For a first order reaction the plot of ‘t’ against log c gives a straight line with a slope equal to :

Options

A. `-2.303/k`
B. `- 2.303/logk`
C. `+ 2.303/k`
D. `+ 2.303/ log k`

Answer

To solve the question regarding the slope of the plot of time (t) against log concentration (log C) for a first-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the First-Order Reaction Rate Law**:
For a first-order reaction, the relationship between concentration and time can be expressed as:
\[ \frac{dC}{dt} = -kC \]
where \( C \) is the concentration, \( t \) is time, and \( k \) is the rate constant.

2. **Integrate the Rate Law**:
Integrating the above equation gives:
\[ \ln C = -kt + \ln C_0 \]
where \( C_0 \) is the initial concentration.

3. **Convert to Logarithm Base 10**:
We can convert the natural logarithm to base 10 using the relationship:
\[ \ln C = 2.303 \log C - 2.303 \log e \]
This can be rearranged to:
\[ \log C = -\frac{k}{2.303}t + \log C_0 \]

4. **Identify the Form of the Equation**:
The equation now resembles the linear form \( y = mx + b \), where:
- \( y \) is \( \log C \)
- \( x \) is \( t \)
- \( m \) (the slope) is \(-\frac{k}{2.303}\)
- \( b \) (the y-intercept) is \( \log C_0 \)

5. **Conclusion**:
Therefore, the slope of the plot of time (t) against log concentration (log C) for a first-order reaction is:
\[ \text{slope} = -\frac{k}{2.303} \]

### Final Answer:
The slope of the plot of 't' against log C for a first-order reaction is \(-\frac{k}{2.303}\).

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