Chapter 6 Lecture Notes

6 2 159 g glucose c 6 h 12 o 6 is burned in an

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2. 1.59 g glucose (C 6 H 12 O 6 ) is burned in an adiabatic calorimeter. The calorimeter constant (calorimeter + contents) is 9.18 kJ/K. In the combustion experiment, the temperature of the calorimeter rises from 21.30 to 24.18 o C. Calculate the heat of combustion, q comb of glucose in a) kJ/g, and b) kJ/mol. Answer . Once again the key to this question is in the statement that it is an adiabatic calorimeter. This means that q combustion = −q calorimeter . q calorimeter = CΔT ΔT = 24.18 − 21.30 = 2.88 o C = +2.88 K. q calorimeter = 9.18(J/K)x2.88(K) = +26.4 kJ → q combustion = −26.4 kJ a) This is for the combustion of 1.59 g glucose. Then for 1 g glucose: q comb = −26.4 kJ/1.59 g glucose = −16.6 kJ/g b) The MM of glucose = 180 g/mol. q comb = −16.6(kJ/g)x180(g/mol) = −2.99x10 3 kJ/mol 3. 25.0 mL 0.500 M H.HCl(aq) is added to 75.0 mL 0.500 M NaOH in an open adiabatic styrofoam “coffee cup calorimeter”. The temperature rises by 1.68 o C. calculate the molar enthalpy of neutralization on HCl. The heat capacity of the Styrofoam cup can be neglected. The heat capacity of 1 mL solution can be assumed to be equal to the heat capacity of 1 mL (1 g) water. Answer . This is a neutralization reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O( l ). We have n HCl = 0.500x0.025 = 0.0125 mol, and n NaOH = 0.500x0.075 = 0.0375 mol, so NaOH is in excess and HCl is the limiting reagent, the HCl is completely neutralized. First we have to find the calorimeter constant C. We are told that the Styrofoam cup has no heat capacity itself, therefore it is only the content, (75 + 25 mL = 100 mL solution = 100 g) that represents the heat capacity of the calorimeter. C = s.m = 4.18(J/K.g)x100(g) = 418 J/K q calorimeter = CΔT = 418x1.68 = 702. J q rxn = −702 J Since it is an open system, pressure is constant, this is q p,reaction = ΔH rxn This is ΔH rxn for 0.0125 mol HCl neutralized by excess NaOH. For the neutralization of 1 mol HCl: ΔH = −702 (J) / 0.0125 mol = −5.62x10 4 J ΔH neutralization = -56.2 kJ/mol 3. Standard enthalpy and Hess’ Law (Chang section 6.6 pp. 246-250) We wrote earlier that for a reaction: Reactants Products ΔH rxn = H(products) − H(reactants). We now have to define what we mean by the enthalpy of a substance. In these calculations we use for H the standard enthalpy of formation , ΔH f o of a compound. 7
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A formation reaction is a reaction (often hypothetical) in which a compound is formed from its constituent elements under standard conditions (T = 25 o C (298K), p = 1 atm), and with the elements in the form in which they naturally occur under standard conditions. The standard enthalpy of formation of a substance, ΔH f o is the enthalpy change when one mole of substance is formed in the formation reaction. Some values of ΔH f o for a number of common compounds are listed in table 6.4 (p. 247), more extensive tables can be found in Chang Appendix 3, pp. A8-A12. By its definition, the standard enthalpy of formation of an element in its natural state at 298K, 1 atm pressure (H 2 (g), O 2 (g), Fe(s), Cl 2 (g), C(s), etc.) is zero. Some examples of formation reactions: For H 2 O( l ): H 2 (g) + ½O 2 (g) → H 2 O( l ) ΔH = ΔH f o (H 2 O, l ) = −285.8 kJ/mol For SO 3 (g): S(s) + 1½O 2 (g) → SO 3 (g) ΔH = ΔH f o (SO 3 , g) = −395.2 kJ/mol For NH 3 (g): ½N 2 (g) + 1½H 2 (g) → NH 3 (g) ΔH = ΔH f o (NH 3 , g) = −46.3 kJ/mol But also: For H 2 O(g): H 2 (g) + ½O 2 (g) → H 2 O(g) ΔH = ΔH f o (H 2 O,g) = −241.8 kJ/mol
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