# Equilibrium constant

## Homework Statement

Calculate the equilibrium constant for the reaction
H2(g) + I2(g) <---> 2HI(g)
at 80oC. 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 80.0oC, what is the minimum amount of hydrogen gas that must be introduced in order to remove all of the solid iodine?

## Homework Equations

ΔrG = ΔfG(products) - ΔfG(reactants)
ln K = -ΔrG / RT

## The Attempt at a Solution

ΔfG for HI(g) = 1.70, for H2(g) = 0 and for I2(g) = 19.33 KJ mol-1
ΔrG = 2(1.7) - 0 - 19.33
ΔrG = -15.93 KJ mol-1
ln K = -ΔrG / 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.

Related Biology and Chemistry Homework Help News on Phys.org
Borek
Mentor
You need enough hydrogen to shift equilibrium right till the pressure of iodine is below 0.0216 bar.

--
methods

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

Borek
Mentor
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.

--
methods

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..

Last edited: