Calculation of reaction enthalpy 4 (thermo-16)

A cold bag is made up of the following ingredients: 25 g NH₄NO₃ (ammonium nitrate) and 75 ml of water. Now the ammonium nitrate is dissolved in water.
Note 1: the (specific) heat capacity c_w of the water is 75.3_nbsp_J/(K·mol) or 4.18 J/(K·kg).
Note 2: Q = m·c·ΔT
Note 3: The ammonium nitrate dissolves in water in its ions, the name indicating which.

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• October 16, 2026 at 4:43 am #9566
    Exercise: thermo-16    

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  A cold bag is made up of the following ingredients: 25 g NH₄NO₃ (ammonium nitrate) and 75 ml of water. Now the ammonium nitrate is dissolved in water.
  Note 1: the (specific) heat capacity c_w of the water is 75.3_nbsp_J/(K·mol) or 4.18 J/(K·kg).
  Note 2: Q = m·c·ΔT
  Note 3: The ammonium nitrate dissolves in water in its ions, the name indicating which.

  ---

  
  

    Exercise: thermo-16    

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

  A cold bag is made up of the following ingredients: 25 g NH₄NO₃ (ammonium nitrate) and 75 ml of water. Now the ammonium nitrate is dissolved in water.
  Note 1: the (specific) heat capacity c_w of the water is 75.3_nbsp_J/(K·mol) or 4.18 J/(K·kg).
  Note 2: Q = m·c·ΔT
  Note 3: The ammonium nitrate dissolves in water in its ions, the name indicating which.

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  October 16, 2026 at 4:43 am #9567
  Solution:

  reaction equation: NH₄NO₃ → NO–3(aq) + NH+4(aq)
  ΔHof(NH₄NO₃) = -366 kJ/mol
  ΔHof(NO–3(aq) = -207 kJ/mol
  ΔHof(NH+4(aq)) = -132 kJ/mol

  1. ΔHoR = ∑ΔHof(Produkte) – ∑ΔHof(Edukte)
  2. ΔHoR = (-207 + -132) – (-366) = + 27 kJ/mol
  3. [translation unavailable]: per Mol ammonium nitrate are 27 kJ 'consumed', to in solution to going.
  4. Given were but not a Mol NH₄NO₃, but only 25 grams. A Mol NH₄NO₃ corresponding to 80 grams, thus are 8.44 kJ (27·25/80) invested, to the solution to cooling.
  5. Q = m·c·ΔT is
     ΔT = Q/(m·c·) = 8440 J / (0.075 kg · 4180 J/(K·kg)) = + 26.9 K
  6. Since the reaction is endothermic (dHR0 > 0), it cools by 26.9 K (resp. 26.9℃ .

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