Question

The following reaction was performed in a sealed vessel at 777 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and I2 were present at concentrations of [H2]=3.50M and [I2]=2.50M. The equilibrium concentration of I2 is 0.0400 M . What is the equilibrium constant, Kc, for the reaction at this temperature? Express your answer numerically.

Answer 1

The following reaction was performed in a sealed vessel at 777 ∘C : H2(g)+I2(g)⇌2HI(g)

Initially, only H2 and I2 were present at concentrations of [H2]=3.50M and [I2]=2.50M.

The equilibrium concentration of I2 is 0.0400 M . What is the equilibrium constant, Kc, for the reaction at this temperature? Express your answer numerically.

Let's make an ICE table for the reaction.

                   H2(g) + I2(g) <=> 2HI(g)
initial . . . . . 3.50 . . . .2.50 . . . . . . .0 . .
change . . . . .-x . . . . . .-x . . . . . . .2x . .
at equ. . . . 3.50-x . . .2.50-x . . . . . 2x . .

Given that; The equilibrium concentration of I2 is 0.0400 M

Therefore =
[I2] = 2.50-x = 0.0400

X= 2.46.

[H2] = 3.50 - x = 3.40 – 2.46

= 0.94 M

[HI] = 2X= 2*2.46 = 4.92 M

Now calculate the equilibrium consist;

Kc = [HI]^2 / ([H2][I2])

Kc= (4.92)^2 / ((0.94)(2.46))

= 10.5