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 $
