SATHEE: Chemical Kinetics Question 304

Chemical Kinetics Question 304

Question: A first order reaction is 50% completed in 20 minutes at $27^{\circ} C$ and in 5 minutes at $47^{\circ} C$ . The energy of activation of the reaction is:

Options:

A) 43.85 kJ/mol

B) 55.14 kJ/mol

C) 11.97 kJ/mol

D) 6.65 kJ/mol

Show Answer

Answer:

Correct Answer: B

Solution:

[b] $k_{1 (300)} = \frac{0.693}{20};$

$k_{2 (320)} = \frac{0.693}{5}$ In $\frac{k_{2 (320)}}{k_{1 (300)}} = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$E_{a} = \frac{2.303 R T_{1} T_{2}}{(T_{2} - T_{1})} \log \frac{k_{2}}{k_{1}}$

$= \frac{2.303 \times 8.314}{20 \times 1000} \times 300 \times 320 \log 4$

$= 55.14 k J / m o l$

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Chemical Kinetics Question 304

Question: A first order reaction is 50% completed in 20 minutes at $27^{\circ} C$ and in 5 minutes at $47^{\circ} C$ . The energy of activation of the reaction is:

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Chemical Kinetics Question 304

Question: A first order reaction is 50% completed in 20 minutes at 27∘C and in 5 minutes at 47∘C . The energy of activation of the reaction is:

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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Question: A first order reaction is 50% completed in 20 minutes at $27^{\circ} C$ and in 5 minutes at $47^{\circ} C$ . The energy of activation of the reaction is: