Question

What does the 0.693 represent in the equation of half-life - \[{{\text{t}}_{{1}/{2}\;}}=\dfrac{0.693}{\text{k}}\].

Answer 1

Hint: Half-life is the time taken by the reactant species to decompose to half of its initial amount. It is inversely proportional to the rate constant for a first-order reaction and it is given as:
\[{{\text{t}}_{{1}/{2}\;}}=\dfrac{\ln 2}{\text{k}}\approx \dfrac{0.693}{\text{k}}\]

Complete answer:
The half-life of a first-order reaction under a given set of reaction conditions is a constant. It is the time taken by the reactant to get used up to its half concentration.
The integrated rate law for a first-order reaction is:
\[\ln \dfrac{\left[ {{\text{A}}_{\text{o}}} \right]}{\left[ \text{A} \right]}=\text{kt}\]
Where $\left[ {{\text{A}}_{\text{o}}} \right]$ is the initial concentration of the reactant and $\left[ \text{A} \right]$ is the concentration of reactant after time t.
k is the rate constant of the first-order reaction.
To find out the half-life, we have to substitute $\left[ \dfrac{{{\text{A}}_{\text{o}}}}{2} \right]$ for $\left[ \text{A} \right]$ and ${{\text{t}}_{{1}/{2}\;}}$ for $\text{t}$ into the above equation, we get:
\[\begin{align}
  & \ln \dfrac{\left[ {{\text{A}}_{\text{o}}} \right]}{\left[ \dfrac{{{\text{A}}_{\text{o}}}}{2} \right]}=\text{k}{{\text{t}}_{{1}/{2}\;}} \\
 & \Rightarrow \ln 2=\text{k}{{\text{t}}_{{1}/{2}\;}} \\
 & \Rightarrow {{\text{t}}_{{1}/{2}\;}}=\dfrac{\ln 2}{\text{k}} \\
\end{align}\]
We can approximate the value of $\ln 2$ to 0.693.
\[\therefore {{\text{t}}_{{1}/{2}\;}}=\dfrac{\ln 2}{\text{k}}\approx \dfrac{0.693}{\text{k}}\]
Hence, 0.693 in the equation of half-life represents the value of $\ln 2$.

Note:
The half-life equation of the first-order reaction indicates that half-life is independent of the concentration of the reactants and thus if we know the rate constant for a first-order reaction, then we can use half-lives to predict how much time is needed for the reaction to reach a certain percent completion.