SATHEE: Equilibrium Question 602

Equilibrium Question 602

Question: In the reversible reaction $A + B$ ⇌ $C + D$ , the concentration of each C and D at equilibrium was 0.8 mole/liter, then the equilibrium constant $K_{c}$ will be [MP PET 1986]

Options:

A) 6.4

B) 0.64

C) 1.6

D) 16.0

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Answer:

Correct Answer: D

Solution:

Suppose 1 mole of A and B each taken then 0.8 mole/litre of C and D each formed remaining concentration of A and B will be (1 - 0.8) = 0.2 mole/litre each. $K c = \frac{[C] [D]}{[A] [B]} = \frac{0.8 \times 0.8}{0.2 \times 0.2} = 16.0$

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Equilibrium Question 602

Question: In the reversible reaction $A + B$ ⇌ $C + D$ , the concentration of each C and D at equilibrium was 0.8 mole/liter, then the equilibrium constant $K_{c}$ will be [MP PET 1986]

Options:

Answer:

Solution:

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एनसीईआरटी व्यायाम वीडियो समाधान

Equilibrium Question 602

Question: In the reversible reaction A+B ⇌ C+D , the concentration of each C and D at equilibrium was 0.8 mole/liter, then the equilibrium constant Kc will be [MP PET 1986]

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: In the reversible reaction $A + B$ ⇌ $C + D$ , the concentration of each C and D at equilibrium was 0.8 mole/liter, then the equilibrium constant $K_{c}$ will be [MP PET 1986]