Answer:
(a)
-1.31 x 10⁴ kJ/mol complete reaction
-7.17 x 10³ kJ/mol incomplete reaction
-4.84 x 10³ kJ/mol soot formation
(b) -1.82 x 10⁴ kJ
(c) -1.70 x 10⁴ kJ
Explanation:
We need to find the change in enthalpy ( which is equivalent to q ) for the following reactions:
(1) C₂₁H₄₄ (s) + 32 O₂ (g) → 21 CO₂ (g) + 22 H₂O (g)
(2) C₂₁H₄₄ (s) + 43/2 O₂ (g) → 21 CO(g) + 22 H₂O(g)
(3) C₂₁H₄₄ (s) + 11 O₂(g) → 21 C(graphite) + 22 H₂O(g)
The enthalpy change for a reaction can be calculated from the heats of formation of reactants and products.
We were only given the heat of formation for C₂₁H₄₄, so we will need to find the rest.
ΔHºf (O₂) = ΔHºf(graphite) = 0 (They are in their elemental form)
ΔHºf (H₂O) = -241.83 kJ/mol (from reference tables)
ΔHºf (CO₂) = -393.52 kJ/mol (from reference tables)
ΔHºf (CO) = -110.53 kJ/mol (from referenfence tables)
Remember that ΔHreaction = ∑ ΔHºf products - ∑ΔHºf reactants
Part (a)
(1) C₂₁H₄₄ (s) + 32 O₂ (g) → 21 CO₂ (g) + 22 H₂O (g)
ΔHcomb₁ = {21(-393.52) +22(-241.83)} - { -476 } = -1.31 x 10⁴ kJ/mol
(2) C₂₁H₄₄ (s) + 43/2 O₂ (g) → 21 CO(g) + 22 H₂O(g)
ΔHcomb₂ = {21(-110.53) + 22(-241.83)} - {-476} = -7.17 x 10³ kJ/mol
(3) C₂₁H₄₄ (s) + 11 O₂(g) → 21 C(graphite) + 22 H₂O(g)
ΔHcomb₃ = {22(-241.83)] - {-476} = -4.84 x 10³ kJ/mol
Part (b)
We now have the heat of combustion for the reaction per mol of C₂₁H₄₄, so it is a simple matter to calculate the heat released for a 413 g candle once we convert it in number of moles.
MW C₂₁H₄₄ = 296.6 kJ/mol
n candle = m/MW = 413 g/296.6 g/mol = 1.39 mol
ΔH = ΔHcomb₁ x n = -1.31 x 10⁴ kJ/mol x 1.39 mol = -1.82 x 10⁴ kJ
Part (c)
For the calculations for this part will use the three reactions for which we know their heats of combustion according to the data given:
5.19 % mass incompletely
6.50 % formation soot
100 - (5.19+6.50) = 88.31 % complete reaction
So for a 413 g candle we have the following number of moles reacting:
n₁ = 0.883 x 413 g/ 296.6 g/mol = 1.23 mol react completely
n₂ = 0.0519 x 413 g / 296.6 g/mol = 0.07 mol incomplete
n₃ = 0.065 x 413 g / 296.6 g/mol = 0.09 mol soot formation
Now we can calculate the heat for the reactions:
ΔH₁ = ΔHcomb₁ x n₁ = -1.31 x 10⁴ kJ/mol x 1.23 mol = -1.61 x 10⁴ kJ
ΔH₂ = ΔHcomb₂ x n₂ = -7.17 x 10³ kJ/mol x 0.07 mol = -5.02 x 10² kJ
ΔH₃ = ΔHcomb₃ x n₃ = -4.84 x 10³ kJ/mol x 0.09 mol = -4.36 x 10² kJ
q = ΔH₁ +ΔH₂ +ΔH₃ = -1.61 x 10⁴ kJ + (-5.02 x 10² kJ) + (-4.36 x 10² kJ)
q = -1.70 x 10⁴ kJ
