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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!!!
Answer:
15.71g
Explanation:
The general combustion equation for all hydrocarbons is
CxHy + (x+y/4) O2 = xCO2 + (y/2) H2O
For octane, C8H18 :
C8H18 + ( 8 + 18/4 ) O2 = 8CO2 + 9H2O
C8H18 + 50/4 O2 = 8CO2 + 9H2O
C8H18 + 25/2 O2 = 8CO2 + 9H2O
2C8H18 + 25 O2 = 16 CO2 + 18H2O (balanced)
From the balanced equation,
2 x 22.4 L of octane produced 16 [ 12 + (16 x 2)] of carbon dioxide
That is,
44.8 L of octane produced 704g of carbon dioxide
So, 1L of octane will produce 1 L x 704g/44.8 L = 15.71g of carbon dioxide
Therefore, 15.71g of carbon dioxide will be produced by the complete combustion of 1 L of octane.
