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Premium member Presentation Transcript 5.5 -Calorimetry : 5.5 -Calorimetry Calorimetry : The value of ΔH can be determined by measuring the magnitude of the temperature change the heat flow produces 5.5 Calorimetry Calorimetry : The measurement of heat flow is calorimetry The device used to measure heat flow is a calorimeter 5.5 Calorimetry Heat Capacity : All substances change temperature when they are heated, but the magnitude of the temperature change produced by a given quantity of heat varies from substance to substance 5.5 Heat Capacity Heat Capacity : The temperature change experienced by an object when it absorbs a certain amount of heat is determined by its heat capacity 5.5 Heat Capacity Heat Capacity : The heat capacity of an object is the amount of heat required to raise its temperature by 1 K The greater the heat capacity, the greater the heat required to produce a given increase in temperature 5.5 Heat Capacity Molar Heat Capacity : The heat capacity of 1 mole of a substance is called its molar heat capacity 5.5 Molar Heat Capacity Specific Heat : The heat capacity of one gram of a substance is called its specific heat capacity, or merely its specific heat 5.5 Specific Heat Specific Heat : Formula q = sxmxΔT q = heat s = specific heat of substance m = mass (grams) Δ T = change in temperature 5.5 Specific Heat Specific Heat : What is the specific heat of water? s = 4.18 J/g-K 5.5 Specific Heat Practice Exercise : Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g-K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if the if temperature increases by 12.0˚C q = s⋅m⋅ΔT = (0.82 J/g⋅K) ⋅ (50,000 g) ⋅ (12 K) = 4.9 x 105 J 5.5 Practice Exercise Practice Exercise : What temperature change would these rocks undergo if they emitted 450 kJ of heat? ∆T = q/(s⋅m) = (-450,000 J) / (0.82 J/g⋅Kx 50,000 g) = 11 K or 11˚C decrease 5.5 Practice Exercise Constant-Pressure Calorimetry : Constant-Pressure Calorimetry Explain the structure and function of a coffee-cup calorimeter. 5.5 Coffee-Cup Calorimetry : There is no physical barrier between the system and the surroundings The reactants and products of the reaction are the system The water in which they are dissolved and the calorimeter are the surroundings 5.5 Coffee-Cup Calorimetry Coffee-Cup Calorimetry : The heat gained by the solution must be produced from the chemical reaction under study For an exothermic reaction, the heat is lost by the reaction and gained by the solution The temperature of the solution rises The opposite happens for an endothermic reaction 5.5 Coffee-Cup Calorimetry Coffee-Cup Calorimetry : The heat gained by the solution, qsoln, is equal in magnitude to qrxn Coffee-Cup Calorimetry Practice Exercise : When 50.0 ml of 0.100 M AgNO3 and 50.0 ml of 0.100 M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.20˚C to 23.11˚C. The temperature increase is caused by the following reaction: AgNO3(aq) + HCl(aq) AgCl(s) + HNO3(aq) Calculate ΔH for this reaction in kJ/mol AgNO3, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.181 J/g-˚C. qrxn = - qsoln = -(s⋅m⋅∆T) = -(4.184 J/g⋅C) ⋅ 100g ⋅ 0.91 K = -380.74 J = -0.380744 kJ = -0.38077 kJ / 0.005 mol AgNO3 -76 kJ/mol 5.5 Practice Exercise Bomb Calorimetry : Bomb Calorimetry Describe the set-up and purpose of a bomb calorimeter 5.5 Bomb Calorimetry : A bomb calorimeter is used to measure the heat accompanying combustion reactions at a constant volume The heat released by the reaction is absorbed by the calorimeter contents, causing a rise in the temperature of the water 5.5 Bomb Calorimetry Bomb Calorimeter : What does the variable Ccal represent? The total heat capacity of the calorimeter 5.5 Bomb Calorimeter Bomb Calorimeter : What is the formula for calculating the heat evolved in a reaction conducted in a bomb calorimeter? qrxn = -Ccal x ΔT 5.5 Bomb Calorimeter Practice Exercise : A 0.5865 g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is 4.812 kJ/˚C. The temperature increases from 23.10˚C to 24.95˚C. Calculate the heat of combustion of lactic acid A) per gram B) per mole 5.5 Practice Exercise Practice Exercise : A) per gram qrxn = -Ccal ⋅ ∆T = - (4.812 kJ/K) ⋅ (1.85 K) = 8.902 kJ / 0.5865 g = -15.2 kJ/g B) per mole = (-15.2 kJ/g) ⋅ (90.0 g/mol) = -1370 kJ/mol 5.5 Practice Exercise You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.