Question

Derive the relation between half life and rate constant for a first order reaction.

Hint:

A key feature of first-order reactions is their constant half-life. This is why radioactive decay, a first-order process, is described by a half-life (e.g., Carbon-14 has a half-life of 5730 years) regardless of how much material you start with.

Answer 1

Step 1: Start with the integrated rate law for a first-order reaction. The integrated rate law relates the initial concentration of a reactant, [A]\(_0\), to its concentration at any time t, [A]\(_t\), for a rate constant k. \[ \ln\frac{[\text{A}]_0}{[\text{A}]_t} = kt \] Alternatively, in base-10 logarithm form: \[ k = \frac{03}{t} \log_{10}\frac{[\text{A}]_0}{[\text{A}]_t} \] Step 2: Define half-life (\(t_{1/2}\)). Half-life is the time required for the concentration of a reactant to decrease to half of its initial value. At \( t = t_{1/2} \), the concentration \( [\text{A}]_t = \frac{1}{2}[\text{A}]_0 \). Step 3: Substitute the half-life condition into the integrated rate law. Using the natural logarithm form: \[ \ln\frac{[\text{A}]_0}{\frac{1}{2}[\text{A}]_0} = k \cdot t_{1/2} \] Step 4: Simplify the equation. The [A]\(_0\) terms cancel out: \[ \ln(2) = k \cdot t_{1/2} \] Step 5: Solve for half-life (\(t_{1/2}\)). Since \(\ln(2) \approx 0.693\): \[ 0.693 = k \cdot t_{1/2} \] \[ t_{1/2} = \frac{0.693}{k} \] This is the relation between the half-life and the rate constant for a first-order reaction. It shows that the half-life is constant and independent of the initial concentration of the reactant.