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If the equilibrium constant for A+2B + 5/2D is 4.0 then what is the value of the equilibrium constant for the reaction 2C+ 5D 2A + 4B at the same temperature?

one year ago

Arun
23032 Points

A + 2B → C + (5/2)D …… K₁ = [C] [D]^(5/2) / {[A] [B]^2} = 4.0
2C + 5D → 2A + 4B …… K₂ = [A]^2 [B]^4 / {[C]^2 [D]^5} = ?

K₁ = 4.0
[C] [D]^(5/2) / {[A] [B]^2} = 4.0

Reciprocal of the above expression :
[A] [B]^2 / {[C] [D]^(5/2)} = 1 / 4.0
[A] [B]^2 / {[C] [D]^(5/2)} = 0.25

Square of the above expression :
[A]^2 [B]^4 / {[C]^2 [D]^5} = 0.25^2
[A]^2 [B]^4 / {[C]^2 [D]^5} = 0.0625
[A]^2 [B]^4 / {[C]^2 [D]^5} ≈ 0.063 (to 2 sig. fig.)

Hence, K₂ = [A]^2 [B]^4 / {[C]^2 [D]^5} = 0.063
one year ago
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