If a 2L reaction vessel initially contains 2 mols of Hydrogen and 2 mols of Iodine,...

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Question

If a 2L reaction vessel initially contains 2 mols of Hydrogen and 2 mols of Iodine,...

If a 2L reaction vessel initially contains 2 mols of Hydrogen and 2 mols of Iodine, what will be the molar concentrations at equilibrium of all species?

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

Kc = 50.5

science chemistry

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Answer 1

Answer #1

Initial concentration of H2 = mol of H2 / volume in L

= 2 moles / 2 L

= 1 M

Initial concentration of I2 = mol of I2 / volume in L

= 2 moles / 2 L

= 1 M

ICE Table:

Equilibrium constant expression is

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

50.5 = (2*x)^2/(1-1*x)^2

sqrt(50.5) = (2*x)/(1-1*x)

7.106 = (2*x)/(1-1*x)

7.106-7.106*x = 2*x

7.106-9.106*x = 0

x = 0.78037

At equilibrium:

[H2] = 1.0-1x = 1.0-1*0.78037 = 0.220 M

[I2] = 1.0-1x = 1.0-1*0.78037 = 0.220 M

[HI] = +2x = +2*0.78037 = 1.56 M

Answer:

[H2] = 0.220 M

[I2] = 0.220 M

[HI] = 1.56 M

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Answer 2

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