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?
I was going to set it up as
2.5HI (1/5.00) (1/127.91HI) but I am not sure if this is correct!
Hmmm. I don't see how it could be correct.
Do you do ICE tables?
I am familiar with how to do the ICE table but I am not sure how to set one up for this question with the 2.5 moles.
To solve this problem, you'll need to set up an expression using the equilibrium constant and the initial concentration of HI.
First, let's convert the initial concentration of HI from moles to molarity (moles per liter):
Initial concentration of HI = 2.50 moles / 5.00 liters = 0.50 M
Now, let's set up the expression using the equilibrium constant (Kc) and the initial concentration of HI:
Kc = [HI]^2 / ([H2][I2])
Knowing that the stoichiometric coefficient for HI is 2, the initial concentration of H2 will be denoted as "[H2]". Since there are no initial concentrations of H2 and I2, the concentrations of these substances at equilibrium will be denoted as "x" for simplicity.
Using the balanced equation, the expression for Kc becomes:
Kc = ([HI] at equilibrium)^2 / ([H2] at equilibrium * [I2] at equilibrium)
Kc = ([0.50 - 2x]^2) / (x * x)
Now, we can solve for "x":
Kc * x^2 = (0.50 - 2x)^2
54.9 * x^2 = (0.50 - 2x)^2
Given the temperature and equilibrium constant, we can solve this quadratic equation to find the value of "x". The positive root will give us the number of moles of H2 at equilibrium.