Question

The velocity constant \(K\) for a first order reaction was found to be \(5.5 \times 10^{-14}\,s^{-1}\). Calculate the half life of this reaction.

Hint:

First order half life \(t_{1/2}=0.693/k\); plug in \(k=5.5\times10^{-14}\,s^{-1}\).

Answer 1

Step 1: Recall the origin. For first order kinetics, integrating gives \(\ln 2 = k\,t_{1/2}\), so \(t_{1/2}=\dfrac{\ln 2}{k}=\dfrac{0.693}{k}\). Note the answer does not depend on how much reactant we start with.
Step 2: Insert data. \(t_{1/2}=\dfrac{0.693}{5.5\times10^{-14}\,s^{-1}}\).
Step 3: Simplify the power of ten. Dividing by \(10^{-14}\) multiplies by \(10^{14}\): \(t_{1/2}=(0.693/5.5)\times10^{14}=0.126\times10^{14}\,s\).
Step 4: Final form. Writing in proper scientific notation, \(t_{1/2}=1.26\times10^{13}\,s\).
\[\boxed{1.26\times10^{13}\,s}\]