Suppose a 250.0 mL flask is filled with 1.3mol of I2 and 1.0mol of HI. The following reaction becomes possible:

H2 (g) + I2 (g) = 2HI (g)

The equilibrium constant K for this reaction is 0.983 at the temperature of the flask.

Calculate the equilibrium molarity of HI. Round your answer to one decimal place.

The molarity of HI at the equilibrium is 2.8M

Explanation:

Step 1: Data given

Volume of the flask = 250.0 mL = 0.250L

Number of moles I2 = 1.3 mol

Number of moles HI = 1.0 mol

Kc = 0.983

Step 2: The balanced equation

H2 (g) + I2 (g) ⇆ 2HI (g)

For 1 mole I2 consumed, we need 1 mole H2 to produce 2 moles HI

Step 3: Calculate initial concentrations

Initial concentration I2 = 1.3mol / 0.25L

Initial concentration I2 = 5.2 M

Initial concentration HI = 1.0 mol / 0.25L

Initial concentration HI = 4.0 M

Step 4: Calculate concentrations at equilibrium

The concentration at equilibrium is:

[I2] = (5.2+x) M

[HI] = (4.0 - x) M

[H2] = xM

Kc = [HI]²/[H2][I2]

0.983 = (4-x) ² / (x * (5.2+x))

0.983 = (4-x) ² / (5.2x + x²)

5.1116x + 0.983 x² = 16 - 8x + x²

-0.017x² + 13.1116x - 16 = 0

x = 1.222 = [H2]

[HI] = 4.0 - 1.222 = 2.778M ≈ 2.8 M

[I2] = 5.2 + 1.222 = 6.422 M ≈ 6.4 M

To control we can calculate:

[2.778]² / [1.222][6.422] = 0.983 = Kc

The molarity of HI at the equilibrium is 2.8M