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

Show Answer

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. $ Kc=\frac{[C][D]}{[A][B]}=\frac{0.8\times 0.8}{0.2\times 0.2}=16.0 $



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