Question

Does the half-life of a first-order reaction depend on the initial concentration of the reaction? Explain.

Hint:

Remember the distinction:

• Zero Order: $t_{1/2} \propto [R]_0$ (Directly proportional).
• First Order: $t_{1/2}$ is independent of $[R]_0$.
• Second Order: $t_{1/2} \propto \frac{1}{[R]_0}$ (Inversely proportional).

Answer 1

Step 1: Rate Constant Equation
For a first-order reaction, the integrated rate equation is: \[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \] Where:

• $[R]_0$ is the initial concentration.
• $[R]$ is the final concentration at time $t$.
• $k$ is the rate constant.

Step 2: Half-Life Derivation
At half-life ($t_{1/2}$), the final concentration $[R]$ becomes half of the initial concentration ($[R]_0/2$). Substituting this into the equation: \[ t_{1/2} = \frac{2.303}{k} \log \frac{[R]_0}{[R]_0 / 2} \] \[ t_{1/2} = \frac{2.303}{k} \log 2 \] Since $\log 2 = 0.3010$: \[ t_{1/2} = \frac{2.303 \times 0.3010}{k} \] \[ t_{1/2} = \frac{0.693}{k} \] Step 3: Conclusion
The formula $t_{1/2} = \frac{0.693}{k}$ contains only constant values and does not include the initial concentration term ($[R]_0$). Therefore, the half-life of a first-order reaction is independent of the initial concentration.