SATHEE: Equilibrium Question 876

Equilibrium Question 876

Question: The rate constant for forward and backward reactions of hydrolysis of ester are $1.1 \times 10^{- 2}$ and $1.5 \times 10^{- 3}$ per minute respectively. Equilibrium constant for the reaction is

$C H_{3} C O O C_{2} H_{5} + H_{2} O$

$\to$ $C H_{3} C O O H$

$+ C_{2} H_{5} O H$

Options:

A) 4.33

B) 5.33

C) 6.33

D) 7.33

Show Answer

Answer:

Correct Answer: D

Solution:

$K_{f} = 1.1 \times 10^{- 2}$ ; $K_{b} = 1.5 \times 10^{- 3}$

$K_{c} = \frac{K_{f}}{K_{b}} = \frac{1.1 \times 10^{- 2}}{1.5 \times 10^{- 3}} = 7.33$ .

पिछला

अगला

Equilibrium Question 876

Question: The rate constant for forward and backward reactions of hydrolysis of ester are $1.1 \times 10^{- 2}$ and $1.5 \times 10^{- 3}$ per minute respectively. Equilibrium constant for 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

Equilibrium Question 876

Question: The rate constant for forward and backward reactions of hydrolysis of ester are 1.1×10−2 and 1.5×10−3 per minute respectively. Equilibrium constant for 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

Menu

Question: The rate constant for forward and backward reactions of hydrolysis of ester are $1.1 \times 10^{- 2}$ and $1.5 \times 10^{- 3}$ per minute respectively. Equilibrium constant for the reaction is