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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
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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