MP Board · Class 12 · Chemistry · Chemical KineticsDerive the integrated rate equation for a first-order reaction. Also, show that the half-life period of a first-order reaction is independent of the initial concentration of the reactants.
Derivation of Integrated Rate Equation for a First-Order Reaction
\nConsider a general first-order reaction: $$\text{Reactant (R)} \rightarrow \text{Products (P)}$$ \nLet the initial concentration of reactant $\text{R}$ at time $t = 0$ be $[R]_0$, and the concentration at time $t$ be $[R]_t$. \nAccording to the rate law for a first-order reaction, the rate is directly proportional to the first power of the concentration of the reactant: $$\text{Rate} = -\frac{d[R]}{dt} = k[R]$$ \nWhere $k$ is the rate constant for the first-order reaction. \nRearranging the equation to separate the variables ($[R]$ and $t$): $$\frac{d[R]}{[R]} = -k \cdot dt$$ \nIntegrating both sides of the equation within the limits: at $t = 0$, $[R] = [R]_0$ and at time $t$, $[R] = [R]t$: $$\int{[R]_0}^{[R]t} \frac{d[R]}{[R]} = -k \int{0}^{t} dt$$ \nUsing the standard integration formula $\int \frac{1}{x} dx = \ln(x)$: $$\ln[R]_t - \ln[R]_0 = -kt$$ \nUsing the logarithmic property $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$: $$\ln\left(\frac{[R]_t}{[R]_0}\right) = -kt$$ \nConverting natural logarithm ($\ln$) to base-10 logarithm ($\log$): $\ln x = 2.303 \log x$: $$2.303 \log\left(\frac{[R]_t}{[R]_0}\right) = -kt$$ \nRearranging for the rate constant $k$ or time $t$: $$k = \frac{2.303}{t} \log\left(\frac{[R]_0}{[R]_t}\right)$$
Half-Life Period ($t_{1/2}$) of a First-Order Reaction
\nThe half-life of a reaction is the time in which the concentration of the reactant is reduced to one-half of its initial concentration. \nThus, at $t = t_{1/2}$, $$[R]_t = \frac{[R]0}{2}$$ \nSubstituting this value into the integrated rate equation: $$k = \frac{2.303}{t{1/2}} \log\left(\frac{[R]_0}{\frac{[R]_0}{2}}\right)$$
$$k = \frac{2.303}{t_{1/2}} \log(2)$$ \nSince $\log(2) = 0.3010$: $$k = \frac{2.303 \times 0.3010}{t_{1/2}}$$
$$k = \frac{0.693}{t_{1/2}}$$ \nRearranging to solve for $t_{1/2}$: $$t_{1/2} = \frac{0.693}{k}$$
Conclusion: The final expression shows that the half-life of a first-order reaction ($t_{1/2} = \frac{0.693}{k}$) depends only on the rate constant $k$ and is completely independent of the initial concentration $[R]_0$ of the reactant.