Question

The integrated rate law and half-life equation for an first order reaction would be :

• A. $[A] = [A]_0 e^{-k_rt}$ & $t_{1/2} = (\ln 2)/K_r$
• B. $[A] = [A]_0 - k_rt$ & $t_{1/2} = (\ln 2)/K_r$
• C. $1/[A] - 1/[A]_0 = k_rt$ & $t_{1/2} = 1/K_r [A]_0$
• D. $[A]_0 = [A]e^{-k_rt}$ & $t_{1/2} = 1/K_r [A]$

Hint:

A key characteristic of first-order reactions is that their half-life is completely independent of the initial concentration $[A]_0$. This is unique to first-order kinetics and is highly useful for identifying reaction orders experimentally.

Answer 1

The Correct Option is A

Step 1: Concept:
The question asks for the mathematical formulas that describe the concentration-time relationship (integrated rate law) and the half-life for a first-order chemical reaction.

Step 2: Key Formula or Approach:
For a first-order reaction $A \rightarrow \text{Products}$, the rate law is given by:
\[ \text{Rate} = -\frac{d[A]}{dt} = k_r[A] \]
Integrating this differential equation yields the concentration profile over time.
The half-life ($t_{1/2}$) is defined as the time required for the concentration of the reactant to drop to exactly half of its initial value ($[A] = [A]_0 / 2$).

Step 3: Step-by-step Explanation:

• Integrated Rate Law: Rearranging the differential rate law gives:
\[ \frac{d[A]}{[A]} = -k_r dt \]
Integrating from $t = 0$ (where $[A] = [A]_0$) to time $t$:
\[ \int_{[A]_0}^{[A]} \frac{1}{[A]} d[A] = -k_r \int_0^t dt \]
\[ \ln[A] - \ln[A]_0 = -k_rt \]
\[ \ln\left(\frac{[A]}{[A]_0}\right) = -k_rt \]
Taking the exponential of both sides gives the standard integrated rate law for a first-order reaction:
\[ [A] = [A]_0 e^{-k_rt} \]

• Half-Life Equation: Substitute $[A] = \frac{[A]_0}{2}$ and $t = t_{1/2}$ into the logarithmic form:
\[ \ln\left(\frac{[A]_0/2}{[A]_0}\right) = -k_r t_{1/2} \]
\[ \ln\left(\frac{1}{2}\right) = -k_r t_{1/2} \]
\[ -\ln(2) = -k_r t_{1/2} \]
\[ t_{1/2} = \frac{\ln 2}{k_r} \]

• Reviewing the options, Option (A) presents both of these correctly derived expressions.

• For completeness: Option (B) mixes zero-order rate law with first-order half-life. Option (C) represents second-order kinetics. Option (D) incorrectly positions the exponential term.

Step 4: Final Answer:
The correct integrated rate law and half-life for a first-order reaction are given in option (A).