The following reaction was performed in a sealed vessel at 798 ∘C : H2(g)+I2(g)⇌2HI(g)?
Initially, only H2 and I2 were present at concentrations of [H2]=3.55M and [I2]=2.90M. The equilibrium concentration of I2 is 0.0300 M . What is the equilibrium constant, Kc, for the reaction at this temperature?
Express your answer numerically.
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Answers & Comments
At equilibrium [I2] = 0.0300
[I2] that reacts: = 2.90 – 0.0300 = 2.87
[H2] that react : 2.87 moles I2 /1 liter x 1 mole H2/1mole I2 = 2.87
{H2] at equilibrium = 3.55 – 2.87 = 0.68
[HI] that is produced:
2.87 moles H2/1liter x 2 moles HI/1mole H2 = 5.74
Kc = [HI]^2 / [H2][I2]
Kc = (5.74)^2 / (0.68)(0.030) = 1.62 x 10^3