Question

In: Chemistry

The following reaction was performed in a sealed vessel at 738 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only...

The following reaction was performed in a sealed vessel at 738 ∘C :

H2(g)+I2(g)⇌2HI(g)

Initially, only H2 and I2 were present at concentrations of [H2]=3.40M and [I2]=2.25M. The equilibrium concentration of I2 is 0.0100 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Expert Solution

H2 + I2 <---------------------> 2HI

3.40 2.25 0 --------------> initial

3.40-x 2.25-x 2x ----------------> equilibrium

but equilibrium concentration of I2 is 0.0100 M is given

so 2.25-x = 0.01

x = 2.24

equilbrium concnetrations :

[H2] = 3.40 - 2.24 = 1.16 M

[I2] = 0.01 M

[HI] = 2x = 2 x 2.24 = 4.48 M

equilibrium constant = Kc = [HI]^2 / [H2][I2]

= (4.48)^2 / (1.16) (0.01)

= 1730

Kc = 1730

queen_honey_blossom answered 2 years ago

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