Chemistry - Chemical Kinetics Question with Solution | TestHub

When half-life = 256 min,

$\text{k} = \frac{\mathit{\text{ ln }}2}{{\text{t}}_{1/2}} = \frac{\text{0.693}}{\text{256}\times 60}{\text{ s}}^{-1}$

$= \text{4.5}\times 10^{-5}{\text{ s}}^{-1}$

$\Rightarrow \frac{1.25\times {10}^4}{T} = 14.34- \left[log 4.5\times {10}^{-5}\right]$

$\Rightarrow \text{ 14}\text{.34 -}\left[\text{-5+0}\text{.65}\right]$

$\Rightarrow 19·34-0·65=18·69$

$\text{T}=\frac{\text{12500}}{\text{18.69}}=\text{669K}$