Chemical Kinetics - Result Question 75
####74. In Arrhenius equation for a certain reaction, the value of $A$ and $E _a$ (activation energy) are $4 \times 10^{13} s^{-1}$ and $98.6 kJ mol^{-1}$ respectively. If the reaction is of first order, at what temperature will its half-life period be $10 min$ ?
(1990, 3M)
Show Answer
Solution:
- Arrhenius equation is :
$$ \begin{gathered} \log k=\log A-\frac{E _a}{2.303 R T} \ \text { when } \begin{aligned} t _{1 / 2}=10 min, k & =\frac{\ln 2}{t _{1 / 2}}=\frac{0.693}{10 \times 60}=1.115 \times 10^{-3} s^{-1} \ \Rightarrow \quad \frac{E _a}{2.303 R T} & =\log A-\log k \ & =\log \frac{A}{k}=\log \frac{4 \times 10^{13}}{1.115 \times 10^{-3}}=16.54 \ \Rightarrow T=\frac{E _a}{2.303 R \times 16.54} & =\frac{98.6 \times 1000}{2.303 \times 16.54 \times 8.314} \ & =311.34 K \end{aligned} \end{gathered} $$
