3. TUDOR VIANU GIURGIU ROMANIA

4. STANISTEANU LAURA ELENA NICULA GEORGETA

5.  PROBLEMATISATION  LEARNING BY GUIDED DISCOVERY  ALGORITHMIC  CHEMICAL EXPERIMENT  INTERDISCIPLINARITY

6.  WE START FROM THE FIRST PRINCIPLE OF THERMODINAMICS:  ΔU = Q + ΔL  ΔU = Q – ΔL ( for chemical reactions)  U – internal energy  L – mechanical work  Q – the heat  We adapt this principle for the chemical reactions:  ΔU = Q – F ΔL  ΔU = Q – F/S * ΔL *S  ΔU = Q – pΔV  F – force V – volume ( capacity)  L – length S - surface  P- pressure

7.  A lot of chemical reactions occur at constant pressure. p = ct => Q = ΔU + ΔpV Q = Δ ( U + pV) U + pV = H → Q = ΔH H – enthalpy ( specific of chemistry)  Where ΔH > 0 → endothermic reaction  ΔH < 0 → exothermic reaction ΔH = ∑ H o f P - ∑ H o f R H o f – formation enthalpy in standard conditions (1 atm, 25 o C) ( the heat emitted or absorbed during the formation of one mol substance from elements) We ask the students to give us some examples of biological reactions which occur with heat release and, respectively, with absorbation of heat

8.  C 6 H 12 O 6 + 6 O 2 → 6 CO 2 + 6 H 2 O + Q - The photosynthesis – endothermic reaction when the plants prepare their food and emit O 2.  nCO 2 + nH 2 O + Q → C n H 2n O n + n O 2 saccharides S. – The combustion of glucose - exothermic reaction where the heat emitted is as energy for our body. We ask the students to calculate which is the energy released when we eat 120 grams chocolate which contains 15% (procent) glucose? Δ H o f C6H12O6 = - 1267,5KJ/ mol, Δ H o f CO2 = -393,5 KJ/mol, Δ H o f H2O = - 285, 83 KJ/ mol

9.  ΔH = ∑H o f P - ∑H o f R  ΔH = 6 H o f CO2 + 6 H o f H2O – ΔH o f C6H12O6 = - 2808, 48KJ/mol  15/100 x 120 = 18 g glucose  n= m/M = 18/ 180 = 0,1 mol glucose 1 mol glucose……………….2808,48 KJ 0,1 mols glucose……………..Q Q = 280,8 KJ = 67,177 Kcal

11.  Procedures : 1. In one calorimeter you introduce 100 ml solution of HCl 1M and establish its temperature (t 1 ). Then add 1g MgO, and after reaction occurs, establish the temperature. (t 2 ). You calculate, then, the variation of temperature (Δ t) 2. In one calorimeter you introduce 100 ml solution of HCl 1M and establish its temperature (t 1 ). Then add 1 g Mg, and after the reaction occurs, establish the temperature (t 2 ). You calculate, then, the variation of temperature (Δt). The students will experimentally establish the enthalpy of formation, in standard conditions, for MgO - Δ H o f MgO

13.  Calculate the number of Kcal ( calorific energy) released in each reaction, knowing that: one caloric is necessary to increase the temperature of one ml solution with 1 o C. Using the obtained results and knowing that: ΔH o f H2O = - 44,92 kcal/ mol, establish the formation enthalpy in standard conditions for MgO (ΔH o fMgO )

14.  1cal……1 o C Q= Δt ( for 1 ml solution)  Q………Δt  Q = 100 Δt ( for 100 ml solution)  n= C M V s = 1* 0,1= 0,1 mols HCl⇒  ΔH= Q/n =100Δt/ 0,1= 1000 Δt cal/mol  Exp 1 ΔH 1 = 1000* Δt 1 cal/mol  t 1 = 21 o C  t 2 = 22 o C ⇒ ΔH 1 = -1000cal/mol  Exp 2 ΔH 2 = 1000* Δt 2 cal/mol  t 1 = 21 o C  t 2 = 71 o C ⇒ ΔH 2 = -50 000cal/mol

15.  (-) 2HCl + MgO → MgCl 2 + H 2 O 2ΔH 1  (+) 2HCl + Mg → MgCl 2 + H 2 2ΔH 2  (+) H 2 + ½ O 2 → H 2 O ΔH 3  ------------------------------------------------------  Mg + ½ O 2 → MgO ΔH  ΔH = -2 ΔH 1 + 2 ΔH 2 + ΔH 3  ΔH = 2*1 – 2* 50 – 44,92 = -142,92kcal/mol = - 142,92 * 4,18= - 597, 40 KJ/mol  H o f MgO = -601KJ/mol

16. In conclusion, the chemical experiment is an active method which allow the students create and practice on their own some operations to observe, study and to measure the results.

17.  This project has been funded with support from the European Commission.  This publication reflects the views only of the author, and the Commission cannot be held responsible for any use which may be made of the information contained therein. 