Chemical Kinetics - Result Question 65
65. The rate constant for the first order decomposition of a certain reaction is described by the equation
$$ \log k\left(s^{-1}\right)=14.34-\dfrac{1.25 \times 10^{4} K}{T} $$
(i) What is the energy of activation for the reaction?
(ii) At what temperature will its half-life be $256 min$ ?
(1997, 5M)
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Answer:
Correct Answer: 65. $\left(3.56 \times 10^{-16} g\right)$
Solution:
- (i) The Arrhenius equation is: k = A e^(-Ea/(RT))
$ \log k=\log A-\dfrac{E _a}{2.303 R T} $
Comparing with the given equation.
$1.25 \times 10^4=\dfrac{E_a}{2.303 R} \Rightarrow E_a=239.33 \mathrm{~kJ} \mathrm{~mol}^{-1}$
(ii) When half-life $=256 \mathrm{~min}$,
$k=\dfrac{\ln 2}{t_{1 / 2}}=\dfrac{0.693}{256 \times 60} \mathrm{~s}^{-1}=4.5 \times 10^{-5} \mathrm{~s}^{-1}$
$\Rightarrow \dfrac{1.25 \times 10^4}{T}=14.34-\log (4.5 \times 10^{-5})=16.68$
$\Rightarrow \quad T=\dfrac{1.25 \times 10^4}{16.68}=669 \mathrm{~K}$
BETA
