Thermochemistry

1. Thermochemistry Sections 17.1, 17.2

2. Observing Heat Flow • Test the temperature difference of a rubber band (by touching to forehead) • Control temp = non-stretched • Test 1 = stretch and hold • Test 2 = stretch and release • Continue tests until certain of observations • When did it feel warm?

3. Energy Changes • The kinds of atoms and their arrangement in the substance determine the amount of energy stored • Gasoline (hydrocarbon) has high chemical potential energy • Controlled explosions to make it run (do WORK) also makes the engine HOT • Energy changes occur either as heat, work, or a combination of the two.

4. Heat • Always flows from a warmer object to a cooler object • Until both temperatures are the same (equilibrium) • System • Surroundings • Law of conservation of energy • Endothermic reaction • System gains (requires) heat from surroundings • Exothermic reaction • System loses (gives) heat to surroundings

5. UNITS! • Common = calorie and joule (SI unit) • 1 Calorie = 1 kilocalorie = 1000 calories • 1 calorie (cal) = 4.184 Joules (J) • 1 J = 0.2390 cal • 1 calorie is defined as the quantity of heat required to raise the temperature of 1 gram of pure H2O by 1°C

6. Heat Capacity • Heat capacity depends on both the mass and chemical composition of the substance (extensive property) • Greater mass = greater heat capacity • (larger pots of water take longer to boil) • Specific heat capacity (specific heat, C ) is the amount of heat needed to raise the temperature of 1 gram of the substance by 1°C (intensive property)

7. Heat Capacity Calculations • Heat = mass x specific heat x change in temperature in Celsius • q = mC ΔT • Units for specific heat= J/(g·°C) OR cal/(g·°C) • Practice: When 435 J of heat is added to 3.4 g olive oil at 21°C, the temperature increases to 85 °C. What is Colive oil? • C=q/m ΔT = 435 J / (3.4g x (85-21°C)) • Colive oil = 1.999 = 2.0 J/(g·°C)

8. More Heat Calculations • How much heat is required to raise the temperature of 250.0 g of Hg by 52ºC? (CHg = 0.14 J/(gºC)) • q = m C ΔT = (250.0g)x(0.14 J/gºC)x(52ºC) = 1820 J = 1.8 kJ

9. Calorimetry • Measures heat flow OUT of a system • Heat released by system = heat absorbed by surroundings • In aq rxns, system = chemicals, surroundings = water • Good b/c know Cwater and Dwater to calc mwater • Calorimeter • Enthalpy (H) Δ = q at constant pressure • qsys = ΔH = -qsurr • - ΔH = exothermic rxn • + ΔH = endothermic rxn

10. Lab Chem skip this slide!!!!Calorimetry Calculations • Practice: • When 50.0 mL of water containing 0.50 mol HCl at 22.5°C is mixed with 50.0 mL of water containing 0.50 mol of NaOH at 22.5 °C in a calorimeter, the temperature of the solution increases to 26.0°C. How much heat (in kJ) was released by this reaction? • Water = surroundings, rxn= system • Mass water = (50.0mL + 50.0mL) x 1.00g/mL = 100.0 g H2O • ΔT = 26.0 – 22.5 = 3.5°C • Hwater= m x C x ΔT = (100.0g)x( 4.184 J/g°C)x(3.5°C) = 1464.4 J = 1.46 kJ (endothermic) • Hsys = - Hwater = -1.46 kJ (exothermic)

11. More Cal Calculations  • A small pebble is heated up and placed in a foam cup calorimeter containing 25.0 mL of water at 25.0°C. The water reaches a max temp of 26.4°C. How many joules of heat were released by the pebble? • System = pebble, surroundings = water • qwater = -qpebble q = mCΔT = (25.0mL x 1.00 g/mL) x (4.184 J/g°C) x (26.4 – 25.0 °C) = (25.0 g) x (4.184 J/g°C) x (1.4°C) = 146.44 J = 146 J • qpebble = - qwater = - (146 J) = -146 J (exothermic)

12. Thermochemical Equations • In a chemical equation, the heat (enthalpy) change for the reaction can be written as either a reactant (endothermic) or product (exothermic). • Thermochemical equation • States MUST be listed since different ΔH • Ex: H2O (l) requires 44 kJ more than H2O (g) to separate into constituent gases • Heat of reaction • The amount of heat absorbed or released during a reaction depends on the number of MOLES of reactants involved. • Heat of combustion • ΔH given in kJ, but understood to be kJ/mol

13. Figure 17.7

14. Thermochemical calculations • When carbon disulfide is formed from its elements, heat is absorbed. Calculate the amount of heat (in kJ) absorbed when 5.66 g of carbon disulfide formed. C(s) + 2S(s)  CS2(l) ΔH = 89.3 kJ 5.66g CS2x 1 mol CS2 /(12.011 + 2x32.066) x 89.3 kJ/mol = 6.638 kJ = 6.64 kJ • Does this make sense?

15. More Thermochemistry Calculations • The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction. How many kilojoules of heat are produced when 3.40 mol Fe2O3 reacts with an excess of CO? Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) + 26.3 kJ 3.40 mol Fe2O3 x 26.3 kJ/mol = 89.4 kJ

16. Heat in Changes of State

17. Phase change calculations • While heating (temp is changing) • q=mcΔT • While changing state (temp is constant) • q= ΔHfus or vap x (moles) • ΔHfus for water = 6.01kJ/mol • Δhvap for water = 40.7kJ/mol

18. How much total heat is needed to convert 10.00 grams of -10°C ice to 120°C steam? • q= mciceΔT • cice = 2.1 J/gC • ΔHfus = 6.01 kJ/mol = 6010 J/mol • q= mcwaterΔT • cwater= 4.184 J/gC • ΔHvap = 40.7 kJ/mol = 40700 J/mol • q= mcsteamΔT • csteam = 1.7 J/gC

20. Molar Heat of Formation • The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products. • Table 17.4 on page 530 in your text book… • ΔH0 = ΔHf0 (products) – ΔHf0 (reactants)

23. Hess’s Law of heat Summation • Hess’s law adds two or more “half” reactions and coordinating changes in enthalpy (ΔH) to get final/total reaction ΔH • Ex: Derive C(graphite) + 2H2 (g)  CH4(g) (1) H2(g) + 1/2 O2(g)  H2O(l) ΔH = -285.8 kJ (2) C(graphite) + O2(g)  CO2(g) ΔH = -293.5 kJ (3) CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ΔH = -890.4 kJ

24. Answer: • C(graphite) + 2H2 (g)  CH4(g) ΔH = -74.7 kJ • HOW? • Multiply reaction 1 by 2, flip reaction 3 and add all together • ΔH = 2(-285.8) + (-393.5) + (+890.4) = -74.7 kJ