Calculation of reaction enthalpy 9 (thermo-21)

In a poster roll (diameter 6 cm, length 72 cm) a few drops of pentane (1/4 ml) are given and ignited with the air-pentan mixture.
(a) Calculate the reaction enthalpy.
(b) It has been shown experimentally that too much pentane does not produce a better explosion than the air-pentan mixture Calculate the maximum amount of pentane.
Background: wonderful teaching experience.
Note: unknown size – molecules self to erude.

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• October 16, 2026 at 4:43 am #9576
    Exercise: thermo-21    

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  In a poster roll (diameter 6 cm, length 72 cm) a few drops of pentane (1/4 ml) are given and ignited with the air-pentan mixture.
  (a) Calculate the reaction enthalpy.
  (b) It has been shown experimentally that too much pentane does not produce a better explosion than the air-pentan mixture Calculate the maximum amount of pentane.
  Background: wonderful teaching experience.
  Note: unknown size – molecules self to erude.

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    Exercise: thermo-21    

  ⓘ    🔄 Reload     🖨️ print           ◀  –   ▶     ?! Question/Answer ?!

  In a poster roll (diameter 6 cm, length 72 cm) a few drops of pentane (1/4 ml) are given and ignited with the air-pentan mixture.
  (a) Calculate the reaction enthalpy.
  (b) It has been shown experimentally that too much pentane does not produce a better explosion than the air-pentan mixture Calculate the maximum amount of pentane.
  Background: wonderful teaching experience.
  Note: unknown size – molecules self to erude.

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  October 16, 2026 at 4:43 am #9577
  Solution:

  Note: [translation unavailable] C₅H₁₂, M(C₅H₁₂), Dichte(C₅H₁₂) = 0.63 g/ml

  1. Assumptions are as follows:
         [translation unavailable] C₅H₁₂, M(C₅H₁₂) = 72 g/mol
         Dichte(C₅H₁₂) = 0.626 g/ml
         dHf0(C₅H12) = -147 kJ/mol
         All substances involved are gaseous
         Experiment at 25℃ (293 K)
  2. (a) reaction equation:
         1·C₅H₁₂ (g) + 8·O₂ (g) ⇄ 5·CO₂(g) + 6·H₂O(g)
  3. ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
     ΔHoR = (5· ΔHof(CO₂,g) + 6· ΔHof(H₂O, g)) – ( ΔHof(C₅H₁₂, g) + 8· ΔHof(O₂,g))
     ΔHoR = (5·-393 + 6·-242) – (-147 + 8·0) = -3270 kJ/mol
  4. actual energy released:
     Dichte(C₅H₁₂) = 0.626 g/ml
     Thus 1/4 ml: 0.15 grams (=0.0021 mol)
     E = – 6.81 kJ
  5. (b) Calculation of the maximum amount of pentane
  6. Volume of the poster roll:
         V = pi · r² · h = 3.141 · 0.03² · 0.72 = 0.00204 m³
         V approx. 2 litres
  7. Ideal gas 1 mol at …
         0℃ (273 K) 22.4 l
         25℃ (298 K): 24 litres
         2 litres (volume of poster roll) is 1/12 mol
  8. Air is 20% O₂, 80% N₂
         Thus 1/60 mol O₂ in the poster roll
         According to the reaction equation 8· less C₅H₁₂ than O₂, i.e. 1/480 mol C₅H₁₂
         With M(C₅H₁₂) = 72 g/mol followed by 0.15 grams C₅H₁₂
         With Dichte(C₅H₁₂) = 0.63 g/ml following: thus approx. 1/4 ml C₅H₁₂

  So it releases 6.81 kJ. The maximum amount of pentane is about 1/4 ml.

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