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 For a first order Reaction \mathrm{R \rightarrow P,} the temperature dependent rate constant is found to follow the expression \mathrm{\log k=-\frac{2000}{2.303 T}+\log 10^6}. The pre-exponenti factor and activation energy are respectively
 

Option: 1

10^6 \mathrm{~s}^{-1}, 8.314 \mathrm{kJmn}^{-1}


Option: 2

6 \mathrm{~s}^{-1}, 16.628 \mathrm{kJmn}^{-1}


Option: 3

10^6 \mathrm{~s}^{-1}, 16.6 \mathrm{kJmol}^{-1}


Option: 4

6 \mathrm{~Hz}, 8.314 \mathrm{kJmol}^{-1}


Answers (1)

best_answer

Given

\begin{aligned} &\mathrm{ \log k=-\frac{2000}{2.303 T}+\log 10^6}\\ &\mathrm{ Multiply~ by ~2.303 , } \end{aligned}
\begin{gathered} \mathrm{2.303 \log k=-\frac{2000}{T}+6 \times 2.303 \log 10 \text { (1) }} \\ \mathrm{k=A e^{-E a / R T} \Rightarrow \ln k=\ln A-\frac{E_a}{R T}(\text { Compare }) }\\ \mathrm{\text { By (1), } \quad \ln k=\ln 10^6 \cdot \frac{2000}{T}} \end{gathered}
\mathrm{\text{By comparison, } A=10^6 \mathrm{~s}^{-1} \&~ E_a=+16.628~ \mathrm{kJmol}^{-1}}

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Gautam harsolia

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