A chemical equilibrium is:
text A end text space left right double arrow 2 text B end text
A volume of 9 L is filled with 6 mol of A with none of B present. At equilibrium 1 mol of A remain.
Calculate the equilibrium constant, Kc.
chemistry question?
2016-10-13 5:41 pm
回答 (3)
2016-10-13 5:56 pm
A ⇌ 2B
Initial concentrations :
[A]ₒ = 6/9 mol/L
[B]ₒ = 0 mol/L
At equilibrium :
[A] = 1/9 mol/L
Decrease in [A] = [(6/9) - (1/9)] mol/L = 5/9 mol/L
Hence, [B] = (5/9) × 2 mol/L = 10/9 mol/L
Equilibrium constant, Kc = [B]² / [A] = (10/9)² / (1/9) = 11 (to 2 sig. fig.)
Initial concentrations :
[A]ₒ = 6/9 mol/L
[B]ₒ = 0 mol/L
At equilibrium :
[A] = 1/9 mol/L
Decrease in [A] = [(6/9) - (1/9)] mol/L = 5/9 mol/L
Hence, [B] = (5/9) × 2 mol/L = 10/9 mol/L
Equilibrium constant, Kc = [B]² / [A] = (10/9)² / (1/9) = 11 (to 2 sig. fig.)
2016-10-13 5:52 pm
So, the reaction is:
A <--> 2 B
Initial [A] = 6 mol/9L = 0.67 M
Final [A] = 1 mol/9L = 0.11 M
Since you consumed 5 moles of A to reach equilibrium, you will have formed 10 moles of B. So
Final [B] = 10 mol / 9L = 1.1 M
Kc = [B]^2/[A] = (1.1)^2 / 0.11 = 11
A <--> 2 B
Initial [A] = 6 mol/9L = 0.67 M
Final [A] = 1 mol/9L = 0.11 M
Since you consumed 5 moles of A to reach equilibrium, you will have formed 10 moles of B. So
Final [B] = 10 mol / 9L = 1.1 M
Kc = [B]^2/[A] = (1.1)^2 / 0.11 = 11
2016-10-13 6:01 pm
thank you
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