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. $669K$
• B. $689K$
• C. $769K$
• D. $469K$

Answer 1

Answer: (A)

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