2.50mols/5.00L = 0.500 M
............H2 + I2 ==> 2HI
initial.....0.....0.....0.500
change......x.....x......-2x
equil........x.....x.....0.500-2x
Kc = (HI)^2/(H2)(I2)
Substitute into the Kc expression and solve for x.
A study of the system, H2(g) + I2(g) == 2 HI(g), was carried
out. Kc = 54.9 at 699.0 K (Kelvin) for this reaction. A system was charged with 2.50 moles of HI in a 5.00 liter vessel as the only component initially. The system was brought up to 699.0 K and allowed to reach equilibrium. How many moles of H2 should there be in the container at that time?
1 answer