Question

Aluminum metal and bromine liquid (red) react violently to make aluminum bromide (white powder). One way to represent this equilibrium is: 2 AlBr3() 2 Al(s) + 3 Br2(1)
We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K. the equilibrium constant for the reaction above.
1) 2 Al(s) + 3 Br2 = 2 AlBrz(s)
2) Al(s) + 3/2 Br2(1) AlBr3(s) K2 =
3) AlBrz(s) = Al(s) + 3/2 Br2(1) K3 =
Drag and drop your selection from the following list to complete the answer: || (1K)12 x 12 1/K Consider the reaction: S(s) + O2(g) SO2(g) Write the equilibrium constant for this reaction in terms of the equilibrium constants, K, and Kb for reactions a and b below:
a) 2 S(s) + 3 O2(g)
b) SO2(g) + 1/2O2(g) = 2 SO3(9) Ka SO3(9) Kb K=

Answer 1

Answer:

Part A

K = (K₂)²

K = (K₃)⁻²

Part B

K = √(Ka/Kb)

Explanation:

Part A

The parent reaction is

2Al(s) + 3Br₂(l) ⇌ 2AlBr₃(s)

The equilibrium constant is given as

K = [AlBr₃]²/[Al]²[Br₂]³

2) Al(s) + (3/2) Br₂(l) ⇌ AlBr₃(s)

K₂ = [AlBr₃]/[Al][Br₂]¹•⁵

It is evident that

K = (K₂)²

3) AlBr₃(s) ⇌ Al(s) + 3/2 Br₂(l)

K₃ = [Al][Br₂]¹•⁵/[AlBr₃]

K = (K₃)⁻²

Part B

Parent reaction

S(s) + O₂(g) ⇌ SO₂(g)

K = [SO₂]/[S][O₂]

a) 2S(s) + 3O₂(g) ⇌ 2SO₃(g)

Ka = [SO₃]²/[S]²[O₂]³

[SO₃]² = Ka × [S]²[O₂]³

b) 2SO₂(g) + O₂(g) ⇌ 2 SO₃(g)

Kb = [SO₃]²/[SO₂]²[O₂]

[SO₃]² = Kb × [SO₂]²[O₂]

[SO₃]² = [SO₃]²

Hence,

Ka × [S]²[O₂]³ = Kb × [SO₂]²[O₂]

(Ka/Kb) = [SO₂]²[O₂]/[S]²[O₂]³

(Ka/Kb) = [SO₂]²/[S]²[O₂]²

(Ka/Kb) = {[SO₂]/[S][O₂]}²

Recall

K = [SO₂]/[S][O₂]

Hence,

(Ka/Kb) = K²

K = √(Ka/Kb)

Hope this Helps!!!