Question

At equilibrium 500ml vessels contains 1.5 m of each a,b,c,d if 0.5 m of c and d expelled out than what would be the kc

Answer 1

Answer : The value of kc is 1.

Explanation :

The given reaction is

$A + B \rightarrow C + D$

We have been given that , at equilibrium we have 1.5 moles of each substance.

The concentration of each substance at equilibrium is

$Concentration (M) =\frac{mol}{L} =\frac{1.5}{0.500L} = 3 M$

Therefore at equilibrium we have,

[A] = 3 M

[B] = 3 M

[C] = 3 M

[D} = 3 M

The equilibrium constant kc for this reaction is calculated as,

$k_{c} = \frac{[C] [D]}{[A] [B]}$

$k_{c} =\frac{(3) (3)}{(3) (3)} = 1$

The value of equilibrium constant kc is 1.

Equilibrium constant depends only on temperature and is unaffected by changes in reactants or products.

This is explained by Le-Chatelier's principle.

According to this, equilibrium is dynamic and when a stress is applied to a system, it moves in such a direction so as to reduce the effect of the applied stress.

When we remove 0.5 moles of C and D from the above equilibrium system, the equilibrium moves towards product side and equal amounts of reactants react in order to keep kc value constant.

Therefore the value of kc is 1.

Answer 2

Answer:

1 is the right ans

Explanation:

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