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Published byGerard Pope Modified over 8 years ago
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Calorimetry How is heat measured? It isn’t Temperature measured T (K) C = heat capacity (J/K) = q (J) C = heat to raise T 1 o x
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heat capacity specific heat capacity q (J) = q system = C (J / K g) x T (K) x mass (g) - molar heat capacity mol g q surroundings J / K (thermometer)(reaction)
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heat capacity K.E. translational heat K.E. rotational compare: waterHelium He C He C H2OH2O a) > b) < c) = - raise T of system + K.E. vibrational + P.E.
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heat capacities substance specific m.w.molar heat capacity (g/mol) heat capacity (J / K g) (J / K mol) Al (s) Fe (s)0.44 0.89 26.9824.0 55.85 24.8 H 2 O(s) 36.5 H 2 O(l) CCl 4 (l) 75.2 133.0
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Calorimetry 1. Measure T (T final - T initial ) 2. Convert to q T (K) x mass (g) =x C (J/K g) q (J)
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50 mL 100 o 100 mL 25 o 150 mL (4.184 J/ o C g) q = C x T x mass q 2 = (4.184 J/ o C g) x q 1 = - q 2 (T f - 100) x (50) = T f = 50 o C q 1 =x (T f - 100)x (50 g) T f Heat transfer experiments (T f - 25) x(100 g) -(T f - 25) x (100)
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Calorimetry 1. Measure T (T final - T initial ) 2. Convert to q T (K) x mass (g) =x C (J/K g) q (J) q is a path function E = q + w q = EE EE - w = + P ext V
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q = E + P ext V H q = E + P V At constant P, H = E + P V q EnthalpyH H = E+ P V+V P P = 0 H = E + PV E + PV p = H = H
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Enthalpy H E + PV Extensive State function H rxn = q rxn = q system =- q surroundings (thermometer)(reaction) T f > T i H rxn ><>< 0 endothermic exothermic
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Enthalpy of reaction H rxn = q rxn coffee cup calorimeter 10.5 g KBr T f = 21 o Calculate H rxn KBr(s) K + (aq) + Br - (aq) q system = = H rxn in 125g waterat 24 o - q surroundings
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H rxn q system = - q surroundings = H rxn q surroundings =C x q surroundings = (4.184 J/g o C)(21(10.5 g = -1756 J q system = - q surroundings = +1756 J = H rxn H is extensive H rxn = 1756 J = 167 J/g= 19873 J/mol TT x mass - 24 o C)+ 125 g) a) endothermic b) exothermic 10.5 g KBr
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E = q + w E = q - P V At constant V, Bomb calorimeter q rxn = q system = -q calorimeter q calorimeter = C (J / o C) x E = q v T ( o C)
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Constant Volume calorimetry 2Fe (s) + 3/2 O 2 (g) Fe 2 O 3 (s) 11.2 g Fe(s), 1 atm O 2 C calorimeter = 2.58 kJ/ o C T calorimeter = + 31.9 o C q rxn = -q calorimeter = E rxn = - (2.58 kJ/ o C) = - 82.2 kJ E rxn = - 822 kJ/mol Fe 2 O 3 / 0.1 mol Fe 2 O 3 (31.9 o C)
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q v v.s. q p q v = E q p = H H = E + PV H = E + PV if n = 0 2Fe (s) + 3/2 O 2 (g) Fe 2 O 3 (s) n = H = H = -826 kJ/mol = E + nRT H = E (0 - 3/2)= - 3/2 - 822 kJ/mol + (- 3/2)(8.314 x 10 -3 kJ)(298)
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