MP Board · Class 12 · Chemistry · Chemical KineticsDerive the integrated rate equation for a zero-order reaction. Also, define half-life period ($t{1/2}$) for a zero-order reaction and show that it is directly proportional to the initial concentration of the reactants.

Derivation of Integrated Rate Equation for a Zero-Order Reaction

\nA zero-order reaction is one whose rate is independent of the concentration of the reactants. \nLet us consider a general zero-order reaction: $$R \rightarrow \text{Products}$$

1. Rate Expression:\nThe rate of the reaction can be written as: $$\text{Rate} = -\frac{d[R]}{dt} = k[R]^0$$\nSince anything raised to the power zero is 1 ($[R]^0 = 1$), the equation simplifies to: $$-\frac{d[R]}{dt} = k$$

2. Rearrangement and Integration:\nRearranging the equation to isolate concentration terms: $$d[R] = -k \cdot dt$$ \nIntegrating both sides within the limits from time $t = 0$ (where concentration is $[R]_0$) to time $t = t$ (where concentration is $[R]t$): $$\int{[R]_0}^{[R]t} d[R] = -k \int{0}^{t} dt$$ \nEvaluating the definite integrals: $$[R]_t - [R]_0 = -k \cdot t$$ \nRearranging for $[R]_t$ gives the integrated rate equation: $$[R]_t = -kt + [R]_0$$

Half-Life Period ($t_{1/2}$) of a Zero-Order Reaction

\nThe half-life period of a reaction is defined as the time during which the concentration of the reactant is reduced to half of its initial concentration.

• At $t = t_{1/2}$, $[R]_t = \frac{[R]0}{2}$ \nSubstitute these values into the integrated rate equation: $$\frac{[R]0}{2} = -kt{1/2} + [R]0$$ \nRearranging to solve for $t{1/2}$: $$kt{1/2} = [R]_0 - \frac{[R]0}{2}$$ $$kt{1/2} = \frac{[R]0}{2}$$ $$t{1/2} = \frac{[R]_0}{2k}$$

Conclusion:\nFrom the final expression, it is clear that the half-life of a zero-order reaction ($t_{1/2}$) is directly proportional to the initial concentration of the reactants ($[R]_0$) and inversely proportional to the rate constant ($k$).