The slope of the straight line obtained by plotting rate versus concentration of reactant in a first order reaction is

Options

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

Answer

To solve the question regarding the slope of the straight line obtained by plotting the rate versus concentration of reactant in a first-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Rate Law**:
- For a reaction where A is the reactant, the rate of the reaction can be expressed as:
\[ \text{Rate} = k \cdot [A]^n \]
- Here, \( k \) is the rate constant, \( [A] \) is the concentration of reactant A, and \( n \) is the order of the reaction.

2. **Identifying the Order of Reaction**:
- Since we are dealing with a first-order reaction, we set \( n = 1 \). Therefore, the rate equation simplifies to:
\[ \text{Rate} = k \cdot [A]^1 \]
- This can be rewritten as:
\[ \text{Rate} = k \cdot [A] \]

3. **Plotting the Graph**:
- In this case, if we plot the rate of the reaction on the y-axis and the concentration of A on the x-axis, we get a linear relationship:
\[ \text{Rate} = k \cdot [A] \]
- This equation represents a straight line where the slope (m) of the line is equal to the rate constant \( k \).

4. **Determining the Slope**:
- From the equation \( \text{Rate} = k \cdot [A] \), we can see that the slope of the line (m) is simply:
\[ m = k \]
- Thus, the slope of the straight line obtained by plotting rate versus concentration of reactant in a first-order reaction is equal to the rate constant \( k \).

### Final Answer:
The slope of the straight line obtained by plotting rate versus concentration of reactant in a first-order reaction is \( k \).

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