For a first order reaction, the reaction, the plot of `log C` against 't' (log C vs 't') gives a straight line with slope equal to:

Options

A. `(k//2.303)`
B. `(-k//2.303)`
C. `(ln k//2.303)`
D. `-k`.

Explanation

To find the slope of the plot of `log C` against time `t` for a first-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the First-Order Reaction**:
The general equation for a first-order reaction can be expressed as:
\[ C_t = C_0 e^{-kt} \]
where \( C_t \) is the concentration at time \( t \), \( C_0 \) is the initial concentration, and \( k \) is the rate constant.

2. **Take the Logarithm**:
To linearize the equation, we take the logarithm (base 10) of both sides:
\[ \log C_t = \log C_0 - kt \]

3. **Rearranging the Equation**:
Rearranging gives us:
\[ \log C_t = -kt + \log C_0 \]

4. **Identify the Linear Form**:
This equation is in the form of \( y = mx + c \), where:
- \( y \) is \( \log C_t \)
- \( m \) (the slope) is \(-k\)
- \( x \) is \( t \)
- \( c \) (the y-intercept) is \( \log C_0 \)

5. **Determine the Slope**:
From the comparison, we can see that the slope \( m \) is:
\[ m = -k \]

6. **Express the Slope in Terms of Logarithm Base 10**:
To express the slope in terms of the logarithm base 10, we can relate it to the natural logarithm:
\[ m = -\frac{k}{2.303} \]
This is because \( k \) in the context of logarithm base 10 can be converted from natural logarithm using the conversion factor \( 2.303 \).

### Final Answer:
The slope of the plot of `log C` against `t` for a first-order reaction is:
\[ \text{slope} = -\frac{k}{2.303} \]