S T A T I O N 1 E X O T H E R M I C / E N D O T H E R M I C P R O C E S S E S

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1 Name Period Date S T A T I O N 1 E X O T H E R M I C / E N D O T H E R M I C P R O C E S S E S Determine if each statement describes an exothermic process (EXO) or endothermic process (ENDO). _EXO EXO_ surroundings get warmer ENDO PE diagram is uphill _EXO_ combustion of propanol ENDO H is positive _EXO_ reactants have more energy than products ENDO CaCO 3(s) + heat CaO(s) + CO 2(g) _EXO_ q < 0 _EXO_ H is negative ENDO q > 0 _EXO_ molecular attractions strengthen ENDO surroundings get colder _EXO_ H 2(g) + ½O 2 (g) H 2O (l) + heat ENDO products have more energy than reactants _EXO_ condensation of water ENDO H 2O (l) H 2O (g) _EXO_ water, when placed in the freezer S T A T I O N 2 H r x n C A L C U L A T I O N S Consider the following balanced equation for the combustion of propane, C 3H 8. Use the Chart of Thermodynamic Values for H f and bond energies. C 3H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2O (l) Products Reactants Method Fill in the values of H f for each substance. Compound Hf (k/mol) C 3H 8 (g) 104 O 2 (g) 0 CO 2 (g) H 2O (l) 286 Bond Energy Method Given the following structural formulas. Substance C 3H 8 O 2 CO 2 H 2O Structures H O H Calculate the H rxn using the H f values. H rxn = [3( 393.5) + 4( 286)] [1( 104) + 5(0)] = = k/mol rxn Calculate the H rxn using bond energies. Break 2 C C bonds +2(347) = C H bonds +8(413) = O=O bonds +5(495) = Form 6 C=O bonds 6(799) = O H bonds 8(467) = 3736 H rxn = 2057 k/mol rxn The two answers above do not match. Which one is more accurate? Why? Products Reactants is more accurate. Bond Energy Method is less accurate because bond energies are averages and substances are in gas phase.

2 S T A T I O N 3 C A L O R I M E T R Y A 100. g aluminum block (C = /g ºC) in boiling water is added to an insulated cup containing 50.0 grams of water (C = /g ºC) at 5.00ºC. Calculate the final temperature of the mixture Aluminum Water m = 100. g m = 50.0 g C = / C = / T = x 100 C T = x 5.00 C q Al = +q H2O (m Al)(C Al)( T Al) = +(m H2O)(C H2O)( T H2O) (100 g)(0.900 /)(x 100 C) = +(50.0 g)(4.184 /)(x 5.00 C) x = 33.6 C S T A T I O N 4 E N E R G Y F L O W Heat always flows from an object of HIGHER temperature to an object of LOWER temperature. Both blocks have a temperature of 22.5 C. Use arrows on the top picture to show the flow of heat and why the ice acts the way it does. Provide a brief explanation for the observed differences. Time Block A Block B Explanation 0 minutes Blocks A and B are different materials. Block B transfers energy to the ice at a faster rate than Block A, so the block of ice on Block B melts faster than that on Block A. 2 minutes

3 S T A T I O N 5 H E A T S O F F U S I O N & V APO R I Z A T I O N For water, H fus = 6.01 k k mol, Hvap = mol, Cice = 2.10, Cwater = 4.18, Csteam = 2.08 What is the value of q when 45.0 g of water freezes at 0 C? n = 45.0 g 1 mol = 2.50 mol q = n Hfus = (2.50 mol)( 6.01 k/mol) = 15.0 k g H = H fus = 6.01 k/mol What mass of water can be vaporized with 75.0 k of energy? q k q = k n = = = 1.84 mol H vap k/mol H = + H vap = k/mol m = 1.84 mol g = 33.2 g 1 mol S T A T I O N 6 H r x n F R O M D A T A Write the balanced thermochemical equation for the combustion of pentane, C 5H 12. C 5H O 2 5 CO H 2O + heat When 10.0 g of C 5H 12 is burned, 453 k of energy is released. What is the H rxn for the combustion of pentane? Molar Mass = 5(12.01) + 12(1.008) = 72.1 g/mol n rxn = 10.0 g C 5H 12 1 mol C 5H g C 5H 12 1 mol rxn = mol rxn 1 mol C 5H 12 H rxn = q 453 k = = 3270 k/mol rxn n rxn mol rxn Write the balanced thermochemical equation for the melting of aluminum. (Include states.) Heat + Al (s) Al (l) When 10.0 grams of aluminum melts, k of energy is required. What is the H fus of Al? Molar Mass = g/mol n rxn = 10.0 g Al 1 mol Al g Al 1 mol rxn = mol 1 mol Al H rxn = q k = = 10.6 k/mol n rxn mol

4 S T A T I O N 7 C O M P A R I N G S U B S T A N C E S Blocks X and Z have the same mass. When 500 of heat is added to each, the temperature of Block X rises 10 ºC while the temperature of Block Z rises 30 ºC. Which block has the larger specific heat, C? ustify your answer without math. Block X. With the same amount of energy, the temperature of Block X rises less than Block Z, It is harder to change the temperature of Block X (it is more resistant to change in temperature) and it has a larger specific heat mole samples of Liquid A and Liquid B were kept at their boiling points. When 400 of energy is added to each, Liquid A completely vaporizes while some of Liquid B remained. Which liquid has a greater H vap? ustify your answer without math. Liquid B. With the same amount of energy, less of Liquid B vaporized than Liquid A. It is harder to vaporize Liquid B, so it has a greater H vap. Sketch the graph when 40.0 g of ice at 30 C is heated to steam at 140 C. Label the axes (with units), and the Freezing and Boiling points. Label the segments (a) through (e). Identify the state(s) of matter for each segment. Temp ( C) 100 S T A T I O N 8 H E A T I N G C U R V E (c) liquid (d) liquid/gas (e) gas 0 (b) solid/liquid (a) solid In which segments: (circle) are molecular attractions weakening? b d is the kinetic energy increasing? a c e are the processes endothermic? a b c d e are the processes exothermic? do you use q = m C T to calculate the heat? a c e do you use q = n H to calculate the heat? b d

5 S T A T I O N 9 H E A T I N G C U R V E C A L C U L A T I O N S For water, H fus = 6.01 k k, Hvap = 40.68, Cice = 2.10 mol mol, Cwater = 4.18, Csteam = 2.08 A 50.0 gram sample of ice at 0.0 ºC is heated until it is liquid at 80.0 ºC. Indicate on the heating curve where the heating process begins and ends. Calculate the heat absorbed by this process. q 2 = (50.0 g) 1 mol g 6.01 k 1 mol 1000 = k q 3 = (50.0 g)(4.184 /)(80 C 0 C) = q = q2 + q3 = 16, ,700 = 33,400