Thermochemistry Unit Chemistry 12

Energy is the capacity to do work Radiant energy comes from the sun and is earth’s primary energy source Thermal energy is the energy associated with the random motion of atoms and molecules Chemical energy is the energy stored within the bonds of chemical substances Nuclear energy is the energy stored within the collection of neutrons and protons in the atom Potential energy is the energy available by virtue of an object’s position

Energy Changes in Chemical Reactions Heat is the transfer of thermal energy between two bodies that are at different temperatures. Temperature is a measure of the thermal energy. Temperature = Thermal Energy 400C 900C greater thermal energy

Thermochemistry is the study of heat change in chemical reactions. The system is the specific part of the universe that is of interest in the study. open closed isolated Exchange: mass & energy energy nothing

2H2 (g) + O2 (g) 2H2O (l) + energy Exothermic process is any process that releases heat – transfers thermal energy from the system to the surroundings. 2H2 (g) + O2 (g) 2H2O (l) + energy H2O (g) H2O (l) + energy Endothermic process is any process in which heat has to be absorbed or supplied to the system from the surroundings. energy + H2O (s) H2O (l) energy + 2HgO (s) 2Hg (l) + O2 (g)

Exothermic Endothermic

, pressure, volume, temperature Thermodynamics is the scientific study of the inter-conversion of heat and other kinds of energy. State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. energy , pressure, volume, temperature ΔE = Efinal - Einitial Δ P = Pfinal - Pinitial Δ V = Vfinal - Vinitial Δ T = Tfinal - Tinitial Potential energy of Hiker 1 and Hiker 2 is the same even though they took different paths.

Exothermic chemical reaction! First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed. ΔEsystem + Δ Esurroundings = 0 or Δ Esystem = - Δ Esurroundings C3H8 + 5O2 3CO2 + 4H2O Exothermic chemical reaction! Chemical energy lost by combustion = Energy gained by the surroundings system surroundings

ΔH = H (products) – H (reactants) Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. ΔH = H (products) – H (reactants) ΔH = heat released or absorbed during a reaction at constant pressure Hproducts > Hreactants Hproducts < Hreactants Δ H > 0 Δ H < 0

Thermochemical Equations Is Δ H negative or positive? System absorbs heat Endothermic Δ H > 0 6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm. H2O (s) H2O (l) Δ H = 6.01 kJ

Thermochemical Equations Is Δ H negative or positive? System gives off heat Exothermic Δ H < 0 890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) Δ H = -890.4 kJ

Thermochemical Equations The stoichiometric coefficients always refer to the number of moles of a substance H2O (s) H2O (l) Δ H = 6.01 kJ If you reverse a reaction, the sign of Δ H changes H2O (l) H2O (s) Δ H = -6.01 kJ If you multiply both sides of the equation by a factor n, then Δ H must change by the same factor n. 2H2O (s) 2H2O (l) Δ H = 2 x 6.01 = 12.0 kJ

Thermochemical Equations The physical states of all reactants and products must be specified in thermochemical equations. H2O (s) H2O (l) H = 6.01 kJ H2O (l) H2O (g) H = 44.0 kJ Calculate H when 266 g of white phosphorus (P4) burn in air? P4 (s) + 5O2 (g) P4O10 (s) Δ H = -3013 kJ 1 mol P4 123.9 g P4 x - 3013 kJ 1 mol P4 x 266 g P4 = - 6470 kJ

Heat (q) absorbed or released: The specific heat (c) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius. Heat (q) absorbed or released: q = m x c x Δ T Δ T = T final - T initial C = m x c q = C x Δ T

Is this reaction exothermic or endothermic? How much heat is given off when an 869 g iron bar cools from 940C to 50C? c of Fe = 0.444 J/g • 0C m = 869 g Δ T = Tfinal – Tinitial = 50C – 940C = - 890C q = mc Δ T = (869 g x 0.444 J/g • 0C)(– 890C) = -34,000 J Is this reaction exothermic or endothermic?

10.0 g of ice was added to 60.0 g of water. The initial temperature Sample Problem: 10.0 g of ice was added to 60.0 g of water. The initial temperature of the water was 26.5oC. The final temperature of the water was 9.70oC. How much heat was lost by the water? cwater= 4.184 J/g • 0C mwater = 60.0 g Δ T = Tfinal – Tnitial = 9.700C – 26.50C = - 16.80C q = mc Δ T = (60.0 g x 4.184 J/g • 0C)(– 16.80C) = - 4217.47J Is this reaction exothermic or endothermic?

Is this reaction exothermic or endothermic? Practice Problem #1: 100.0 g of ethanol at 25.0oC is heated until it reaches 50.0oC. How much heat does the ethanol gain? Before you begin, you need to determine the specific heat capacity of ethanol using the table on page 632. cethanol= 2.46 J/g • 0C m = 100.0 g Δ T = Tfinal – Tnitial = 50.00C – 25.00C = 25.00C q = mc Δ T = (100.0 g x 2.46 J/g • 0C)(25.00C) = 6150 J Is this reaction exothermic or endothermic?

Is this reaction exothermic or endothermic? Practice Problem #2: Ice is added to 120.0 g of water. The Ti = 26.5oC and the Tf = 17.4oC. Calculate the heat lost by the water. Before you begin, you need to determine the specific heat capacity of water using the table on page 632. cwater= 4.184 J/g • 0C mwater = 120.0 g Δ T = Tfinal – Tnitial = 17.40C – 26.50C = -9.100C q = mc Δ T = (120.0 g x 4.184 J/g • 0C)(– 9.100C) = - 4568.93J Is this reaction exothermic or endothermic?

Practice Problem #3: Solve for “q” for both equations and compare the values. cwater= 4.184 J/g • 0C mA = 50.0 g mB= 100.0 g Δ T = 10.0oC qA= mc Δ T = (50.0 g x 4.184 J/g • 0C)(10.00C) = 2092 J qB =mc Δ T =(100.0 g x 4.184 J/g • 0C)(10.00C) = 4184 J

Sample Problem (p.635): Calculate the specific heat capacity of canola oil using the following information. ccanola oil= ? J/g • 0C mcanola oil = 60.0 g qcanola oil = - 4.00 X 103 J Δ T = Tfinal – Tinitial= 5.200C – 35.00C = -29.80C c = q / m Δ T = (- 4.00 X 103 J) / (60.0 g X -29.80C) = 2.24 J / goC

Practice Problem #7: How much heat is required to raise the temperature of 789.0 g of liquid ammonia from 25.0oC to 82.7oC. Before you begin, you need to determine the specific heat capacity of ammonia using the table on page 632. cammonia= 4.70 J/g • 0C m = 789.0 g Δ T = Tfinal – Tnitial = 82.70C – 25.00C = 57.70C q = mc Δ T = (789.0 g x 4.70 J/g • 0C)(– 57.70C) = 213 968.91 J or 2.14 X 105 kJ

According to the table, the substance is granite. Practice Problem #8: A solid substance has a mass of 250.0 g. It is cooled by 25.0oC and loses 4937.50 J of heat. What is its specific heat capacity? Using the table on page 632, identify the substance. cunknown= ? J/g • 0C m = 250.0 g qunknown = - 4937.50 J Δ T = - 25.00C c = q / m Δ T = (- 4937.50J) / (250.0 g X -25.00C) = 0.79 J / goC According to the table, the substance is granite.

Assume all heat released by the metal us absorbed by the water. Practice Problem #9: A piece of metal with a mass of 14.9 g is heated to 98.0oC. When the metal is placed in 75.0 g of water at 20.0oC, the temperature of the water rises by 28.5oC. What is the specific heat capacity of the metal? Assume all heat released by the metal us absorbed by the water. cmetal= ? J/g • 0C mmetal = 14.9 g Δ Tmetal = Tfinal – Tinitial = 48.50C – 98.00C = -49.50C cwater= 4.184 J/g • 0C Tf metal = Tf water Tf water = ΔT + Ti = 20oC + 28.5oC mwater = 75.0 g Δ Twater = 28.50C

- qmetal= qwater - mc∆T = mc∆T - [(14.9 g)(c)(-49.5oC)] = [(75.0 g)(4.184 J/g • oC)(28.50C) c = 8943.3 J / 737.55 g • oC c = 12.13 J / g • oC

Heat Capacity C = mc Where: C = heat capacity (KJ or J/oC) m = mass (in g or kg) c = specific heat capacity (KJ/KgoC or J/goC) Recall that: q = mcΔT q = C ΔT

Practice Problem #11: A bathtub contains 100.0 kg of water. a) What is the heat capacity, C, of the water in the bathtub? cwater= 4.184 J/g • 0C or 4.184 kJ/kg • 0C m = 100.0 g C = mc = (100 g)(4.184 kJ/kg • 0C) = 418.4 KJ/oC b) q = mcΔT = (418.4 KJ/oC)(20oC – 60oC) = - 16736 KJ or -1.67 X 10-4 KJ c) Method 2: Where: m = 100 kg = 100000 g c = 4.184 J/g • 0C ΔT = Tf –Ti = 20oC – 60oC = - 40oC q = mcΔT = - 163736000 J = -16736 KJ