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

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