15.2 Heat Calorimetry

A. Calorimetry qgained = - qlost Calorimeter – an insulated device used to measure heat absorbed or released during a reaction Measures the heat flow into or out of a system Heat released by the system is equal to heat absorbed by the surroundings qgained = - qlost

A. Calorimetry We will use simple calorimeters of foam cups open to atmospheric pressure Chemicals are placed in inner cup Temperature will change until both chemicals reach same temperature Coffee cup Calorimeter

Practice A. Calorimetry GIVEN: WORK: qwater = ? qwater = m c ΔT A sample of an unknown metal is placed in a calorimeter that contains 125 g of water at 25.60oC. The 50.0 g metal is heated to 115.0oC and placed in the calorimeter. The metal transfers heat to the water until they both reach 29.30oC. GIVEN: qwater = ? m = 125 g c = 4.184 J/goC T = 29.30 – 25.60 = 3.70oC WORK: qwater = m c ΔT qwater = (125)(4.184)(3.70) qwater = 1940 J CONTINUE

Practice A. Calorimetry GIVEN: WORK: qwater = 1940 J qwater = - qmetal A sample of an unknown metal is placed in a calorimeter that contains 125 g of water at 25.60oC. The 50.0 g metal is heated to 115.0oC and placed in the calorimeter. The metal transfers heat to the water until they both reach 29.30oC. GIVEN: qwater = 1940 J qmetal = -1940 J m = 50.0 g c = ? T = 29.30 – 115.0 = - 85.7oC WORK: qwater = - qmetal - qmetal = m c ΔT - 1940 = (50.0) c (- 85.7) cmetal = 0.453 J/goC

Practice A. Calorimetry Are you ready for the lab tomorrow? 150.0 g sample of a metal at 75.0oC is added to 150.0 g H2O at 15.0oC. The temperature of the water rises to 18.3oC. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.

B. Enthalpy Negative enthalpy = exothermic ENTHALPY: (H) the heat content of a system at constant pressure The terms heat and enthalpy change are interchangeable q = ΔH qsys = ΔH = - m x C x ΔT Negative enthalpy = exothermic Positive enthalpy = endothermic

B. Enthalpy ΔHrxn = Hfinal – Hinitial ΔHrxn = Hproducts – Hreactants Enthalpy of a reaction: (∆Hrxn) the change in enthalpy for a reaction ∆Hrxn is the heat absorbed or released in a chemical reaction ∆Hrxn = difference in the enthalpies that exist between substances at the end of a reaction and the substances at the beginning of a reaction ΔHrxn = Hfinal – Hinitial ΔHrxn = Hproducts – Hreactants

B. Enthalpy Enthalpy changes for exothermic reactions are always negative 4Fe(s) + 3O2(g)  2Fe2O3(s) Hrxn = -1625 kJ Enthalpy changes for endothermic reactions are always positive NH4NO3(s)  NH4+(aq) + NO3-(aq) Hrxn = 27 kJ

B. Enthalpy

B. Enthalpy

15.3 Thermochemical Equations Pages 529 – 533

A. Thermochemical Equations A thermochemical equation is a balanced chemical equation that includes the physical states of all reactants and products, and energy change. In a thermochemical equation, the enthalpy of change for the reaction can be written as either a reactant or a product

A. Thermochemical Equations In a thermochemical equation, the enthalpy of change for the reaction can be written as either a reactant or a product Endothermic (positive ΔH) 2NaHCO3 + 129kJ Na2CO3 + H2O + CO2 Exothermic (negative ΔH) CaO + H2O Ca(OH)2 + 65.2kJ

B. Thermochemical Equation Practice Write the thermochemical equation for the addition of oxygen to iron (III) if the ΔHrxn= -1652 kJ/mol Fe(s) + O2(g)→ Fe2O3(s) + 1652 kJ Write the thermochemical equation for the decomposition of sulfur trioxide into sulfur dioxide and oxygen if the ΔHrxn= 198 kJ/mol SO3(g) + 198 kJ → SO2 (g) + O2 (g) 4 3 2 2 2

15.4 Hess’s Law Pages 534 – 540

A. Hess’s Law Hess’s law states that if you can add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction 2S(s) + 3O2(g)  2SO3(g) H = ?

A. Hess’s Law

A. Hess’s Law N2(s) + 2O2(g)  2NO2(g) H = ? Build your equation: Known: N2 (g) + O2(g)  2NO(g) H = -180 kJ 2NO(g) + O2(g)  2NO2(g) H = -112 kJ Build your equation:

A. Hess’s Law 2H2O2(l)  2H2O(l) + O2(g) H = ? Build your equation: Known: 2H2(g) + O2(g)  2H2O(l) H = -572 kJ H2(g) + O2(g)  H2O2(l) H = -188 kJ Build your equation:

A. Hess’s Law 2S(s) + 3O2(g)  2SO3(g) H = ? Known: S(s) + O2(g)  SO2(g) H = -297 kJ 2SO3(g)  2SO2(g) + O2(g) H = 198 kJ Build your equation:

A. Hess’s Law Challenge 2B(s) + 3H2(g)  B2H6(g) H = ? Known: 2B (s) + 3/2O2(g)  B2O3(s) H = -1273 kJ B2H6(g) + 3O2(g)  B2O3 (s) + 3H2O (g) H = -2035 kJ H2 (g) + ½ O2(g)  H2O (l) H = -286 kJ H2O (l)  H2O (g) H = 44 kJ Build your equation:

B. Standard Heats of Formation Standard heat of formation, ΔHf, of a compound is the change in enthalpy that accompanies the formation of one mole of the compound from its elements All substances at standard pressure and 25oC ΔHf of free elements (and diatomic molecules) in standard state is set at 0

B. Standard Heats of Formation Write the thermochemical equation for the formation of sodium oxide Na(s) + O2(g)  Na2O(s) -416 kJ 2 4 ½ 2 ΔHf =

B. Standard Heats of Formation Provides an alternative to Hess’s Law of calculating heat of reaction indirectly For reactions occurring at standard conditions, calculate heat of reaction, ΔHo , using standard heats of formation (given to you on appendix copies) ΔHo = ΔHf (products) – ΔHf (reactants)

B. Standard Heats of Formation What is the standard heat of reaction of CO(g) with O2(g) to form CO2 (g)? 2CO(g) + O2(g) → 2CO2(g) ΔH = ? 2(-393.5) – [2(-110.5) + 0] ΔH =-566.0 kJ/mol

B. Standard Heats of Formation CaCO3(s) → CaO(s) + CO2 (g) ΔH = ? (ΔHfCaO + ΔHfCO2) – (ΔHfCaCO3) = (-635 + -393.5) – (-1207) ΔH =179 kJ/mol