CHEM 105 Problem Set 31 PDF

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Problem Set 31 - Calorimetry Chem 105 1. (a) What is calorimetry? Experimental method wherein changes in temperature are used to determine the energy released or absorbed by a process or material. Calorie  energy, metry  measure: Calorimetry = Measuring energy (b) What is a calorimeter? Device typically used to encapsulate a material or reaction in order to measure the absorption or release of energy during a physical or chemical process; in essence, it is a container that can be used to measure energy absorbed/released by something inside it (c) What is a bomb calorimeter? Device designed to maintain a constant volume while measuring the energy released during a combustion reaction conducted inside it 2. (a) Why is it necessary to know the heat capacity of a calorimeter? If our goal is to meaure the energy absorbed or released by a material or reaction inside a container by measuring the temp of the container, we need to know how much a given amount of energy will cause the container to change its temp. Mathematically, we’re measuring change in t of the container in order to determine q of the rxn which is equal to -q of the container (the calorimeter) which is equal tot CAT thus we need to already know C to calculate q. (b) Why do we use Cp (heat capacity of an object) instead of specific heat or molar heat capacity to describe the heat capacity of the calorimeter? Calorimeter consist of many components, all of which exchange energy with the material/reaction inside the calorimeter. Thus, we need to know the heat capacity of all the components together, which cannot be described by specific heat or molar heat capacity, which are used to describe pure substances (c) When measuring the heat of combustion of a very small amount of material, would it be better to use a calorimeter having a heat capacity that is small or large? Small. A small amount f material releases a small amount of energy. To detect a small amount of energy, we would want to use a calorimeter that readily registers a change in temp. Therefore, a calorimeter with a small heat capacity would be best since it takes less energy to change the temp than with a large-heat-capacity calorimeter. 3. Calculate the heat capacity of a calorimeter if the combustion of 4.663 g of benzoic acid produces an increase in temperature of 7.149°C. Combustion of exactly 1 g of benzoic acid releases 26.38 kJ to its surroundings. Since 4.663 benzoic acid were combusted here, the heat released had to be q = 26.38kJ/ g x 4.663 g = 123.0 kJ The heat capacity of the calorimeter, C, is related to the heat absorbed, q as q = CAT the temp change was 7.149 so rearranging the equation algebraically to solve for C: C = q/ change t = 123.okJ / 7.149 = 17.21 kJ degrees Celsius 4. The aromatic hydrocarbon cymene (C10H14) is found in nearly 100 spices and fragrances, including coriander, anise, and thyme. The complete combustion of 1.608 g of cymene in a bomb calorimeter (Ccalorimeter = 3.640 kJ/°C) produced an increase in temperature of 19.35°C. How much thermal energy is produced during the complete combustion of one mole of cymene? In a bomb calorimeter q system = alpha e combustion but since the PV work is usually small, alpha e combustion = alpha H combustion So we can assume alpha H combustion = -q calorimeter since qrxn = q calorimeter we can find alpha h combustion through alpha h combustion = -C calorimeter alpha T: - 3,60kJ x 19.35 = 70.43 kJ this is the thermal energy produced for the combustion of 1.608 g of cyme. To find the energy produced per mole, convert this mass to moles and divide the energy

by that number of moles: molar alpha H combustion 70.43kJ/ 1.608 g x 1/134.2 g = -5878 kJ/ mol 5. The flavor of anise is due to anethole, a compound with the molecular forumal C10H12O. Combustion of one mole of anethole produces 5541 kJ of thermal energy. If 0.950 g of anethole is combusted in a bomb calorimeter whose heat capacity (Ccalorimeter) is 7.854 kJ/°C, what is the change in temperature of the calorimeter? The alpha H combustion for 0.950 g of anethole (148.20 g/mol) is Alpha h combustion = 0.950 g x 1 mol/ 148.20 g x -5541 kJ/ mol = -35.5 kJ Since the temp change of the calorimeter is related to the heat of combustion as Alpha H combustion = -C calorimeter alpha T we can solve algebraically for alpha T as follows: Alpha T = - alpha h combustion / c calorimeter = 35.5kJ / 7.854 kJ = 4.52 degree Celsius 6. When 4.00 g NH4NO3 (80.04g/mol) – the active ingredient in some chemical cold packs – is dissolved in 96.0 g H2O, the temperature of the resulting solution is 3.07°C colder than the water and ammonium nitrate were before they were mixed together. What is the value of ∆H for the following dissolution process? NH4NO3(s)  NH4NO3(aq) ∆H = ? Hint: Because the amount of ammonium nitrate dissolved is relatively small, you can assume that the heat capacity of the final solution is close to that of pure water (4.184 J/g°C). The heat q necessary for the dissolution reaction all came from the solution. We can find q through q = mcat where m is the mass of the solution (h2O + ammonium nitrate) : Q = mcat – (96.o g h2O + 4.00 g NH4NO3) x 4.184 J / g x 3.07 degrees Celsius = 1280 J Because this energy change resulted solely from the dissolution of 4.00 g of ammonium nitrate, and because we can assume pressure was constate, q = alpha H dissolution to get the enthalpy per mole, we can convert the 4.00 g to moles and divide the energy by those moles: Alpha H dissolution = 1280 J/ 4.00 g NH4NO3 x 1 mol/ 8004 g = 25600 J/ mol = 25.6 kJ/ mol...