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CALORIMETRY ΔH of a chemical rxn can experimentally be determined by measuring the heat flow accompanying the rxn at constant pressure. When heat flows.

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Presentation on theme: "CALORIMETRY ΔH of a chemical rxn can experimentally be determined by measuring the heat flow accompanying the rxn at constant pressure. When heat flows."— Presentation transcript:

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2 CALORIMETRY ΔH of a chemical rxn can experimentally be determined by measuring the heat flow accompanying the rxn at constant pressure. When heat flows into/out of a substance, its temperature changes. The heat flow is experimentally determined by using the temperature change produced.

3 The measurement of heat flow is called “calorimetry” and the apparatus used to measure the heat flow is called “a calorimeter.” CALORIMETRY

4 Heat capacity (C or m.c) of an object is the amount of heat required to raise its temperature by 1 K or 1 °C. The greater the heat capacity, the greater the heat required to produce a certain rise in temperature. CALORIMETRY

5 Specific heat capacity or specific heat (s or c) is the heat capacity of 1 g of a substance.  Specific heat of H2O(l) is the amount of energy required to change temperature of 1 g of water by 1°C. Therefore, it is 4.184 J/g-K or 1 cal/g –K. CALORIMETRY

6 substanceSpecific heat ( J/g-K) N2 (g)1.04 Al(s).90 Fe(s).45 H2O(l)4.18 Specific heat of water is quite higher than those of other substances. It’s very important for Earth’s climate since it makes oceans resistant to temperature changes.

7 The amount of heat gained /lost by a substance: CALORIMETRY q=(grams of substance)x(specific heat)x ΔT Q=mcΔT !!!ΔT in K = ΔT in °C

8 When a substance gains heat - its temperature rises. When a substance loses heat, - Its temperature lowers. CALORIMETRY

9 CALORIMETERS 1) CONSTANT-PRESSURE CALORIMETER - A coffee-cup calorimeter - Because the calorimeter isn’t sealed, the rxn happens under constant pressure of the atmosphere.

10 CALORIMETERS 1) CONSTANT-PRESSURE CALORIMETER - Since the calorimeter has a very low thermal conductivity & heat capacity, we assume that; 1. The heat absorbed/gained during the rxn doesn’t escape the coffee cup. 2. The calorimeter itself doesn’t absorb/release heat.

11 Coffee cup calorimeter is usually used to determine the “heat of solution” and “heat of neutralization.” CALORIMETERS 1) CONSTANT-PRESSURE CALORIMETER

12 - Heat exchange happens only between the solution and the chemicals reacting in the calorimeter. Therefore; In exothermic rxns: q lost by the rxn = - q gained by the solution In endothermic rxns: q gained by the rxn = - q lost by the solution - q solution = -(specific heat of solution)x(grams of soln)xΔT=q rxn

13 CALORIMETERS 1) CONSTANT-PRESSURE CALORIMETER q rxn = - q solution ΔH rxn = q rxn / (number of moles of the acid/base reacted ) ΔH rxn = q rxn / (number of moles of the solid dissolved )

14 For dilute aqueous solutions, the specific heat of solution will be approximately the same as that of water. CALORIMETERS 1) CONSTANT-PRESSURE CALORIMETER

15 Example 1 When a student mixes 50. mL of 1.0 M HCl and 50. mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 °C to 27.5 °C. Calculate the enthalpy change for the rxn, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100 mL, that its density is 1.0 g/mL, and that its specific heat is 4.18 J/ g-K.

16 Solution -q solution = -(specific heat of solution)x(grams of soln)xΔT=q rxn -[( 4.18 J/ g-K) x (50 g+50 g)x (27.5-21.0) ] =q rxn -2717 J =q rxn - 2.717 kJ =q rxn M= n/V=> n= MV => 1.0 x 0.050 = 0.05 mol HCl 1.0 x 0.050 = 0.05 mol NaOH NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) 1: 1 ratio between NaOH and HCl in the balanced equation

17 Solution - 2.717 kJ =q rxn NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) 1: 1 ratio between NaOH and HCl in the balanced equation 0.05 mol HCl reacted w/ 0.05 mol NaOH ΔH rxn = q rxn / number of moles of the acid/base reacted ΔH rxn = - 2.717 kJ / 0.05 mol ΔH rxn = - 54.34 kJ/mol

18 2)BOMB CALORIMETER(CONSTANT- VOLUME) It’s usually used to determine “molar heat of combustion (ΔH° comb )” of substances. molar heat of combustion is the enthalpy change when 1 mole of the substance undergoes a complete combustion in excess oxygen under standard conditions. It’s always negative in sign.

19 Qrxn= - [Qwater + Qmetal] Qrxn = - [(m w.c w.ΔT) + (m metal. c metal. ΔT)] OR Qrxn = - [(m w.c w ) + (m metal. c metal )]. ΔT If the calorimeter cup is made of a different material (e.g. a metal): Q rxn = - C cal x ΔT

20 2)BOMB CALORIMETER(CONSTANT- VOLUME) We calculate the heat evolved by the rxn with: Q rxn = - C cal x ΔT

21 Example 2 data above is from an experiment used to measure the enthalpy change for the combustion of 1 mole of glucose (C 6 H 12 O 6(s) ). The time-temperature data was taken from a data-logging software programme.

22 Mass of sample of glucose, m = 1.389 g Heat capacity of the system, Csystem = 12.224 kJ K–1 ( C : 12 ; H: 1 ; O : 16 ) (a)Calculate ΔT, for the water, surrounding the chamber in the calorimeter. (b)Determine the amount, in moles, of glucose. (c)Calculate the enthalpy change for the combustion of 1 mole of glucose.

23 solution A) ΔT= 23.78-22.01=1.77°C B) n=m/M n= 1.389/ 180=0.007717mol=0.008mol C) Qrxn= -C ΔT= -12.224x1.77= -21.63648 kJ ΔH comb =-21.63648 kJ/0.007717mol= - 2803.7 kJ/mol ΔH comb = -3000 kJ/mol

24 Example 3 Methyl hydrazine (CH 6 N 2 ) is commonly used as a liquid rocket fuel. The combustion of methyl hydrazine w/ oxygen produces N 2 (g), CO 2 (g), and H 2 O(l). When 4.00 g of methyl hydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 °C to 39.50°C. In a separate experiment the heat capacity of the calorimeter is measured to be 7.794 kJ/°C. What is heat of reaction for the combustion of a mole of methyl hydrazine in this calorimeter? (N: 14.01 g/mol, H: 1.01g/mol, C: 12.01 g/mol)

25 Solution - (heat capacity of the calorimeter)xΔT=q rxn - (7.794 kJ/°C) x (39.50 °C-25.00 °C) - 113.013 kJ =q rxn Molar mass of CH 6 N 2 = (1x12.01+ 6x1.01+ 2x14.01)= 46.09 g/mol n=mass/molar mass=> n=4.00g / 46.09 gmol -1 n= 0.0868 mol 0.0868 mol CH 6 N 2 combusts - 113.013 kJ is released 1 mol CH 6 N 2 combusts ? ? = -1302.19 kJ/mol


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