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If initially the concentration of A and B are both at equilibrium ,concentration of D will be twice of that of A , then what is the equilibrium constant of the reaction? the reaction is shown as below; A +B sign of equilibrium C+D

If initially the  concentration of A and B are both at equilibrium ,concentration of D will be twice of that of A , then what is the equilibrium constant of the reaction? the reaction is shown as below; A +B sign of  equilibrium C+D

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1 Answers

Priyansh Goel
21 Points
11 years ago

Let there be a moles of A and B each at time =  0.

At equilibrium,

 

[A] =  a-x

[B] = a-x

[C] = x

[D] =  x

Keq =  x2/(a-x)2 ........................................  (1)

Given

[D]  = 2[A] so, x =  2(a-x)

Put this in (1)

Keq  = 4

If there is any error , plz do tell me.

 

If it is correct plz approve my answer.

 

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