Question

1) The equilibrium constant Kc for the reaction N 2(g) + O 2(g) 2NO(g) at 1200 C is 1.00x 10^-5. Calculate the molar concentration of NO, N2 and O2 in equilibrium at 1200 C in a 1.00L container that initially had 0.114 mol of N2 and 0.114 mol of O2

2) A 2.0 mmol sample of Cl2 was closed inside a 2.0 L reaction vessel and heated to 1000k to study its dissociation into Cl atoms, Kc= 1.2x10^- 7 (a) Calculate the composition of the mixture in equilibrium. What is the percentage of decomposition of Cl2? (b) If 2.0 mmol of F2, Kc= 1.2x10^-4, is placed inside the container instead of chlorine, what will be its equilibrium composition at 1000k? Use your results from (a) and (b) to determine which is the most stable with respect to your atoms, Cl2 or F2, at 1000k

Answer 1

Step-by-step explanation:

1) N₂ + O₂ → 2 NO

Kc = [NO]² / ([N₂] [O₂])

Set up an ICE table:

`\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\N_(2)&0.114&-x&0.114-x\\O_(2)&0.114&-x&0.114-x\\NO&0&+2x&2x\end{array}\right]`

Plug into the equilibrium equation and solve for x.

1.00×10⁻⁵ = (2x)² / ((0.114 − x) (0.114 − x))

1.00×10⁻⁵ = (2x)² / (0.114 − x)²

√(1.00×10⁻⁵) = 2x / (0.114 − x)

0.00316 = 2x / (0.114 − x)

0.00361 − 0.00316x = 2x

0.00361 = 2.00316x

x = 0.00018

The volume is 1.00 L, so the concentrations at equilibrium are:

[N₂] = 0.114 − x = 0.11382

[O₂] = 0.114 − x = 0.11382

[NO] = 2x = 0.00036

2(a) Cl₂ → 2 Cl

Kc = [Cl]² / [Cl₂]

`\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\Cl_(2)&2.0&-x&2.0-x\\Cl&0&+2x&2x\end{array}\right]`

1.2×10⁻⁷ = (2x)² / (2 − x)

1.2×10⁻⁷ (2 − x) = 4x²

2.4×10⁻⁷ − 1.2×10⁻⁷ x = 4x²

2.4×10⁻⁷ ≈ 4x²

x² ≈ 6×10⁻⁸

x ≈ 0.000245

2x ≈ 0.00049

2(b) F₂ → 2 F

Kc = [F]² / [F₂]

`\left[\begin{array}{cccc}&Initial&Change&Equilibrium\\F_(2)&2.0&-x&2.0-x\\F&0&+2x&2x\end{array}\right]`

1.2×10⁻⁴ = (2x)² / (2 − x)

1.2×10⁻⁴ (2 − x) = 4x²

2.4×10⁻⁴ − 1.2×10⁻⁴ x = 4x²

2.4×10⁻⁴ ≈ 4x²

x² ≈ 6×10⁻⁵

x ≈ 0.00775

2x ≈ 0.0155

F₂ dissociates more, so Cl₂ is more stable at 1000 K.