Thermochemistry Coffee cup calorimetry Energy content of fuels In addition to this presentation, before coming to lab or attempting the prelab quiz you must also:  Watch the prelab videos (#4 and 5) for this experiment  Read the introduction to the lab in the coursepack

What’s the point? See practical applications of thermochemistrySee practical applications of thermochemistry hot packshot packs fuelsfuels Perform coffee cup calorimetryPerform coffee cup calorimetry Practice calorimetry calculations ( Chapter 5 Brown, LeMay, Bursten)Practice calorimetry calculations ( Chapter 5 Brown, LeMay, Bursten)

Background ThermochemistryThermochemistry study of energy flow during chemical reactionsstudy of energy flow during chemical reactions when heat is produced from a reaction, the reaction is called exothermicwhen heat is produced from a reaction, the reaction is called exothermic when heat is required for a reaction to proceed, the reaction is endothermicwhen heat is required for a reaction to proceed, the reaction is endothermic Our experiments focus on the production of energy in the form of heat as a result ofOur experiments focus on the production of energy in the form of heat as a result of dissolving a salt in waterdissolving a salt in water combusting a fuelcombusting a fuel

Heat (q) An energy transfer between system and surroundings due to temperature differenceAn energy transfer between system and surroundings due to temperature difference Heat flows from “hot” to “cold”Heat flows from “hot” to “cold” but heat flow cannot be measured directlybut heat flow cannot be measured directly instead, the effects of heat flow are measuredinstead, the effects of heat flow are measured relate temperature change and heat flow via specific heat capacityrelate temperature change and heat flow via specific heat capacity

Mathematically, q = m s  TMathematically, q = m s  T q = heat (joules J)q = heat (joules J) m = mass (grams)m = mass (grams) s = specific heat capacity (J g -1 o C -1 )s = specific heat capacity (J g -1 o C -1 )  the heat required to raise 1 gram 1 o C  T = change in temperature ( o C)  T = change in temperature ( o C)   T = T final – T initial

Coffee Cup Calorimetry We will measure the temperature change that occurs upon dissolving CaCl 2 in water: CaCl 2 (s)  Ca 2+ (aq) + 2Cl - (aq)We will measure the temperature change that occurs upon dissolving CaCl 2 in water: CaCl 2 (s)  Ca 2+ (aq) + 2Cl - (aq) Since this is at constant pressure, we will be measuring the enthalpy (  H = q P )Since this is at constant pressure, we will be measuring the enthalpy (  H = q P ) Since this is a dissolution reaction, we call this enthalpy the “enthalpy of solution”Since this is a dissolution reaction, we call this enthalpy the “enthalpy of solution”

Estimating the Maximum  T To make an accurate measure of the temperature change, you must have a well insulated container or heat will be lost to the air during the experimentTo make an accurate measure of the temperature change, you must have a well insulated container or heat will be lost to the air during the experiment You also want to have efficient stirring so that the salt dissolves quickly. This heats the solution quickly, which also helps minimize heat lossYou also want to have efficient stirring so that the salt dissolves quickly. This heats the solution quickly, which also helps minimize heat loss Well insulated Good stirring Well insulated Poor stirring Poorly insulated Poor stirring

Estimating the Maximum  T We will “interpolate” the maximum temperature by drawing lines through the heating and cooling portion of the curves as shown below (T max ~ 84 o C).We will “interpolate” the maximum temperature by drawing lines through the heating and cooling portion of the curves as shown below (T max ~ 84 o C). Linear portion of heating curve Linear portion of cooling curve Estimated maximum temperature for perfectly insulated and stirred calorimeter

Sample Calculation Calculating  H requires that you knowCalculating  H requires that you know the temperature change of solutionthe temperature change of solution the mass of solutionthe mass of solution the specific heat capacity of solutionthe specific heat capacity of solution the moles of salt reactedthe moles of salt reacted When 7.85 g of NaCl is dissolved in 50.0 g of water, the solution temperature drops by 2.07 o C? If the solution has a specific heat capacity of 4.18 J g -1 o C -1, what is the enthalpy of solution for NaCl?

First, find the heat of reaction: q soln = m soln s soln  T soln = (50.0 g g)( 4.18 J g -1 o C -1 )(-2.07 o C ) = Jq soln = m soln s soln  T soln = (50.0 g g)( 4.18 J g -1 o C -1 )(-2.07 o C ) = J q rxn = -q soln = +500 J Now, express this per mole of NaCl: n NaCl = (7.85 g)(1 mol / g) = mol n NaCl = (7.85 g)(1 mol / g) = mol  H = (+500. J) / (0.100 mol) = 5.00 x 10 3 J mol - 1

Combustion Reactions Combustion of fuel is an exothermic process hydrocarbons are very good at releasing energy Involves reaction with O 2 to form CO 2 and H 2 O For example, combustion of propane (C 3 H 8 ): C 3 H 8 (g) + 5 O 2  3 CO H 2 O + heat Note: combustion reactions are typically written for 1 mole of “fuel” this may require fractional coefficients for O 2, CO 2 and/or H 2 O

Energy Content in Fuels In this experiment, we will see how much heat given masses of fuel produce (i.e., J per gram) Heat flow is determined by measuring a mass of water recording the temperature of the water heating the water over the burning fuel so that at least a 20 o C increase occurs recording the new temperature Because of Conservation of Energy, the heat lost by the fuel is gained by H 2 O

set-up: T H2O glass rod fuel burner

Sample Calculation To raise the temperature of 100 grams of water by 20 o C, it is necessary to burn g of propane (C 3 H 8 ). Water has s = 4.18 J g -1 o C -1. What is the enthalpy of combustion of propane in kJ g -1 ? kJ mol -1 ?To raise the temperature of 100 grams of water by 20 o C, it is necessary to burn g of propane (C 3 H 8 ). Water has s = 4.18 J g -1 o C -1. What is the enthalpy of combustion of propane in kJ g -1 ? kJ mol -1 ? q H2O = -q combustion energy conservation q H2O = m H2O s H2O  T H2O = (100 g)(4.18 J g -1 o C -1 )(20 o C) = (100 g)(4.18 J g -1 o C -1 )(20 o C) = 8,360 J = 8,360 J

Sample Calculation (cont.) q combustion = -q H2O = -8,360 J Per gram:(-8360 J) / (0.307 g) = x 10 4 J g kJ g kJ g -1 Per mol: (-27.2 kJ g -1 )(44.09 g / mol) = -1,200 kJ mol -1

Safety Lab goggles must be worn at all times The CaCl 2 solution can be caustic Any spilled CaCl 2 must be cleaned off the balance immediately Be careful using the fuel lamps tie back long hair do not light lamps until ready