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The first order rate constant for decomposition of ethyl iodide is 1.60 x 10^-5 s^-1 at 600 K with Ea = 209 kJ/mol. What is the rate constant at 700 K? (the calculated log k2 = -2.197)
A6.36 x 10^-3 s^-1
B1.60 x 10^-3 s^-1
C6.36 x 10^-4 s^-1
D3.18 x 10^-2 s^-1
Answer & Solution
Correct answer: A. 6.36 x 10^-3 s^-1
1. $\log k_2 = \log k_1 + \dfrac{E_a}{2.303R}\left[\dfrac{1}{T_1} - \dfrac{1}{T_2}\right]$.
2. $\log k_1 = \log(1.60\times10^{-5}) = -4.796$.
3. The temperature term evaluates to $+2.599$, so $\log k_2 = -4.796 + 2.599 = -2.197$.
4. $k_2 = \text{antilog}(-2.197)$.
5. $= 6.36\times10^{-3}\ \text{s}^{-1}$.
6. Distractor C reads the antilog one decade too small; distractor B keeps the wrong mantissa.
_Source: NCERT Class 12 Chemistry Ch 3 "Chemical Kinetics", p.21_
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