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!

Chemistry(Please help, thank you!) - bobpursley, Wednesday, March 14, 2012 at 9:18pm
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. I do not know what the initial values would be.

To solve this problem using an ICE table, we need to set up the initial, change, and equilibrium values for the concentrations of H2, I2, and HI.

The initial concentration of H2 is 0 moles, as there is no H2 initially in the container.

The initial concentration of I2 is also 0 moles.

The initial concentration of HI is given as 2.50 moles in a 5.00 liter vessel. Therefore, the initial concentration of HI is 2.50 moles / 5.00 liters = 0.50 M.

Since the stoichiometry of the reaction is 1:1 for H2 and HI, the change in concentration of H2 can be expressed as -x, where x is the number of moles of H2 that reacted.

The change in concentration of I2 can also be expressed as -x, as the stoichiometry of the reaction is 1:1 for I2 and HI.

The change in concentration of HI will be +2x, as 2 moles of HI are formed for every mole of H2 and I2 that react.

The equilibrium concentrations will be the initial concentrations plus the changes.

Therefore, the equilibrium concentration of H2 is 0 M - x, which can be simplified to -x.

The equilibrium concentration of I2 is 0 M - x, which can also be simplified to -x.

The equilibrium concentration of HI is 0.50 M + 2x.

Now, we can use the equilibrium constant expression to set up an equation.

Kc = [HI]^2 / ([H2] * [I2])

Substituting in the values we have:

54.9 = (0.50 + 2x)^2 / (-x * -x)

To simplify the equation, we can multiply both sides by (-x * -x):

54.9 * (-x^2) = (0.50 + 2x)^2

Now, let's solve the equation for x to find the moles of H2 at equilibrium:

-54.9x^2 = 0.25 + 2x + 4x^2

Combining like terms:

54.9x^2 + 4x^2 + 2x - 0.25 = 0

Simplifying further:

58.9x^2 + 2x - 0.25 = 0

We can now solve this quadratic equation for x using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

a = 58.9, b = 2, c = -0.25

x = (-2 ± √(2^2 - 4 * 58.9 * -0.25)) / (2 * 58.9)

Calculating x using the quadratic formula:

x ≈ -0.00320 or 0.00341

Since we are interested in moles, we will consider the positive value for x.

Therefore, the moles of H2 at equilibrium are approximately 0.00341 moles.

To solve this question using an ICE table, we need to set up the initial, change, and equilibrium values for each of the species involved in the reaction.

Let's start by setting up the initial values:

Initial:
H2: 0 moles (since it is not initially present in the container)
I2: 0 moles (since it is not initially present in the container)
HI: 2.50 moles

Now, let's define the change as x moles. This represents the change in the number of moles of H2 and I2 from the initial state to equilibrium.

Change:
H2: x moles
I2: x moles
HI: -2x moles (since 2 moles of HI are consumed for every 1 mole of H2 and 1 mole of I2 produced)

Next, let's calculate the equilibrium values:

Equilibrium:
H2: x moles
I2: x moles
HI: 2.50 - 2x moles (assuming the reaction goes to completion)

Now, we can use the expression for the equilibrium constant, Kc, to solve for x.

Kc = [HI]^2 / ([H2][I2])

Plugging in the equilibrium values, we get:

54.9 = (2.50 - 2x)^2 / (x * x)

Simplifying this equation, we get:

54.9 * x^2 = (2.50 - 2x)^2

Now, we can solve this quadratic equation to find the value of x. Once we have the value of x, we can determine the moles of H2 in the container at equilibrium by substituting x into the equilibrium expression for H2.

Hope this helps! Let me know if you have any further questions.