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The following reaction was performed in a sealed vessel at 742 ∘C : H2(g)+I2(g)⇌2HI(g) Initially, only H2 and...

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Chemistry

The following reaction was performed in a sealed vessel at 742∘C :

H2(g)+I2(g)⇌2HI(g)

Initially, only H2 and I2 were present at concentrations of[H2]=3.50M and [I2]=2.00M. The equilibriumconcentration of I2 is 0.0900 M . What is the equilibriumconstant, Kc, for the reaction at this temperature?

Answer & Explanation Solved by verified expert
4.3 Ratings (713 Votes)

H2(g)     +   I2(g) <---------------------------> 2HI(g)

3.50            2.00                                            0 ----------------------> initial

3.50-x          2.00-x                                        2x ----------------------> equilibrium

but equilibrium concentration of I2 is 0.0900 M is given so

2.00 - x = 0.09

x = 1.91

equilibrium concentrations :

[H2] = 3.50 - x = 3.50 -1.91 = 1.59 M

[I2] = 0.09 M

[HI] = 2x = 2 x 1.91 = 3.82 M

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

Kc = (3.82)^2 / (1.59) (0.09)

Kc = 102
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