COMEDK · Chemistry · 19. Chemical Kinetics
The temperature \((\mathrm{T})\) and rate constant \((\mathrm{k})\) for a first order reaction \(\mathrm{R} \rightarrow \mathrm{P}\), was found to follow the equation \(\log \mathrm{k}=-(2000) \dfrac{1}{\mathrm{~T}}+8.0\). The pre-exponential factor '\(\mathrm{A}\)' and activation energy \(\mathrm{E}_{\mathrm{a}}\), respectively are: [Given: \(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)]
- A \(6.0 \mathrm{~s}^{-1} \text { and } 10^8 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- B \(1 \times 10^8 \mathrm{~s}^{-1} \text { and } 38.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- C \(1 \times 10^{-8} \mathrm{~s}^{-1} \text { and } 16.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- D \(8.0 \mathrm{~s}^{-1} \text { and } 16.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(B) \(1 \times 10^8 \mathrm{~s}^{-1} \text { and } 38.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
The Arrhenius equation is given by \(k = A e^{-E_a / (RT)}\). Taking the logarithm on both sides, we get \(\log k = \log A - \dfrac{E_a}{2.303 RT}\). Comparing this with the given equation \(\log k = -2000 \left( \dfrac{1}{T} \right) + 8.0\), we identify the slope and the…
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