SATHEE: Chemical Kinetics Question 79

Chemical Kinetics Question 79

Question: In a first order reaction the concentration of reactant decreases from $800, m o l / d m^{3}$ to $50, m o l / d m^{3}$ is $2 \times 10^{2} \sec$ . The rate constant of reaction in $\sec^{- 1}$ is [IIT-JEE (Screening) 2003]

Options:

A) $2 \times 10^{4}$

B) $3.45 \times 10^{- 5}$

C) $1.386 \times 10^{- 2}$

D) $2 \times 10^{- 4}$

Show Answer

Answer:

Correct Answer: C

Solution:

$k = \frac{2.303}{t} \log_{10} \frac{a}{a - x}$ ; $t = 2 \times 10^{2}, a = 800, a - x = 50$

$k = \frac{2.303}{2 \times 10^{2}} \log_{10} \frac{800}{50} = \frac{2.303}{2 \times 10^{2}} \log_{10} 16$

$= \frac{2.303}{2 \times 10^{2}} \log_{10} 2^{4} = \frac{2.303}{2 \times 10^{4}} \times 4 \times 0.301$

$= 1.38 \times 10^{- 2} s^{- 1}$

पिछला

अगला

Chemical Kinetics Question 79

Question: In a first order reaction the concentration of reactant decreases from $800, m o l / d m^{3}$ to $50, m o l / d m^{3}$ is $2 \times 10^{2} \sec$ . The rate constant of reaction in $\sec^{- 1}$ is [IIT-JEE (Screening) 2003]

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 79

Question: In a first order reaction the concentration of reactant decreases from 800,mol/dm3 to 50,mol/dm3 is 2×102sec . The rate constant of reaction in sec−1 is [IIT-JEE (Screening) 2003]

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

Menu

Question: In a first order reaction the concentration of reactant decreases from $800, m o l / d m^{3}$ to $50, m o l / d m^{3}$ is $2 \times 10^{2} \sec$ . The rate constant of reaction in $\sec^{- 1}$ is [IIT-JEE (Screening) 2003]