Answer:
1.9 mol.
Explanation:
H₂ + I₂ → 2HI,
The initial concentrations:
The equilibrium constant (K) = [HI]²/[H₂][I₂],
Write down the equation and the start concentrations. Let (x) be the finish concentrations:
H₂ + I₂ → 2HI
start: 4.0 M 2.0 M 0 M
finish: 4.0-x 2.0-x 2x
∴ The equilibrium constant (K) = [HI]²/[H₂][I₂] = (2x)²/(4.0 - x)(2.0 - x).
∴ 50 = (2x)²/(4.0 - x)(2.0 - x).
∴ 50 = (4x²)/(8.0 - 6.0 x + x²).
∴ 400 - 300 x + 50 x² = 4x².
∴ 46x² - 300 x + 400 = 0 (standard form)
Use the quadratic equation to solve for (x):
x = (- b ± √(b² - 4ac))/(2a),
In this case a = +46, b = -300 and c = +400.
x = (+ 300) ± √((- 300)² - 4 (46)(400))/(2 (46))
x = (+ 300) ± √(9000 - 73600)/(92)
x = (+ 300) ± 127)/(92)
∴ x = 4.7 and 1.9
When x = 4.7 is not possible, because 4.0 - 4.7 (the concentration of H2 would be negative).
So, x = 1.9.
∴ [HI] = 2x
If x = 1.9,
∴ [HI] = 2(1.9) = 3.8 mol/L.
∵ [HI] = no. of moles/volume of the container.
∴ no. of moles of (HI) = [HI]*volume of the container = (3.8 mol/L)(0.5 L) = 1.9 mol.
