Question

homework 2 : the decomposition of ethane (c₂h₆) to methyl radicals is a first - order reaction with a rate constant of 5.36 × 10⁻⁴ s⁻¹ at 700°c: c₂h₆(g) ⟶ 2ch₃·(g) calculate the half - life of the reaction in minutes.

Explanation:

Step1: Recall first-order half-life formula

For a first - order reaction, the half - life formula is \(t_{1/2}=\frac{\ln2}{k}\), where \(t_{1/2}\) is the half - life and \(k\) is the rate constant.

Step2: Substitute the value of \(k\)

We are given \(k = 5.36\times10^{-4}\ s^{-1}\) and \(\ln2\approx0.693\). So, \(t_{1/2}=\frac{0.693}{5.36\times10^{-4}\ s^{-1}}\).
Calculate \(\frac{0.693}{5.36\times10^{-4}}\approx1292.91\ s\).

Step3: Convert seconds to minutes

Since \(1\ minute = 60\ seconds\), to convert seconds to minutes, we use the conversion factor \(\frac{1\ minute}{60\ seconds}\).
\(t_{1/2}=\frac{1292.91\ s}{60\ s/min}\approx21.55\ minutes\).

Answer:

The half - life of the reaction is approximately \(\boldsymbol{21.6}\) minutes (or more precisely around 21.55 minutes).