In a first ordr reaction, reactant concentration 'C' varies with time 't' as

Options

A. C decreases with `(1)/(t)`
B. log C decreases with `(1)/(t)`
C. `(1)/(c)` increases linearly with t
D. log C decreases lineary with t

Explanation

To solve the problem regarding how the reactant concentration 'C' varies with time 't' in a first-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the First-Order Reaction**:
In a first-order reaction, the rate of reaction is directly proportional to the concentration of one reactant. The general form of the rate equation is:
\[ \text{Rate} = -\frac{dC}{dt} = kC \]
where \( k \) is the rate constant.

2. **Integrate the Rate Equation**:
Rearranging the rate equation gives:
\[ \frac{dC}{C} = -k \, dt \]
Integrating both sides, we get:
\[ \int \frac{dC}{C} = -k \int dt \]
This results in:
\[ \ln C = -kt + \ln C_0 \]
where \( C_0 \) is the initial concentration at time \( t = 0 \).

3. **Rearranging the Equation**:
Exponentiating both sides gives:
\[ C = C_0 e^{-kt} \]
Alternatively, we can express it in logarithmic form:
\[ \ln C = \ln C_0 - kt \]

4. **Expressing in Logarithmic Form**:
We can convert the natural logarithm to base 10 logarithm using the conversion factor \( \ln x = 2.303 \log_{10} x \):
\[ \log_{10} C = \log_{10} C_0 - \frac{kt}{2.303} \]

5. **Identifying the Linear Relationship**:
This equation can be rearranged to fit the linear equation format \( y = mx + b \):
\[ \log_{10} C = -\frac{k}{2.303} t + \log_{10} C_0 \]
Here, \( y = \log_{10} C \), \( m = -\frac{k}{2.303} \), \( x = t \), and \( b = \log_{10} C_0 \).

6. **Conclusion**:
From the linear relationship, we can conclude that as time \( t \) increases, \( \log_{10} C \) decreases linearly. Therefore, the concentration \( C \) decreases exponentially with time.

### Final Answer:
The correct answer is that \( \log_{10} C \) decreases linearly with time \( t \).