For a first order reaction, the concentration of reactant :

Options

A. is independent of time
B. varies linearly with time
C. varies exponentially with time
D. None

Answer

To solve the question regarding the concentration of a reactant in a first-order reaction, let's break down the steps systematically.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- We have a first-order reaction represented as \( A \rightarrow B \).
- The initial concentration of reactant \( A \) is denoted as \( C_0 \).

2. **Defining Concentration at Time \( t \)**:
- At time \( t = 0 \), the concentration is \( C_0 \).
- After some time \( t \), the concentration of \( A \) is \( C_t \), which can be expressed as \( C_t = C_0 - x \), where \( x \) is the amount of reactant that has reacted.

3. **Writing the Rate Law**:
- For a first-order reaction, the rate of reaction is given by:
\[ -\frac{d[A]}{dt} = k[A] \]
- Here, \( k \) is the rate constant and \( [A] \) is the concentration of reactant \( A \).

4. **Integrating the Rate Law**:
- Rearranging the rate equation gives:
\[ -\frac{d[A]}{[A]} = k \, dt \]
- We integrate both sides from \( t = 0 \) to \( t = T \) and from \( [A] = C_0 \) to \( [A] = C_t \):
\[ \int_{C_0}^{C_t} -\frac{d[A]}{[A]} = \int_{0}^{T} k \, dt \]

5. **Performing the Integration**:
- The left side integrates to:
\[ -\ln\left(\frac{C_t}{C_0}\right) = kT \]
- Rearranging gives:
\[ \ln\left(\frac{C_0}{C_t}\right) = kT \]

6. **Exponential Form**:
- Exponentiating both sides results in:
\[ \frac{C_t}{C_0} = e^{-kT} \]
- Thus, we can express \( C_t \) as:
\[ C_t = C_0 e^{-kT} \]

7. **Conclusion**:
- The concentration \( C_t \) varies exponentially with time \( T \), which indicates that the concentration of the reactant decreases exponentially as the reaction progresses.

### Final Answer:
The concentration of reactant in a first-order reaction varies exponentially with time.