1. calculate the equilibrium constant for the reaction
H2 (g) + I2 (g) ⇋ 2 HI (g)
at 80.0 °C. The vapor pressure of solid iodine is 0.0216 bar at that temperature. If 42.7 g of solid iodine are placed in a 10.0 L vessel at 80.0 °C, what is the minimum amount of hydrogen gas that must be introduced in order to remove all the solid iodine? (fH°298 (I2(g)) = 62.44 kJ mol-1, S°298 (I2(g)) = 260.69 J mol-1K-1)
2. Relevant equations
ΔrG = ΔfG(products) - ΔfG(reactants)
ln K = -ΔrG / RT
K= [HI]^2 / [H_2]*[I_2]
ln K = -ΔH/RT + ΔS/T
3. 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
K= [HI]^2 / [H_2]*[I_2]
lnK = -62.44jk/mol/ (8.314 J/kmol)*(80oC) + 260.69 J/molK / 8.314 J/kmol
K= 2.22*10^4 for converting solid iodine to gas
now what?
i tried using mass balance with 42.7g Iodine
i got 0.168 mol I2 and H2
and 0.336 mol of HI
but i dont think its right because i dont know how to use the 10.0 L or 0.0216 bar
do i use PV=nRT??
0.0216 bar * 10.0 L = 0.168 mol I2 * 0.08206 LatmK-1mol-1 * 80oC??? no unknowns???
and they dont even equal