Bomb calorimetry

The chemical energy of a system is changed as a result of a reaction. Calorimetry. Heat of combustion . Calculation of caloric content of sucrose or food. Combustion reaction.

1. Bomb Calorimetry Dr. K. Shahzad Baig Memorial University of Newfoundland (MUN) Canada Petrucci , et al. 2011. General Chemistry: Principles and Modern Applications. Pearson Canada Inc., Toronto, Ontario. Tro , N.J. 2010. Principles of Chemistry. : A molecular approach. Pearson Education, Inc.

2. Heats of Reaction and Calorimetry A chemical reaction is a process in which some chemical bonds are broken and others are formed, in general, the chemical energy of a system is changed as a result of a reaction. A heat of reaction , is the quantity of heat exchanged between a system and its surroundings when a chemical reaction occurs within the system at constant temperature . One of the most common reactions studied is the combustion reaction . This is such a common reaction that we often refer to the heat of combustion when describing the heat released by a combustion reaction

3. An endothermic reaction , is the one a temperature decrease in an isolated system or a gain of heat from the surroundings by a nonisolated system, heat of reaction is a q rxn > 0 An endothermic reaction Ba (OH) 2 . 8H 2 O (s) and NH 4 Cl (s) and are mixed at room temperature, and the temperature falls to 5.8 o C in the reaction. An exothermic reaction is one that produces a temperature increase in an isolated system or, gives off heat to the surroundings , in a non-isolated system, the heat of reaction is (q rxn < 0) An exothermic reaction. Slaked lime, Ca (OH) 2 is produced by the action of H 2 O on quicklime, ( CaO ). The reactants are mixed at room temperature, but the temperature of the mixture rises to 40.5 o C    

4. Bomb Calorimetry Used to measure the heat evolved in a combustion reaction. The system is everything within the double-walled outer jacket of the calorimeter. This includes the bomb and its contents, the water in which the bomb is immersed, the thermometer, the stirrer, and so on. The system is isolated from its surroundings. .   q calorim = heat capacity of calorium * ∆T When the combustion reaction occurs, chemical energy is converted to thermal energy , and the temperature of the system rises

5. Using Bomb Calorimetry Data to Determine a Heat of Reaction Example 7-3 The combustion of 1.010 g sucrose, in a bomb calorimeter causes the temperature to rise from 24.92 to28.33 o C. The heat capacity of the calorimeter assembly is 4 .90 kJ/°C What is the heat of combustion of sucrose expressed in kilojoules per mole of C 12 H 22 O 11 (b) Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g) contains only19 Calories . Solution (a) Calculate q calorim   q rxn = -q calorim = -16.7 kJ

6. This is the heat of combustion of the 1.010 g sample. Per gram C 12 H 22 O 11   Per mole     (b) To determine the caloric content of sucrose , we can use the heat of combustion per gram of sucrose determined in part (a), together with a factor to convert from kilojoules to kilocalories. (Because 1 cal = 4.184 J)  

7. Problem statement A 1.000 g sample of octane (C 8 H 18 ) is burned in a bomb calorimeter containing 1200 grams of water at an initial temperature of 25.00ºC. After the reaction, the final temperature of the water is 33.20ºC. The heat capacity of the calorimeter (also known as the “calorimeter constant”) is 837 J/ºC. The specific heat of water is 4.184 J/g ºC. Calculate the heat of combustion of octane in kJ/mol. Solution Since this is a combustion reaction , heat flows from the system to the surroundings- thus, it is exothermic. The heat released by the reaction will be absorbed by two things: (a) the water in the calorimeter and (b) the calorimeter itself. The temperature change of the calorimeter is the same as the temperature change for water.

8. a. Calculate the heat absorbed by the water (q water ) b. Calculate the heat absorbed by the calorimeter (q cal )

9. The TOTAL heat absorbed by the water and the calorimeter = (a) +(b): =41.2 + 6.86 = + 48.1 kJ. ( Remember, q is positive because the heat is being absorbed ). The amount of heat released by the reaction = the amount of heat absorbed by the water and the calorimeter . q reaction = – 48.1 kJ 1.000 gram of octane was burned, the heat of combustion for octane = to – 48.1 kJ/gram What is the heat of combustion in kJ/ mol ? = -48.1 kJ/g x 114 g/ mol = – 5483 kJ/mol.

10. Work Work involved in the expansion or compression of gases is called pressure volume work . This type of work is performed: i ) by explosives, and ii) by the gases formed in the combustion of gasoline in an automobile engine . Consider the decomposition of potassium chlorate to potassium chloride and oxygen. The pressure inside the reaction vessel exceeds the atmospheric pressure and the piston is lifted, the system does work on the surroundings . The work can be calculated by   work (w) = force (M * g) * distance (∆h) = -M * g * ∆h  

11. Bomb Calorimetry Used to measure the heat evolved in a combustion reaction. The system is everything within the double-walled outer jacket of the calorimeter. This includes the bomb and its contents, the water in which the bomb is immersed, the thermometer, the stirrer, and so on. The system is isolated from its surroundings. .   q calorim = heat capacity of calorium * ∆T When the combustion reaction occurs, chemical energy is converted to thermal energy , and the temperature of the system rises Determination of Δ U and Δ H for Chemical Reactions

12. In a bomb calorimeter, the reaction is carried out at constant volume. The motivation for doing so is that if dV = 0, Δ U = q V . Therefore, a measurement of the heat flow normalized to 1 mole of the specified reaction provides a direct measurement of Δ U R . no heat flow will occur between the system and surroundings, and q = 0. Because the combustion experiment takes place at constant volume, w = 0. Therefore, Δ U= 0. So, it is an isolated system of finite size

13. Consider the system as consisting of three subsystems: the reactants in the calorimeter, the calorimeter vessel, the inner water bath. These three subsystems are separated by rigid diathermal walls and are in thermal equilibrium. Energy is redistributed among the subsystems as reactants are converted to products, the temperature of the inner water bath changes, and the temperature of the calorimeter changes. Δ T, the change in the temperature of the three subsystems. The mass of water in the inner bath, m H2O ; its molecular weight, M H2O ; its heat capacity, C P,m (H 2 O); the mass of the sample, m S ; and its molecular weight, M S , are known. Δ U combustion is defined per mole of the combustion reaction, but because the reaction includes exactly 1 mole of reactant, the factor m S / M S in Equation (4.21) is appropriate.

14. Example 4.3 When 0.972 g of cyclohexane undergoes complete combustion in a bomb calorimeter, of the inner water bath is 2.98°C. For cyclohexane, Δ U combustion is –3913 kJ mol -1 . Given this result, what is the value for Δ U combustion for the combustion of benzene if Δ T is 2.36°C when 0.857 g of benzene undergoes complete combustion in the same calorimeter? The mass of the water in the inner bath is 1.812 x 10 3 g and the C P,m of water is 75.3 J K -1 mol -1 . Solution To calculate the calorimeter constant through the combustion of cyclohexane, we write Equation (4.21)

15. In calculating Δ U combustion for benzene, we use the value for Ccalorimeter : Once Δ U combustion has been determined, Δ H combustion can be determined

16. For reactions involving only solids and liquids, Δ U >> Δ (PV) and Δ H ≈ Δ U . If some of the reactants or products are gases, the small change in the temperature that is measured in a calorimetric experiment can generally be ignored and Δ H combustion = Δ U combustion = Δ nRT Δ H combustion = Δ U combustion + Δ nRT Δ n is the change in the number of moles of gas in the overall reaction. and Δ n = –3. Note that at T = 298.15 K, the most stable form of C 6 H 12 and H 2 O is liquid. For this reaction, Δ U combustion and Δ H combustion differ by only 0.2%.

17. at constant P using a constant pressure calorimeter. is directly determined because Δ H = q P . Equation (4.21) takes the following form for constant pressure calorimetry involving the solution of a salt in water: Δ H° solution is defined per mole of the solution reaction, but because the reaction includes exactly 1 mole of reactant, the factor in Equation (4.21) is appropriate. Because Δ (PV) is negligibly small for the solution of a salt in a solvent, Δ U° solution = Δ H° solution The solution must be stirred to ensure that equilibrium is attained before Δ T is measured.

18. Using Bomb Calorimetry Data to Determine a Heat of Reaction Example The combustion of 1.010 g sucrose, in a bomb calorimeter causes the temperature to rise from 24.92 to28.33 o C. The heat capacity of the calorimeter assembly is 4 .90 kJ/°C What is the heat of combustion of sucrose expressed in kilojoules per mole of C 12 H 22 O 11 (b) Verify the claim of sugar producers that one teaspoon of sugar (about 4.8 g) contains only19 Calories . Solution (a) Calculate q calorim   q rxn = -q calorim = -16.7 kJ

19. This is the heat of combustion of the 1.010 g sample. Per gram C 12 H 22 O 11   Per mole     (b) To determine the caloric content of sucrose , we can use the heat of combustion per gram of sucrose determined in part (a), together with a factor to convert from kilojoules to kilocalories. (Because 1 cal = 4.184 J)  

20. Problem statement A 1.000 g sample of octane (C 8 H 18 ) is burned in a bomb calorimeter containing 1200 grams of water at an initial temperature of 25.00ºC. After the reaction, the final temperature of the water is 33.20ºC. The heat capacity of the calorimeter (also known as the “calorimeter constant”) is 837 J/ºC. The specific heat of water is 4.184 J/g ºC. Calculate the heat of combustion of octane in kJ/mol. Solution Since this is a combustion reaction , heat flows from the system to the surroundings- thus, it is exothermic. The heat released by the reaction will be absorbed by two things: (a) the water in the calorimeter and (b) the calorimeter itself. The temperature change of the calorimeter is the same as the temperature change for water.

21. a. Calculate the heat absorbed by the water (q water ) b. Calculate the heat absorbed by the calorimeter (q cal )

22. The TOTAL heat absorbed by the water and the calorimeter = (a) +(b): =41.2 + 6.86 = + 48.1 kJ. ( Remember, q is positive because the heat is being absorbed ). The amount of heat released by the reaction = the amount of heat absorbed by the water and the calorimeter . q reaction = – 48.1 kJ 1.000 gram of octane was burned, the heat of combustion for octane = to – 48.1 kJ/gram What is the heat of combustion in kJ/ mol ? = -48.1 kJ/g x 114 g/ mol = – 5483 kJ/mol.