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Unit 3: Thermochemistry. Chemistry 3202. Unit Outline. Temperature and Kinetic Energy Heat/Enthalpy Calculation Temperature changes (q = mc∆T) Phase changes (q = n∆H) Heating and Cooling Curves Calorimetry (q = C∆T & above formulas). Unit Outline. Chemical Reactions
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Unit 3: Thermochemistry Chemistry 3202
Unit Outline • Temperature and Kinetic Energy • Heat/Enthalpy Calculation • Temperature changes (q = mc∆T) • Phase changes (q = n∆H) • Heating and Cooling Curves • Calorimetry (q = C∆T & above formulas)
Unit Outline • Chemical Reactions • PE Diagrams • Thermochemical Equations • Hess’s Law • Bond Energy • STSE: What Fuels You?
Temperature and Kinetic Energy Thermochemistryis the study of energy changes in chemical and physical changes eg. dissolving burning phase changes
Temperature - a measure of the average kinetic energy of particles in a substance - a change in temperature means particles are moving at different speeds - measured in either Celsius degrees or degrees Kelvin Kelvin = Celsius + 273.15
The Celsius scale is based on the freezing and boiling point of water The Kelvin scale is based on absolute zero- the temperature at which particles in a substance have zero kinetic energy.
300 K # of particles 500 K Kinetic Energy
Heat/Enthalpy Calculations system - the part of the universe being studied and observed surroundings - everything else in the universe open system - a system that can exchange matter and energy with the surroundings eg. an open beaker of water a candle burning closed system - allows energy transfer but is closed to the flow of matter.
isolated system – a system completely closed to the flow of matter and energy heat - refers to the transfer of kinetic energy from a system of higher temperature to a system of lower temperature. - the symbol for heat is q WorkSheet: Thermochemistry #1
Heat/Enthalpy Calculations specific heat capacity – the energy , in Joules (J), needed to change the temperature of one gram (g) of a substance by one degree Celsius (°C). • The symbol for specific heat capacity is a lowercase c
A substance with a large value of c can absorb or release more energy than a substance with a small value of c. ie. For two substances, the substance with the larger c will undergo a smaller temperature change with the same loss or gain of heat.
q = mc∆T q = heat (J) m = mass (g) c = specific heat capacity ∆T = temperature change = T2 – T1 = Tf – Ti FORMULA
eg. How much heat is needed to raise the temperature of 500.0 g of water from 20.0 °C to 45.0 °C? Solve q = m c ∆T for c, m, ∆T, T2 & T1 • p. 634 #’s 1 – 4 • p. 636 #’s 5 – 8 WorkSheet: Thermochemistry #2
heat capacity- the quantity of energy , in Joules (J), needed to change the temperature of a substance by one degree Celsius (°C) • The symbol for heat capacity is uppercase C • The unit is J/ °C or kJ/ °C
C = mc q = C ∆T C = heat capacity c = specific heat capacity m = mass ∆T = T2 – T1 FORMULA Your Turn p.637 #’s 11-14 WorkSheet: Thermochemistry #3
Enthalpy Changes enthalpy change - the difference between the potential energy of the reactants and the products during a physical or chemical change AKA: Heat of Reaction or ∆H
Products ∆H Reactants PE Endothermic Reaction Reaction Progress
Products ∆H Reactants PE Endothermic Reaction Enthalpy ∆H Reaction Progress
Products Reactants Enthalpy ∆H is + Endothermic
products ∆H is - reactants Enthalpy Exothermic
Enthalpy Changes in Reactions • All chemical reactions require bond breaking in reactants followed by bond making to form products • Bond breaking requires energy (endothermic) while bond formation releases energy (exothermic) see p. 639
Enthalpy Changes in Reactions endothermic reaction - the energy required to break bonds is greater than the energy released when bonds form. ie. energy is absorbed exothermic reaction - the energy required to break bonds is less than the energy released when bonds form. ie. energy is produced
Enthalpy Changes in Reactions ∆H can represent the enthalpy change for a number of processes • Chemical reactions ∆Hrxn – enthalpy of reaction ∆Hcomb – enthalpy of combustion (see p. 643)
Formation of compounds from elements ∆Hof– standard enthalpy of formation The standard molar enthalpy of formation is the energy released or absorbed when one mole of a compound is formed directly from the elements in their standard states. (see p. 642) eg. C(s) + ½ O2(g) → CO(g) ΔHfo = -110.5 kJ/mol
Use the equations below to determine the ΔHfo for CH3OH(l) and CaCO3(s) 2 C(s) + 4 H2(g) + O2(g) → 2 CH3OH(l) + 477.2 kJ 2 CaCO3(s) + 2413.8kJ → 2 Ca(s) + 2 C(s) + 3 O2(g)
Phase Changes (p.647) ∆Hvap – enthalpy of vaporization ∆Hfus – enthalpy of melting ∆Hcond – enthalpy of condensation ∆Hfre – enthalpy of freezing eg. H2O(l) H2O(g) ΔHvap = Hg(l) Hg(s) ΔHfre = +40.7 kJ/mol -23.4 kJ/mol
Solution Formation(p.647, 648) ∆Hsoln– enthalpy of solution • eg. • ΔHsoln, of ammonium nitrate is +25.7 kJ/mol. • NH4NO3(s) + 25.7 kJ → NH4NO3(aq) • ΔHsoln, of calcium chloride is −82.8 kJ/mol. • CaCl2(s) → CaCl2(aq) + 82.8 kJ
Three ways to represent an enthalpy change: 1. thermochemical equation - the energy term written into the equation. 2. enthalpy term is written as a separate expression beside the equation. 3. enthalpy diagram.
eg. the formation of water from the elements produces 285.8 kJ of energy. 1. H2(g) + ½ O2(g) → H2O(l) + 285.8 kJ 2. H2(g) + ½ O2(g) → H2O(l) ∆Hf = -285.8 kJ/mol thermochemical equation
H2(g) + ½ O2(g) ∆Hf = -285.8 kJ/mol H2O(l) enthalpy diagram 3. Enthalpy (H) examples: pp. 641-643 questions p. 643 #’s 15-18 WorkSheet: Thermochemistry #4
FORMULA: q = n∆H q = heat (kJ) n = # of moles ∆H = molar enthalpy (kJ/mol) Calculating Enthalpy Changes
eg. How much heat is released when 50.0 g of CH4forms from C and H ? (p. 642) q = nΔH = (3.115 mol)(-74.6 kJ/mol) = -232 kJ
eg. How much heat is released when 50.00 g of CH4 undergoes complete combustion? (p. 643) q = nΔH = (3.115 mol)(-965.1 kJ/mol) = -3006 kJ
eg. How much energy is needed to change 20.0 g of H2O(l) at 100 °C to steam at 100 °C ? Mwater = 18.02 g/mol ΔHvap = +40.7 kJ/mol q = nΔH = (1.110 mol)(+40.7 kJ/mol) = +45.2 kJ
∆Hfre and ∆Hcond have the opposite sign of the above values.
eg. The molar enthalpy of solution for ammonium nitrate is +25.7 kJ/mol. How much energy is absorbed when 40.0 g of ammonium nitrate dissolves? q = nΔH = (0.4996 mol)(+25.7 kJ/mol) = +12.8 kJ
What mass of ethane, C2H6, must be burned to produce 405 kJ of heat? ΔH = -1250.9 kJ q = - 405 kJ q = nΔH n = 0.3238 mol m = n x M = (0.3238 mol)(30.08 g/mol) = 9.74 g
Complete: p. 643 #’s 15 - 18 p. 645; #’s 19 – 23 pp. 648 – 649; #’s 24 – 29 19. (a) -8.468 kJ (b) -7.165 kJ 20. -1.37 x103 kJ 21. (a) -2.896 x 103 kJ 21. (b) -6.81 x104 kJ 21. (c) -1.186 x 106 kJ 22. -0.230 kJ 23. 3.14 x103 g
24. 2.74 kJ 25.(a) 33.4 kJ (b) 33.4 kJ 26.(a) absorbed (b) 0.096 kJ 27.(a) NaCl(s) + 3.9 kJ/mol → NaCl(aq) (b) 1.69 kJ (c) cool; heat absorbed from water 28. 819.2 g 29. 3.10 x 104 kJ
p. 638 #’ 4 – 8 pp. 649, 650 #’s 3 – 8 p. 657, 658 #’s 9 - 18 WorkSheet: Thermochemistry #5
Heating and Cooling Curves Demo: Cooling of p-dichlorobenzene
KE PE KE Cooling curve for p-dichlorobenzene 80 Temp. (°C ) liquid 50 freezing solid 20 Time
KE KE PE Heating curve for p-dichlorobenzene 80 Temp. (°C ) 50 20 Time
What did we learn from this demo?? • During a phase change temperature remains constant and PE changes • Changes in temperature during heating or cooling means the KE of particles is changing