# Thermochemistry. Energy. Heat. Energy is the capacity to do work. Forms of energy: The SI unit for energy is J (joule).

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1 Thermochemistry Study of heat change in chemical reactions Chapter 6 Energy Energy is the capacity to do work. Forms of energy: kinetic energy (energy of motion) potential energy (stored energy energy of an object due to its position, condition, or composition). The SI unit for energy is J (joule). Heat Heat is a form of energy associated with the motion of atoms or molecules. Heat is capable of being transmitted through solid & fluid media by conduction through fluid media by convection through empty space by radiation. 1

2 Thermodynamic Concepts System Substances involved in the chemical & physical changes that we are studying Surroundings Everything in the system s environment Universe System plus its surroundings Exothermic/Endothermic Exothermic processes give off heat burning of natural gas Endothermic processes absorb heat melting of ice Thermodynamic State of the System Set of conditions that completely describe the properties of the system Temperature Pressure Composition (what and how many moles) Physical state 2

3 State Properties Properties of the system such as P, V, T are called State Functions The value of the state function depends only on the state of the system and not the way in which the system came to be in that state A change in a state function describes the difference between the 2 states Independent of the pathway of the change First Law of Thermodynamics The total amount of energy in the universe is constant Energy can be converted to one form or another, but can not be created or destroyed d Energy is neither created or destroyed in ordinary chemical reactions and physical changes Changes in Internal Energy of a System From a Chemical Reaction ΔE = heat + work = q + w = heat absorbed by the system + work done on the system The first law of thermodynamics says that energy can not be created or destroyed, so it has to go somewhere either heat or work 3

4 Sign Convention q is positive Heat is absorbed by the system from its surroundings (endothermic) q is negative Heat is released by the system to the surrounding (exothermic) w is positive Work is done on the system by the surroundings w is negative Work is done by the system on the surroundings 1 st Law of Thermodynamics Whenever energy is added or removed from a system either as heat or work the energy of the surroundings changes by that same amount Another way of expressing the 1 st Law of Thermodynamics The total amount of energy in the universe is constant Pressure-Volume Work The most common type of work involved in chemical reactions and physical changes is pressure-volume work Pressure is Force/unit area ( distance squared) - d 2 Volume is distance cubed d 3 WorFd3kw Fd= = d2= 4

5 Pressure-Volume Work In a chemical reaction, when a gas is produced at constant pressure The gas does work as it expands against the pressure In a chemical reaction, when a gas is consumed the atmosphere (pressure) does work on the reacting system Pressure-Volume Work Work done on or by the system depends upon the external pressure and the volume Work done by the system therefore e e negative At constant pressure (such as atmospheric pressure) Work = - P ΔV = - P (V Prods V Rxts ) Units for PV work would be L-atm 1L-atm = J Pressure-Volume Work Calculate the amount of work done if a gas expands from 1.0 L to 10.0 L, when external pressure is 1.0 atm (at constant temperature). W = - P V = - P (V final V initial ) 1 L-atm = J -912 J 5

6 Pressure-Volume Work ΔE = q + w ΔE = q - PΔV (at constant pressure) When the change in volume is zero (equal number of moles of gases are produced and consumed) ΔE = q The amount of heat absorbed or released by the system Pressure Volume Work ΔE = q - PΔV (at constant pressure) Ideal gas Law says PΔV = ΔnRT R = J/K-mol T = K ΔE = q - ΔnRT (at constant pressure) Δn = no. of moles of gaseous products - no. of moles of gaseous reactants Pressure Volume Work (PΔV) When the number of moles of gaseous products is greater than the number of moles of gaseous reactants (Δn is positive), w is negative and work is done by the system on the surroundings When the number of moles of gaseous products is less than the number of moles of gaseous reactants (Δn is negative), w is positive and work is done on the system by the surroundings 6

7 Enthalpy H Enthalpy (H) of a system is defined as: H = E + PV Where E is the internal energy of the system and PV is Pressure-Volume System t Heat Enthalpy has energy units J We can t measure the absolute value of H directly, but we can measure a change in enthalpy (ΔH) ΔH = ΔE + P Δ V (at constant T,P) Change in Enthalpy ΔH Enthalpy change ΔH The amount of heat transferred into or out of a system as it undergoes a chemical or physical change at constant pressure ΔH = H final - H initial ΔH = H products - H reactants Because ΔH is only dependent upon the initial and final state of the system, it is state function ΔH By Convention Endothermic processes have a positive ΔH (ΔH > 0) - they absorb b heat Exothermic processes have a negative ΔH (ΔH < 0) - they release heat 7

8 Thermochemical Equations C 2 H 5 OH (l) + 3O 2(g) 2CO 2(g) + 3H 2 O (l) kj 1367 kj per mole of reaction 1367 kj per mole of C 2 H 5 OH 1367 kj per 3 mole of O 2 C 2 H 5 OH (l) + 3O 2(g) 2CO 2(g) + 3H 2 O (l) ΔH = kj Thermochemical Equations CH 3 OCH 3(l) + 3 O 2(g) 2 CO 2(g) + 3 H 2 O (l) When 2.61 g of CH 3 OCH 3 is burned at constant pressure, 82.5 kj of heat is evolved. Calculate the ΔH for the reaction. Find out how much heat for one mole of CH 3 OCH 3. FW = 46.0 g/mole Add the negative sign, since it is exothermic Thermochemical Equations 4 Al (s) + 3 O 2(g) 2 Al 2 O 3(s) ΔH = kj/rxn mol How much heat is evolved if 24.2 g of Al is used? 8

9 Thermochemical Equations 4 Al (s) + 3 O 2(g) 2 Al 2 O 3(s) ΔH = kj/rxn mol How much heat is evolved if 24.2 g of Al is used? 24.2 g Al 1 mol Al 1 rxn mol kj g Al 4 mol Al 1 rxn mol = 752 kj Standard States Thermodynamic Standard State The most stable, pure form under standard pressure (1 atm) and at some specified temperature (25 C unless otherwise stated) O 2 (g) gaseous diatomic oxygen Na (s) solid sodium metal C (s) graphite solid H 2 O (l) liquid water Standard State Conventions 1. For a pure substance in the liquid or solid phase, the standard state is the pure liquid or solid 2. For a gas, the standard state is the gas at a pressure of one atmosphere; In a mixture the partial pressure must be 1 atmosphere 3. For a substance in solution, the standard state refers to 1 M concentration 9

10 Standard Enthalpy Change ΔH rxn Change in enthalpy (ΔH) when a specified number of moles of reactants, t all at standard d states, t are converted completely to the specified number of moles of products, all at standard states ΔH f Standard molar enthalpy of formation ΔH for the reaction in which one mole substance in a specified state is formed from its elements in their standard states By convention, ΔH f value of any element in its standard state is zero Thermodynamic Formation Reactions H 2(g) + ½ O 2(g) H 2 O (l) ΔH o f = KJ/mol Cu (s) + ½ O 2(g) CuO (s) ΔH o f = -157 KJ/mol 3C (graphite) + 4H 2(g) C 3 H 8(g) ΔH o f = KJ/mol Hg (l) + Cl 2(g) HgCl 2(s) ΔH o f = -224 KJ/mol See Table 6-4 of Text to obtain ΔH o f values 10

11 Standard Molar Enthalpies at 298 K Substance ΔH f (kj/mole) Br 2 (l) 0 Br 2 (g) C (diamond) C (graphite) 0 CH 4 (g) C 2 H 4 (g) CaCO 3 (s) C 2 H 5 OH(l) Hess Law The change of enthalpy for a chemical reaction is the same whether it occurs by one step or by any series of steps Enthalpy is State Function Independent of pathway ΔH o rxn = ΔH o a + ΔH o b + ΔH o c +. Hess s Law C (graphite) + 1/2 O 2(g) CO (g) ΔH o rxn =? C + o (graphite) O 2(g) CO 2(g) H rxn = kj/mol CO 2(g) CO (g) + 1/2 O 2(g) H o rxn = kj/mol C (graphite) + 1/2 O 2(g) CO (g) H o rxn = kj/mol Heats of reactions experimentally determined in the lab 11

12 Hess s Law Elements ΔH o rxn = Σ n ΔH o f products Σn ΔH o f products Σn ΔH o f reactants - Σ n ΔH o f reactants Energy Reactants ΔH o f rxn Products Hess s Law Application NH 4 NO 3(s) N 2 O (g) + 2H 2 O (g) ΔH o rxn =? NH 4 NO 3(s) ΔH o f = kj/mol N 2 O (g) ΔH o f = 82.0 kj/mol H 2 O (g) ΔH o f = kj/mol ΔH o rxn = kj/mol ΔH f o Hess s Law: ΔH rxn =? SiH 4(g) + 2 O 2(g) SiO 2(s) + 2H 2 O ( l ) kj/mol Use ΔH fo and Hess s law to calculate the heat of reaction. 12

13 C 2 H 4(g) + H 2 O (l) C 2 H 5 OH (l) ΔH rxn =? Known Equations: C 2 H 5 OH (l) + 3 O 2(g) 2 CO 2(g) + 3 H 2 O (l) C 2 H 4(g) + 3 O 2(g) 2 CO 2(g) + 2 H 2 O (l) ΔH = kj ΔH = kj ΔH = -44 kj Specific Heat (s) The amount of heat required to raise the temperature 1 g of a substance 1 C. The units of specific heat are J/g- C. This is an intrinsic property of a pure material. Specific Heat How much heat is required to raise liquid 10.0 g of water 10.0 C. The specific heat of liquid water is J/g- C. q= m s ΔT 418 J 13

14 Heat Capacity Heat Capacity (C) is the amount of heat required to raise the temperature of a given quantity of substance one degree Celsius C = m s m = mass s = specific heat q= C ΔT Calorimetry A calorimeter is an instrument used to measure the energy change associated with a physical or chemical change Based on observing the temperature change when a system absorbs or releases energy in the form of heat The temperature change of a known substance (usually water) of known specific heat (J/g- o C) is measured The volume of the calorimeter is constant (ΔE = q) Picture by Dr. Gary Bertrand - Calorimetry Amount of Heat released/absorbed by the reaction = Amount of heat gained/lost by the calorimeter ( constant ) plus (+) Amount of heat gained/lost by the solution This is just the first law of thermodynamics. 14

15 Calorimetry C 2 H 5 OH (l) + 2O 2(g) 2CO 2(g) + 3H 2 O (l) g of C 2 H 5 OH (MW =46.07 g/mol) Specific heat of Calorimeter (calorimeter constant) = 2.71 kj/ o C The calorimeter contains 3000 g of H 2 0. The specific heat of water is J/g- o C Starting Temperature = o C Final Temperature = o C What is H (kj/mol) for this reaction? Calorimeter Constant We add kj of heat to a calorimeter that contains g of water. The temperature of the water and the calorimeter, originally at C, increases to C. Calculate the heat capacity of the calorimeter in J/ C. The specific heat of water is J/g C. Calorimetry 50.0 ml of H 2 O at 53.0 C is add to 50.0 ml of H 2 O at 22.0 C in a coffee cup calorimeter. The final temperature of the mixture was 37.0 C. Specific Heat of water = J/g- C. How much heat is lost by hot water (J)? How much heat is gained by room temperature water (J)? What is the calorimeter constant (heat capacity of the calorimeter), J/ C? 13.9 J/ C 15

16 Calorimetry ml of 1.00 M HCl and ml of 1.05 M NaOH both at C are mixed in the coffee cup calorimeter The final temperature of the resulting mixture is C The density of resulting NaCl solution is 1.04 g/ml The specific heat of the NaCl solution is 3.89 J/g C The calorimeter heat capacity is 13.9 J/ C Calculate the molar heat of reaction ΔH = KJ/mol H H of Physical Processes Change in enthalpy ( H) can be measured for physical processes, not just chemical reactions Phase Changes Evaporation absorbs heat ( H positive) Condensation releases heat ( H negative) Melting? Freezing? Heat of Solution Dissolution of ionic solids into solutions results in heat absorption or heat release Energy must be input to overcome the lattice energy which holds the solid together Energy is released upon dissolution of ions The energy balance determines whether the overall process is endothermic or exothermic 16

17 Heat of Dilution Dilution of solutions into solutions results in heat absorption or heat release If the dissolution process is endothermic, then the dilution process will absorb more heat (get cooler) If the dissolution process is exothermic, then the dilution process will release more heat (get hotter) Heat of Dilution The dilution process for sulfuric acid (H 2 SO 4 ) is extremely exothermic Because of this, you should never add water to sulfuric acid to dilute it To dilute, always add sulfuric acid to water 17

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