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Q3.13(ii):
Calculate the half-life of a first order reaction from their rate constants given below: (ii) \(2 min^{-1}\)

Solution :

Step 1: Theoretical Foundation of First-Order Kinetics

For a reaction following first-order kinetics, the rate of reaction is directly proportional to the concentration of the reactant. The integrated rate law is expressed as $kt = \ln\frac{[R]_0}{[R]}$. By definition, the half-life ($t_{1/2}$) is the time required for the concentration of the reactant to be reduced to half of its initial value, i.e., $[R] = \frac{[R]_0}{2}$ at $t = t_{1/2}$.

Step 2: Derivation of the Half-Life Formula

Substituting the half-life conditions into the integrated rate equation:

$k \cdot t_{1/2} = \ln\frac{[R]_0}{[R]_0 / 2}$

$k \cdot t_{1/2} = \ln(2)$

Since $\ln(2) \approx 0.693$ [Per logarithmic constant properties]:

$t_{1/2} = \frac{0.693}{k}$

Step 3: Identification of Given Variables

  • Rate constant ($k$) = $2 \text{ min}^{-1}$

Step 4: Calculation of Half-Life

Applying the values to the derived formula:

$t_{1/2} = \frac{0.693}{2 \text{ min}^{-1}}$

$t_{1/2} = 0.3465 \text{ min}$

Visual Representation of First-Order Decay

[R] t t = 0.3465 min

Analytical Summary

Parameter Value
Rate Constant ($k$) $2 \text{ min}^{-1}$
Half-life ($t_{1/2}$) $0.3465 \text{ min}$

Final Solution: The half-life of the first-order reaction is 0.3465 min.


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