Question

The rate constant for the first order decomposition of a certain reaction is described by the equation:
$\log k \left( s ^{-1}\right)=14.34-\frac{1.25 \times 10^{4} K }{ T }$
At what temperature will its half-life period be $\text{256\, min}$ ?

• A. $669\, K$
• B. $689\, K$
• C. $76\,9 K$
• D. $469\, K$

Answer 1

Answer: (A)

When half-life $=256\, \min$,
$k =\frac{\ln 2}{ t _{1 / 2}}=\frac{0.693}{256 \times 60} s ^{-1} $
$=4.5 \times 10^{-5} s ^{-1}$
$\Rightarrow \frac{1.25 \times 10^4}{T}=14.34-\left[\log 4.5 \times 10^{-5}\right]$
$\Rightarrow \frac{1.25 \times 10^4}{T}=14.34-[-5+0.65] $
$\Rightarrow \frac{1.25 \times 10^4}{T}=19 \cdot 34-0 \cdot 65=18 \cdot 69$
$ T =\frac{12500}{18.69}=669\, K$