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  • At 407 K, the rate constant of a chemical reaction is <img alt="9.5 \times 10^{-5} \mathrm{~s}^{-1}" src="https://entrancecorner.oncodecogs.com/gif.latex?9.5%20%5Ctimes%2010%5E%7B-5%7D%20%5Cmathrm%

At 407 K, the rate constant of a chemical reaction is 9.5 \times 10^{-5} \mathrm{~s}^{-1} and at 420 K, the rate constant is 1.9 \times 10^{-4} \mathrm{~s}^{-1}. Find the value of log (A) where A is the frequency factor of the reaction.
Take: log (9.5)=0.97.

Option: 1

5.68


Option: 2

8.34


Option: 3

13.1


Option: 4

11.01


Answers (1)

best_answer

Expression for rate constant at two different temperature is given by,
\mathrm{\log _{10} \frac{k_2}{k_1}=\frac{E_a}{2.303 \times R}\left[\frac{T_2-T_1}{T_1 T_2}\right]}
Given: \mathrm{k_1=9.5 \times 10^{-5} s^{-1} ; k_2=1.9 \times 10^{-4} s^{-1}};
\mathrm{R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}};
\mathrm{T_1=407 \mathrm{~K} {~and} {~T_2=420 \mathrm{~K}}}
Substituting the values in above equation
\mathrm{ \log _{10} \frac{1.9 \times 10^{-4}}{9.5 \times 10^{-5}}=\frac{E_a}{2.303 \times 8.314}\left[\frac{420-407}{420 \times 407}\right] }
\mathrm{E_a=75531.05 \mathrm{~J} \mathrm{~mol}^{-1}}
Arrhenius equation of a reaction at 407 K
\mathrm{ k_1=A e^{-E_a / R T_1} }
Taking log on both sides,
\mathrm{\log k_1=\log A-\frac{E_a}{2.303 R T_1}}
Substituting the values,
\mathrm{ \log \left(9.5 \times 10^{-5}\right)=\log A-\frac{75531}{2.303 \times 8.314 \times 407} }

\mathrm{ \Rightarrow-4.02=\log A-9.7}

\mathrm{ \log A=5.68}

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avinash.dongre

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