Answer:
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
<u>Step 3:</u> 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
