Question

The rate constant for the first order decomposition of a certain reaction is described by the equation:

$\text{ log }\text{k }\left({\left(\text{s}\right)}^{-1}\right)=\text{14.34}-\frac{\text{1.25}\times {\left(\text{10}\right)}^4\text{ K}}{\text{T}}$

At what temperature will its half-life period be $\text{256 min}$ ?

• A. 239.33 kJ ; 669 K
• B. 259.33 kJ
• C. 329.33 kJ
• D. 539.33 kJ

Answer 1

Answer: (A)

When half-life $=256min$

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