Calculating Equilibrium Constants The following reaction was performed in a sealed vessel at 752 ∘C :...

Calculating Equilibrium Constants

The following reaction was performed in a sealed vessel at 752 ∘C :
H2(g)+I2(g)⇌2HI(g)
Initially, only H2 and I2 were present at concentrations of [H2]=3.75M and [I2]=3.00M. The equilibrium concentration of I2 is 0.0900 M . What is the equilibrium constant, Kc, for the reaction at this temperature?

Solutions

Expert Solution

H2(g) + I2(g) ⇌ 2HI(g)

Initial 3.75 M 3.00 M 0

at equilibrium 3.75- x 3-x 2 x

So,

equilibrium concentration of I2 = 3-x

But given that equilibrium concentration of I2 = 0.0900 M

Then,

3-x = 0.09

x = 2.91 M

Therefore, equilibrium concentrations are

[H2] = 3-75 -x = 3.75 - 2.91 = 0.84 M

[I2] = 0.09 M

[HI]= 2 x= 2 x 2.91 = 5.82 M

Kc = [HI]²/ [H2] [I2]

= (5.82)² / (0.84) (0.09)

= 448.05

Therefore, equilibrium constant Kc = 448.05

queen_honey_blossom answered 1 year ago

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