How Do You Calculate the Equilibrium Constant for a Reaction at 80°C?

[johnny b]

Homework Statement

Calculate the equilibrium constant for the reaction
H₂(g) + I₂(g) <---> 2HI(g)
at 80^oC. The vapor pressure of solid iodine is 0.0216 bar at that temperature. If 52.7g of solid iodine are placed in a 10.0 L vessel at 80.0^oC, what is the minimum amount of hydrogen gas that must be introduced in order to remove all of the solid iodine?

Homework Equations

Δ_rG = Δ_fG(products) - Δ_fG(reactants)
ln K = -Δ_rG / RT

The Attempt at a Solution

Δ_fG for HI(g) = 1.70, for H₂(g) = 0 and for I₂(g) = 19.33 KJ mol^-1
Δ_rG = 2(1.7) - 0 - 19.33
Δ_rG = -15.93 KJ mol^-1
ln K = -Δ_rG / RT
ln k = -(-15.93E3) / (8/314)(353) = 5.42
k = 225.9

Now I need to find the minimum amount of hydrogen gas and I have no idea where to start. Thanks for any help.

 

[Borek]

You need enough hydrogen to shift equilibrium right till the pressure of iodine is below 0.0216 bar.

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methods

 

[johnny b]

I'm not sure how to do that.. do I use the equilibrium constant? How would I incorporate the amount of iodine given?

 

[Borek]

General approach is to write three equations - one will be the equilibrium constant, one iodine mass balance (mass of iodine present in HI and present as unreacted I₂) and mass balance for hydrogen (again - in HI and unreacted). Fourth equation is condition mentioned above. Solve for total number of moles of hydrogen.

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methods

 

[johnny b]

EDIT
I'm not sure what a mass balance is..
mass of iodine is 126.9 g/mol mass of H is 1.008g/mol
mass of iodine in HI is 126.9 g/mol
mass of iodine in I₂ is 253.8 g/mol and there are 2 moles so 507.6 g
mass of H in HI is 1.008 g
mass of H in H₂ is 2.016
k=[C]^c/[A]^ab
Sorry I'm still really confused.. where I need to include the 10.0L as the volume of the vessel? And the 52.7g of iodine..