Chemistry - Chemical Kinetics Question with Solution | TestHub

For a first-order reaction, the concentration after $n$ half-lives is given by $[A{]}_t=\frac{[A{]}_0}{2^n}$.

Given $[A{]}_0=1.0\text{M}$ and $[A{]}_t=0.25\text{M}$, we have:

$0.25=\frac{1.0}{2^n}\Rightarrow 2^n=4\Rightarrow n=2$

Thus, two half-lives have passed in $276\text{s}$.

$2t_{1/2}=276\text{s}\Rightarrow t_{1/2}=138\text{s}$

The rate constant $k$ is given by:

$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{138}$

$k\approx 0.0050{\text{s}}^{-1}$