reactant A, and \( k \) is the rate constant.

3. **Integrating the Rate Equation**:
Rearranging the rate law gives:
\[ \frac{d[A]}{[A]} = -k \, dt \]
Integrating both sides, we get:
\[ \int \frac{d[A]}{[A]} = -k \int dt \]
This results in:
\[ \ln[A] = -kt + C \]
where \( C \) is the integration constant.

4. **Finding the Integration Constant**:
To find \( C \), we use the initial condition. At \( t = 0 \), let \( [A]_0 \) be the initial concentration of A:
\[ \ln[A]_0 = C \]
Therefore, we can rewrite the equation as:
\[ \ln[A] = -kt + \ln[A]_0 \]

5. **Exponentiating to Solve for Concentration**:
By exponentiating both sides, we have:
\[ [A] = [A]_0 e^{-kt} \]

6. **Calculating the Fraction of Reactant**:
The fraction of the total concentration of the reactant at time \( t \) is given by:
\[ \text{Fraction} = \frac{[A]}{[A]_0} = \frac{[A]_0 e^{-kt}}{[A]_0} = e^{-kt} \]

### Final Answer:
The fraction of the total concentration of the reactant that varies with time \( t \) in a first-order reaction is:
\[ \text{Fraction} = e^{-kt} \]