At a given temperature, 1.43 atm of H2 and 3.15 atm of I2 are mixed and...

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Chemistry

At a given temperature, 1.43 atm of H2 and 3.15 atm of I2 aremixed and allowed to come to equilibrium. The equilibrium pressureof HI is found to be 1.844 atm. Calculate Kp for the reaction atthis temperature. H2(g) + I2(g) <=> 2 HI(g).

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4.0 Ratings (809 Votes)

construct the ICE table

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

I         1.43      3.15                0

C         -x          -x                +2x

E        1.43-x       3.15-x        +2x

Kp = [HI]2 / [H2][I2]

but he has alrready given equilibrium pressure which is 1.844 atm

[HI] = 2x = 1.844 atm

x = 1.844 / 2 = 0.922 atm

[H2] = 1.43-x = 1.43-0.922 = 0.508 atm

[I2] = 3.15-x = 3.15 - 0.922 = 2.228 atm

now we have all equilibrium concentratiions

Kp = [1.844]2 / [0.508][2.288]

Kp = 3.0043


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