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Thermochemistry Intro to thermochem - Discuss HEAT v. TEMPERATURE.

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Presentation on theme: "Thermochemistry Intro to thermochem - Discuss HEAT v. TEMPERATURE."— Presentation transcript:

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2 Thermochemistry Intro to thermochem - Discuss HEAT v. TEMPERATURE

3 Thermochemistry  Thermochemistry: the study of energy (in the form of heat) changes that accompany physical & chemical changes  Heat flows from high to low (hot to cool)

4 Thermochemistry  endothermic reactions: absorb energy in the form of heat; show a positive value for quantity of heat (q > 0)  exothermic reactions: release energy in the form of heat; show a negative value for quantity of heat (q < 0)

5 Magnitude of Heat Flow  Units of heat energy:  1 kcal = 1,000 cal = 1 Cal (nutritional)  1 kJ = 1,000 J  1 calorie = J  1 kcal = kJ

6 Magnitude of Heat Flow  The relationship between magnitude of heat flow, q, and temperature change, T, is: q = C  T  q = magnitude of heat  C = heat capacity (J/C) of a calorimeter  T = change in temp

7 Magnitude of Heat Flow  For a pure substance of mass m (constant state), the expression of q can be written as: q = m  c  T  q = magnitude of heat  m = mass  c = specific heat of substance (J/gC)  T = change in temp

8 Specific Heat  Specific heat = the amount of heat that must be added to a stated mass of liquid to raise its temp. by 1C, with no change in state.  Specific heat values (in J/gC):  CO 2 (g) = J/gC  Cu(s) = J/gC  Fe(s) = J/gC  H 2 O (l) = J/gC

9 Example 1 How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0C? q = mcΔT q = (50.0 g)(0.382 J/gC)(50.0 C C) q = -573 J q = (50.0 g)(0.382 J/gC)(-30.0 C) (heat is given off)

10 Example 2 Iron has a specific heat of J/gC. When a 7.55 g piece of iron absorbs J of heat, what is the change in temperature? If it was originally at room temp. (22.0C), what is the final temperature? q = mcΔT J = (7.55 g)(0.446 J/gC)(T) T f = 25.1 C T = 3.07 C = T f – 22.0 C

11 Example 3 The specific heat of copper is J/gC. How much heat is absorbed by a copper plate with a mass of g to raise its temperature from 25.0C to oven temperature (420F)? q = mcΔT q = (135.5 g)(0.382 J/gC)(190.6 C) q = 9860 J = 9.86 kJ F = (1.8)(C) = (1.8)(T f ) + 32 T f = C

12 Calorimetry To measure the heat flow in a reaction, it is carried out in a calorimeter. q rxn = - q calorimeter It is possible to calculate the amount of heat absorbed or evolved by the reaction if you know the heat capacity, C cal, and the temp. change, ΔT, of the calorimeter: q rxn = - C cal  ΔT

13 Coffee Cup Calorimeter The cup is filled with water, which absorbs the heat evolved by the reaction, so: q ice = -m w c w T

14 Coffee Cup Calorimeter Example When 1.00 g of ammonium nitrate, NH4NO3, is added to 50.0 g of water in a coffee cup calorimeter, it dissolves, NH 4 NO 3 (s)  NH 4 + (aq) + NO 3 - (aq), and the temperature of the water drops from 25.00C to 23.32C. Calculate q for the reaction system. q cal = m w c w T q cal = (50.0g)(4.18 J/gC)(-1.68 C) q cal = -351 J (water releases heat) q rxn = -q cal = -(-351 J) = 351 J (endothermic)

15 Bomb Calorimeter

16 EXAMPLE 1: The reaction between hydrogen and chlorine, H 2 + Cl 2  2HCl, can be studied in a bomb calorimeter. It is found that when a 1.00 g sample of H 2 reacts completely, the temp. rises from  C to  C. Taking the heat capacity of the calorimeter to be 9.33 kJ/  C, calculate the amount of heat evolved in the reaction. q cal = C cal  ΔT q cal = (9.33 kJ/C )(9.82 C ) q cal = 91.6 kJ q rxn = -q cal = kJ

17 EXAMPLE 2: When 1.00 mol of caffeine (C8H10N4O2) is burned in air, 4.96 x 10 3 kJ of heat is evolved. Five grams of caffeine is burned in a bomb calorimeter. The temperature is observed to increase by  C. What is the heat capacity of the calorimeter in J/  C?

18 EXAMPLE 3: When twenty milliliters of ethyl ether, C 4 H 10 O. (d=0.714 g/mL) is burned in a bomb calorimeter, the temperature rises from 24.7  C to 88.9  C. The calorimeter heat capacity is kJ/  C. (a) What is q for the calorimeter? (b) What is q when 20.0 mL of ether is burned? (c) What is q for the combustion of one mole of ethyl ether? q cal = C cal  ΔT q cal = (10.34 kJ/C )(64.2 C ) q cal = 664 kJ

19 EXAMPLE 3: When twenty milliliters of ethyl ether, C 4 H 10 O. (d=0.714 g/mL) is burned in a bomb calorimeter, the temperature rises from 24.7  C to 88.9  C. The calorimeter heat capacity is kJ/  C. (a) What is q for the calorimeter? (b) What is q when 20.0 mL of ether is burned? (c) What is q for the combustion of one mole of ethyl ether? q rxn = -q cal q rxn = -664 kJ

20 EXAMPLE 3: When twenty milliliters of ethyl ether, C 4 H 10 O. (d=0.714 g/mL) is burned in a bomb calorimeter, the temperature rises from 24.7  C to 88.9  C. The calorimeter heat capacity is kJ/  C. (a) What is q for the calorimeter? (b) What is q when 20.0 mL of ether is burned? (c) What is q for the combustion of one mole of ethyl ether?


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