1 Thermochemistry -Energy of Chemical Reactions -OR- -The study of heat changes that occur during chemical reactions and physical changes of state

2 Heat Transfer and Changes of State Changes of state involve energy Ice  Water 333 J/g Heat of Fusion (Heat of Fusion) Water  Vapor 2260 J/g (Heat of vaporization)

3 Heating/Cooling Curve for Water Heat water Evaporate water Melt ice

4 Energy and Chemistry ENERGY is the capacity to do work or transfer heat. can be: light, electrical, kinetic, potential, chemical HEAT (represented by q) is the form of energy that flows between 2 samples because of a difference in temperature – will always go from a warmer object to a cooler one. WORK is the form of energy that results in a macroscopic displacement of matter such as gas expansion or motion of an object (force x distance) LAW OF CONSERVATION OF ENERGY energy is neither created nor destroyed

5 Energy and Chemistry CALORIE quantity of heat that raises the temperature of 1g of pure water 1°C 1 Cal or 1 kcal = 1000 cal JOULE SI unit of heat and energy (1kcal = 4186J, 1 cal = J and 1J = cal) HEAT CAPACITY the amount of heat it takes to change an object’s temperature by exactly 1°C Depends on an object’s mass Ex. A cup of water has a greater heat capacity than a drop of water.

CHEMICAL ENERGY Chemical Potential Energy: Energy stored in chemicals because of their compositions Chemical bonds are a source of energy BOND BREAKING - requires energy BOND MAKING - releases energy In a chemical reaction : if more energy is released in forming bonds than is used in breaking bonds then EXOTHERMIC... reaction is EXOTHERMIC if more energy is used in breaking bonds than is released in forming bonds then... reaction is ENDOTHERMIC Energy is released as HEAT, LIGHT, SOUND, WORK Energy can be provided by - LIGHT - photochemistry - WORK - electrochemistry - COOLING of surroundings

7 Specific Heat Capacity 1°C (measured in J/(g x °C) A.K.A. Specific Heat: (represented by C) the amount of heat it takes to raise temperature of 1g of a substance 1°C (measured in J/(g x °C) A difference in temperature leads to energy transfer. Specific heat capacity= heat lost or gained by substance (J) (mass, g) (T change, K)

8 Specific Heat Capacity SubstanceSpec. Heat (J/gK) H 2 O4.184 Al0.902 glass0.84 Aluminum Water

Specific Heat q = mC  T q = heat (j) m = mass (g) C = specific heat (j/g  C)  T = change in temperature = Tf-Ti (  C)

10 Specific Heat Capacity - an example If 25.0 g of Al cool from 310 K to 37 K, how many joules of heat energy are lost by the Al? where  T = T final - T initial = = -273 K q = (0.902 J/gK)(25.0 g)(-273 K) q = J negative sign of q  heat is “lost by” or transferred from Al J/g.K =  heat gain/lost = q = (specific heat)(mass)(  T)

Specific Heat A 10.0 g sample of iron changes temperature from 25.0  C to 50.4  C while releasing 114 joules of heat. Calculate the specific heat of iron.

Example q = mc  T 114 = (10.0)c( ) c = j/g  C

Another example If the temperature of 34.4 g of ethanol increases from 25.0  C to 78.8  C how much heat will be absorbed if the specific heat of the ethanol is 2.44 j/g  C

Another example q = mc  T q = (34.4)(2.44)( ) q = 4.53 x j

Yet another example 4.50 g of a gold nugget absorbs 276 J of heat. What is the final temperature of the gold if the initial temperature was 25.0  C & the specific heat of the gold is 0.129j/g  C

Yet another example q = mc  T 276 = (4.50)(0.129)  T  T = 475  C  T = Tf-Ti 475 = Tf-25 Tf = 500  C

17  H = H final - H initial If H final > H initial then  H is positive Process is ENDOTHERMIC If H final > H initial then  H is positive Process is ENDOTHERMIC If H final < H initial then  H is negative Process is EXOTHERMIC If H final < H initial then  H is negative Process is EXOTHERMICENTHALPYENTHALPY For systems at constant pressure, the heat content is the same as the Enthalpy, or H, of the system. Heat changes for reactions carried out at constant pressure are the same as changes in enthalpy (∆H)

18 Standard Enthalpy Values Most  H values are labeled  H o P = 1 atmosphere ( = 760 torr = kPa) Concentration = 1 mol/L T = usually 25 o C with all species in standard states e.g., C = graphite and O 2 = gas o means measured under standard conditions

19 Endo- and Exothermic Surroundings Heat System ENDOTHERMICENDOTHERMIC Heat Surroundings EXOTHERMIC

20 But the reverse reaction, the decomposition of water : H 2 O(g) kJ ---> H 2 (g) + 1/2 O 2 (g) Endothermic reaction — heat is a “reactant”,  H = +242 kJ. This does not occur spontaneously. Consider the combustion of H 2 to form water.. H 2 (g) + 1/2 O 2 (g) ---> H 2 O(g)  242 kJ Exothermic reaction — heat is a “product”.  H = -242 kJ. This is spontaneous and proceeds readily once initiated. USING ENTHALPY

21 liquid Making H 2 from liquid H 2 O involves two steps. H 2 O(liq) + 44 kJ  H 2 O(g) H 2 O(g) kJ  H 2 (g) + 1/2 O 2 (g) H 2 O(liq) kJ  H 2 (g) + 1/2 O 2 (g) This is an example of HESS’S LAW — If a reaction is the sum of 2 or more others, the net  H is the sum of the  H’s of the other rxns. If a reaction is the sum of 2 or more others, the net  H is the sum of the  H’s of the other rxns.

Heat and Changes of State Heat of combustion (∆H)= the heat of reaction for the complete burning of one mole of a substance Molar heat of fusion (∆H fus )= the heat absorbed by one mole of a substance in melting from a solid to a liquid at a constant temperature Molar heat of solidification (∆H solid )= heat lost when one mole of a liquid freezes to a solid at a constant temperature (equal to the negative heat of fusion) Molar heat of vaporization (∆H vap )= the heat absorbed by one mole of a substance in vaporizing from liquid to a gas Molar heat of condensation (∆H cond )= heat released by one mole of a vapor as it condenses

Example (Heat of combustion) The standard heat of combustion (∆H° rxn ) for glucose (C 6 H 12 O 6 ) is kJ/mol. If you eat and burn 70.g of glucose in one day, how much energy are you getting from the glucose? –Step one: convert g of glucose to moles 70. g glucose 1 mol = 0.28 mol glucose 246 g –Step two: Use (∆H° rxn ) to find amount of kJ gained 0.28 mol glucose x 2808 kJ = 790 kJ gained (+ b/c gained not lost) 1 1 mol

Example(Other ∆H’s) You have a sample of H 2 O with a mass 23.0 grams at a temperature of °C. ΔHfus= 6.01 kJ/mol ΔHvap= 40.7 kJ/mol How many kilojoules of heat energy are necessary to carry out each step? Heat the ice to 0 °C? –Which equation do you need? q = mCΔT q = (23g)(4.184 J/g °C)(0 °C – ( - 46) °C) q = 4427 J = 4.43 kJ Boil the water? –Which equation do you need? ΔH = mol x ΔHvap ΔH = 23 g x 1 mol x 40.7kJ 18.02g 1 mol ΔH = 52 kJ