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Question

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

This question was previously asked in
CUET PG 2026 Agri-Business Management Question Paper (25-Mar-2026) (Shift 2)
The correct answer is
$[A] = [A]_0 e^{-k_r t} \text{ \& } t_{1/2} = (\ln 2) / K_r$

First Order Reaction Equations

A first-order reaction is a chemical reaction where the rate of reaction depends on the concentration of only one reactant. The question asks for the specific mathematical expressions governing its concentration over time and its half-life.

Integrated Rate Law for First Order

The integrated rate law describes how the concentration of a reactant, $[A]$, changes over time, $t$. For a first-order reaction A $\to$ Products, the relationship is derived from the rate law Rate = $k_r[A]$.

The integrated form is:

$[A] = [A]_0 e^{-k_r t}$

Where:

  • $[A]$ is the concentration of reactant A at time $t$.
  • $[A]_0$ is the initial concentration of reactant A at time $t=0$.
  • $k_r$ is the first-order rate constant.
  • $t$ is the time.
  • $e$ is the base of the natural logarithm.

Half-Life Equation for First Order

The half-life ($t_{1/2}$) is the time required for the concentration of a reactant to decrease to half of its initial value ($[A] = [A]_0 / 2$).

Substituting $[A] = [A]_0 / 2$ into the integrated rate law yields the half-life equation for a first-order reaction:

$t_{1/2} = \frac{\ln 2}{k_r}$

Note that the half-life for a first-order reaction is constant and independent of the initial concentration.

Conclusion

Comparing these derived equations with the given options, Option A provides the correct integrated rate law ($[A] = [A]_0 e^{-k_r t}$) and the correct half-life equation ($t_{1/2} = (\ln 2) / K_r$, assuming $K_r$ represents $k_r$).

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