reaction enthalpy Calculation 4 (thermo-16) One of them KBags for old people exists from the following Ingredients: 25 g NH4NO3 (Ammonium nitrate) and 75 Milligrams Water. Well, you know what? will be This Ammonium nitrate in Water dissolved. In order to How a lot Degree Centigrade cooling The Commission The Mixed from? Notes 1: The (specific) WPoor capacity cw of the Water is 75.3_nbspJ/(K·mol) Responsibility. 4.18 J/(K·kg). Notes 2: Q = m·c·ΔT Notes 3: This Ammonium nitrate Resolves The Commission in Water in his Ions on, The Name is given One Notes on this which.

  Aufgabe: thermo-16    

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One of them KBags for old people exists from the following Ingredients: 25 g NH₄NO₃ (Ammonium nitrate) and 75 Milligrams Water. Well, you know what? will be This Ammonium nitrate in Water dissolved. In order to How a lot Degree Centigrade cooling The Commission The Mixed from?
Notes 1: The (specific) WPoor capacity c_w of the Water is 75.3_nbsp_J/(K·mol) Responsibility. 4.18 J/(K·kg).
Notes 2: Q = m·c·ΔT
Notes 3: This Ammonium nitrate Resolves The Commission in Water in his Ions on, The Name is given One Notes on this which.

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Lösung:

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. Interpretation: for each Mol Ammonium nitrate will be 27 KJ 'Consumed', to in Lthe island to Going.
4. Given were But not a Mol NH₄NO₃, but It 's just … 25 Grams. One of them Mol NH₄NO₃ corresponding to 80 Grams, This means that will be 8.44 KJ (27·25/80) invested, to The Lthe island to cooling.
5. With Q = m·c·ΔT is given The Commission
   ΔT = Q/(m·c·) = 8440 J / (0.075 kg · 4180 J/(K·kg)) = + 26.9 K
6. There you go. The reaction Endotherm is (dHR0 Gt0), cooling They The Commission to 26.9 K (Responsibility. 26.9℃ ) from.

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