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Sagot :
Answer:
Yes
Explanation:
By definition, the equilibrium constanct, Kc, for the reaction A ⇒ 2B is
= [A]^1 / [B]^2
Substitute [A] = 4 and [B] = 2 in the equation,
[A]^1 / [B]^2
= 4^1 / 2^2
= 1
= Kc
So yes the reaction is at equilibrium.
Let's find K_c
[tex]\\ \rm\Rrightarrow K_c=\dfrac{[B^]^2}{[A]^1}[/tex]
[tex]\\ \rm\Rrightarrow 1=\dfrac{2^2}{4}[/tex]
[tex]\\ \rm\Rrightarrow 1=\dfrac{4}{4}[/tex]
[tex]\\ \rm\Rrightarrow 1=1[/tex]
Yes it's at equilibrium
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