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## Homework Statement

Calculate the equilibrium constant for the reaction

H

_{2}(g) + I

_{2}(g) <---> 2HI(g)

at 80

^{o}C. 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

^{o}C, what is the minimum amount of hydrogen gas that must be introduced in order to remove all of the solid iodine?

## Homework Equations

Δ

_{r}G = Δ

_{f}G(products) - Δ

_{f}G(reactants)

ln K = -Δ

_{r}G / RT

## The Attempt at a Solution

Δ

_{f}G for HI(g) = 1.70, for H

_{2}(g) = 0 and for I

_{2}(g) = 19.33 KJ mol

^{-1}

Δ

_{r}G = 2(1.7) - 0 - 19.33

Δ

_{r}G = -15.93 KJ mol

^{-1}

ln K = -Δ

_{r}G / 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.