measure ΔT in water to get q = q surroundings and use q system = q surroundings

Transcription

1 example using water: Calculate the amount of energy required to heat 95.0 g of water from 22.5 C to 95.5 C. q = s m ΔT ( C (4.184 J g 1 C 1 ) (95.0 g) (73.0 = = 2.90 x 10 4 J or 29.0 kj Constant Pressure Calorimetry (q p = ΔH) use a styrofoam cup calorimeter neglect the heat capacity of the cup measure heat exchanged between system (reactants/products) and surroundings (water) for: -- Heat of neutralization -- Heat of solution -- Heat of dilution -- Specific heat of a substance measure ΔT in water to get q = q surroundings and use q system = q surroundings

2 A metal pellet with mass grams and temperature 92.5 C is dropped into a calorimeter containing grams of water at temperature 23.1 C. The final temperature of the water and the pellet is 26.8 C after heat has been exchanged. Calculate the heat capacity and the specific heat of the metal. q water = msδt = ( g) (4.184 J g 1 C 1 ) (3.7 C) = x 10 3 J (water) = x 10 3 J (pellet released energy) q metal Heat capacity of pellet: q = CΔT C = q/δt = x 10 3 J/( C) = 35 J C 1 specific heat = s = (35 J C 1 ) / (85.00 g) = 0.41 J g 1 C 1 Constant Volume bomb Calorimetry (q v = ΔU) Useful for determining heats of combustion, which can tell us how much energy is in a particular food. place sample in steel container, fill with O 2 (g) place container in water ignite sample electrically measure water temperature

3 example: A snack chip with a mass of 2.35 g was burned in a bomb calorimeter. The heat capacity of the calorimeter is kj C 1. During the combustion the water temperature rose by 2.70 C. Calculate the energy in kj g 1 for the chip. q rxn = C cal ΔT = (38.57 kj C 1 ) (2.70 C) = 104 kj (heat given off by reaction) energy per gram = ( 104 kj) / (2.35 g) = 44.1 kj g 1 Hess s Law = conservation of energy for chemical reactions If two or more chemical equations are added together to give another chemical equation, the corresponding enthalpies of reaction must be added. Useful for calculating enthalpy for a reaction that cannot be carried out directly -- calorimetry used to measure ΔH values -- Hess s Law used to predict new ΔH values

4 Given the following equations: H 3 BO 3 (aq) HBO 2 (aq) + H 2 O (l) H 2 B 4 O 7 (aq) + H 2 O(l) 4 HBO 2 (aq) H 2 B 4 O 7 (aq) 2 B 2 O 3 (s) + H 2 O (l) Find ΔH for the following reaction: 2 H 3 BO 3 (aq) B 2 O 3 (s) + 3 H 2 O (l) ΔH rxn = 0.02 kj ΔH rxn = 11.3 kj ΔH rxn = 17.5 kj Calculate the ΔH for the reaction: C (s) + 2 H 2 (g) CH 4 (g) Given the following: C (s) + O 2 (g) CO 2 (g) H 2 (g) + ½ O 2 (g) H 2 O (l) CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) ΔH = kj/mol ΔH = kj/mol ΔH = kj/mol

5 Standard Enthalpies of Formation ΔH rxn = ΔH products ΔH reactants Almost impossible to measure the value of H for a substance. Instead, use an arbitrary reference point, as follows: Standard heat of formation = ΔH f o = heat change that results when 1 mole of a compound is formed from its elements in their standard states The standard state of a compound is the most stable state found at 1 atm (and usually at 25 C = 298 K) Therefore ΔH f o for elements in their most stable form = 0 examples: O 2 (g) C(graphite) This leads to the concept of formation reactions: ½ N 2 (g) + 3 / 2 H 2 (g) NH 3 (g) ΔH f o [NH 3 (g)] = 46.3 kj mol 1 2 Na(s) + C(graphite) + 3 / 2 O 2 (g) Na 2 CO 3 (s) ΔH f o [Na 2 CO 3 (s)] = kj mol 1 We can use standard enthalpies of formation to calculate the ΔH o rxn (standard enthalpy of reaction) ΔH o rxn = Σ [nδh f o ] (products) Σ [mδh f o ] (reactants) Σ = sum n and m = stoichometric coefficients from the balanced chemical equation

6 example: Calculate the ΔH o rxn for the following reaction CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) ΔH o rxn = [( 393.5) + 2( 285.8)] [( 74.85) + 2(0)] = kj mol 1 (exothermic) Calculate the ΔH o rxn for the following reaction: 2 NaHCO 3 (s) Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) Acetylene burns in air according to the following equation: C 2 H 2 (g) + 5 / 2 O 2 (g) 2 CO 2 (g) + H 2 O(g) ΔH o rxn = kj What is ΔH o f of C 2 H 2 (g)?

7 Bond Enthalpies and Hess s Law Consider the formation and breakage of bonds form a bond = exothermic ΔH < 0 break a bond = endothermic ΔH > 0 combine the two: CH O 2 CO H 2 O the reaction requires: breakage of 4 C H bonds breakage of 2 O=O bonds formation of 2 C=O bonds formation of 4 O H bonds Is energy needed or released for this reaction?

8 If we had ΔH for bonds, they could be used to calculate ΔH rxn Cl 2 (g) Cl (g) + Cl (g) HCl (g) H (g) + Cl (g) O 2 (g) O (g) + O (g) N 2 (g) N (g) + N (g) ΔH = kj = ΔH (Cl Cl) ΔH = kj = ΔH (H Cl) ΔH = kj = ΔH (O=O) ΔH = kj = ΔH (N N) For bonds in larger molecules, we take an average CH 4 CH 3 + H ΔH = 435 kj = ΔH (C H) C 2 H 6 C 2 H 5 + H ΔH = 410 kj = ΔH (C H) + more, can compute ΔH avg for (C H) bond across a representative series of compounds. example using bond enthalpies (BE): ΔH rxn = ΣBE(bonds broken) ΣBE(bonds formed) atoms enthalpy BE(r) reactant s - BE(p) product s

9 Calculate the enthalpy of reaction for: CH 4 (g) + Br 2 (g) CH 3 Br(g) + HBr(g) Approach 1: pay attention only to bonds broken/formed: ΔH rxn = [ΔH (C H) + ΔH (Br Br) ] [ΔH (C Br) + ΔH (H Br) ] = 36 kj/mol

10 Approach 2: break/form ALL bonds: ΔH rxn = [4(ΔH (C H)) + ΔH (Br Br) ] [3(ΔH (C H) + ΔH (C Br) + ΔH (H Br) ] = 36 kj/mol Lattice Energy LE lets us compute the stability of ionic compounds; experimentally we cannot measure LE directly. The way we get the LE is called the Born-Haber Cycle.

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