Question

Derive the integrated rate equation for a first-order reaction.

Hint:

{First-order reaction:} \(\ln \frac{[A]_0}{[A]} = kt\) Straight-line plot of \(\ln [A]\) vs time.

Answer 1

Concept: A first-order reaction is one in which the rate of reaction is directly proportional to the concentration of a single reactant. The integrated rate equation relates concentration with time and helps determine reaction kinetics.
Step 1: Rate law for first-order reaction Consider a reaction: \[ A \rightarrow \text{Products} \] For a first-order reaction, \[ \text{Rate} = -\frac{d[A]}{dt} = k[A] \] where:

• \( [A] \) = concentration of reactant at time \( t \)
• \( k \) = rate constant

Step 2: Rearranging the equation \[ \frac{d[A]}{[A]} = -k\,dt \]
Step 3: Integration Integrate both sides: \[ \int \frac{d[A]}{[A]} = -k \int dt \] \[ \ln [A] = -kt + C \] where \( C \) is the integration constant.
Step 4: Applying initial condition At \( t = 0 \), concentration \( [A] = [A]_0 \) \[ \ln [A]_0 = C \] Substitute back: \[ \ln [A] = -kt + \ln [A]_0 \]
Step 5: Final integrated rate equation \[ \ln \frac{[A]_0}{[A]} = kt \] Alternative forms: Using base-10 logarithm: \[ \log \frac{[A]_0}{[A]} = \frac{kt}{2.303} \] Conclusion: Thus, the integrated rate equation for a first-order reaction is: \[ \ln \frac{[A]_0}{[A]} = kt \] which shows that concentration decreases exponentially with time.