The slope in the plot of `ln[R]` vs. time gives in first order reaction

Options

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

Explanation

To find the slope in the plot of `ln[R]` versus time for a first-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction Order**:
- A first-order reaction is characterized by a rate that is directly proportional to the concentration of one reactant. The rate law can be expressed as:
\[ R = -\frac{d[R]}{dt} = k[R] \]
where \( R \) is the rate of reaction, \( [R] \) is the concentration of the reactant, and \( k \) is the rate constant.

**Hint**: Identify the order of the reaction and write the rate law.

2. **Rearranging the Rate Equation**:
- Rearranging the rate equation gives:
\[ \frac{d[R]}{[R]} = -k \, dt \]

**Hint**: Isolate the variables to prepare for integration.

3. **Integrate Both Sides**:
- Integrating both sides from the initial concentration \([R]_0\) at time \( t = 0 \) to \([R]\) at time \( t \):
\[ \int_{[R]_0}^{[R]} \frac{d[R]}{[R]} = -k \int_0^t dt \]
- This results in:
\[ \ln[R] - \ln[R]_0 = -kt \]

**Hint**: Remember to apply the limits of integration correctly.

4. **Simplifying the Equation**:
- The equation can be simplified using properties of logarithms:
\[ \ln\left(\frac{[R]}{[R]_0}\right) = -kt \]
- Rearranging gives:
\[ \ln[R] = \ln[R]_0 - kt \]

**Hint**: Use logarithmic properties to combine terms.

5. **Identifying the Linear Form**:
- The equation \(\ln[R] = \ln[R]_0 - kt\) can be rewritten in the form of \(y = mx + c\):
- Here, \(y = \ln[R]\), \(x = t\), \(m = -k\) (the slope), and \(c = \ln[R]_0\).

**Hint**: Recognize the linear relationship in the equation.

6. **Conclusion**:
- From the linear equation, we see that the slope of the plot of \(\ln[R]\) versus time \(t\) is:
\[ \text{slope} = -k \]

**Hint**: The slope represents the negative of the rate constant for a first-order reaction.

### Final Answer:
The slope in the plot of `ln[R]` versus time for a first-order reaction is \(-k\).