For a first order reaction `R to P` the rate constant is k . If the initial concentration of R is `[R_(0)]` the concentration of R at any time 't' is given by the expression

Options

A. `[R_(0)]e^(kt)`
B. `[R_(0)](1-e^(-kt))`
C. `[R_(0)]e^(-kt)`
D. `[R_(0)](1-e^(kt))`

Answer

To derive the expression for the concentration of reactant \( R \) at any time \( t \) for a first-order reaction \( R \to P \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**:
We have a first-order reaction where \( R \) is converting to \( P \). The rate of the reaction can be expressed in terms of the concentration of \( R \).

2. **Write the Rate Law**:
The rate of a first-order reaction is given by:
\[ -\frac{d[R]}{dt} = k[R] \]
Here, \( k \) is the rate constant, and \( [R] \) is the concentration of \( R \).

3. **Rearrange the Equation**:
Rearranging the rate law gives:
\[ -\frac{d[R]}{[R]} = k \, dt \]

4. **Integrate Both Sides**:
We will integrate both sides. The left side will be integrated with respect to \( [R] \) from the initial concentration \( [R_0] \) to \( [R] \), and the right side will be integrated with respect to time from \( 0 \) to \( t \):
\[ \int_{[R_0]}^{[R]} -\frac{d[R]}{[R]} = \int_{0}^{t} k \, dt \]

5. **Perform the Integration**:
The left side integrates to:
\[ -\ln[R] + \ln[R_0] = -k t \]
This can be rewritten as:
\[ \ln[R] - \ln[R_0] = -kt \]

6. **Use Properties of Logarithms**:
By using the property of logarithms \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \), we can express this as:
\[ \ln\left(\frac{[R]}{[R_0]}\right) = -kt \]

7. **Exponentiate Both Sides**:
To remove the natural logarithm, we exponentiate both sides:
\[ \frac{[R]}{[R_0]} = e^{-kt} \]

8. **Solve for [R]**:
Finally, multiplying both sides by \( [R_0] \) gives us the expression for the concentration of \( R \) at time \( t \):
\[ [R] = [R_0] e^{-kt} \]

### Final Expression:
The concentration of \( R \) at any time \( t \) is given by:
\[ [R] = [R_0] e^{-kt} \]