For the reaction I2 + Br2 <-> 2IBr Kc=280 at 150C . Suppose that 0.500 mol IBr in a 2.00-L flask is allowed to reach equilibrium at 150C .

Question

For the reaction I2 + Br2 <-> 2IBr Kc=280 at 150C . Suppose that 0.500 mol IBr in a 2.00-L flask is allowed to reach equilibrium at 150C .

What is the equilibrium partial pressure of IBr?
What is the equilibrium partial pressure of I2?
What is the equilibrium partial pressure of Br2?

Using an ICE Chart to demonstrate this

Answer 1

To find the equilibrium partial pressures of IBr, I2, and Br2 for the given reaction, we can use an ICE (Initial, Change, Equilibrium) chart and the equilibrium constant expression.

Step 1: Write the balanced equation for the reaction.
I2 + Br2 ↔ 2IBr

Step 2: Set up the ICE chart.

I2 + Br2 ↔ 2IBr
Initial: 0 + 0 ↔ 0
Change: -x -x +2x
Equilibrium: -x -x +2x

Step 3: Write the equilibrium constant expression.

Kc = ([IBr]^2) / ([I2][Br2])

Remember that the concentration of a gas is proportional to its partial pressure.

Kc = (P(IBr)^2) / (P(I2) * P(Br2))

Step 4: Substitute the given values into the equation.

Kc = (P(IBr)^2) / (P(I2) * P(Br2))
280 = (P(IBr)^2) / (P(I2) * P(Br2))

Step 5: Since we have the initial amount of IBr (0.500 mol) and the total volume of the flask (2.00 L), we can calculate its initial concentration and partial pressure.

Concentration of IBr (initial) = (Number of moles of IBr) / (Total volume in liters)
= 0.500 mol / 2.00 L
= 0.250 M

Since the volume and the stoichiometric coefficient are in a 1:1 ratio, the initial partial pressure of IBr would be the same as its initial concentration.

Partial pressure of IBr (initial) = 0.250 atm

Step 6: Plug in the known values into the Kc expression and solve for the unknowns.

280 = (P(IBr)^2) / (P(I2) * P(Br2))

We know that the initial partial pressure of IBr is 0.250 atm. So, let's substitute it into the equation:

280 = (0.250^2) / (P(I2) * P(Br2))

Rearranging the equation, we get:

(P(I2) * P(Br2)) = (0.250^2) / 280

(P(I2) * P(Br2)) = 0.000625 / 280

(P(I2) * P(Br2)) = 0.0000022321

Now, we need to solve for the product of the partial pressures of I2 and Br2. Since we only have one equation and two unknowns, we need another piece of information to solve for both partial pressures separately.

Note: If another piece of information is provided, such as the molar ratio of the reactants and products at equilibrium, we can solve for the partial pressures of I2 and Br2. Without that additional information, we cannot determine their individual values.