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 't' against log 'c' for a first-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the First-Order Reaction Rate Law**:
A first-order reaction can be described by the equation:
\[ \text{Rate} = k[C] \]
where \( k \) is the rate constant and \( [C] \) is the concentration of the reactant.

2. **Integrated Rate Law for First-Order Reactions**:
The integrated rate law for a first-order reaction is given by:
\[ \ln[C] = -kt + \ln[A] \]
where \( [A] \) is the initial concentration and \( [C] \) is the concentration at time \( t \).

3. **Convert to Logarithm Base 10**:
To express this in terms of base 10 logarithm, we can use the conversion:
\[ \ln[C] = 2.303 \log[C] \quad \text{(since } \ln x = 2.303 \log x\text{)} \]
Thus, we can rewrite the equation as:
\[ \log[C] = \log[A] - \frac{kt}{2.303} \]

4. **Rearranging the Equation**:
Rearranging the equation gives:
\[ \log[C] = -\frac{k}{2.303}t + \log[A] \]
This is in the form of the straight line equation \( y = mx + c \), where:
- \( y = \log[C] \)
- \( x = t \)
- \( m = -\frac{k}{2.303} \) (the slope)
- \( c = \log[A] \) (the y-intercept)

5. **Identifying the Slope**:
From the rearranged equation, we can see that the slope of the plot of \( t \) against \( \log[C] \) is:
\[ \text{slope} = -\frac{k}{2.303} \]

6. **Final Answer**:
Therefore, the slope of the plot of \( t \) against \( \log[C] \) for a first-order reaction is:
\[ -\frac{k}{2.303} \]