1 Chap 6: Thermochemistry Bushra Javed

Contents 1.Introduction to Thermochemistry 2.Energy and it’s units 3.Understanding Heats of Reaction 2. Enthalpy and Enthalpy Changes 3. Thermochemical Equations 4. Applying Stoichiometry to Heats of Reaction 5. Measuring Heats of Reaction 6. Using Heats of Reaction; Hess’s Law 7. Standard Enthalpies of Formation 2

6 | 3 Thermodynamics The science of the relationship between heat and other forms of energy. Thermochemistry An area of thermodynamics that concerns the study of the heat absorbed or evolved by a chemical reaction.

Energy The potential or capacity to move matter. One form of energy can be converted to another form of energy: electromagnetic, mechanical, electrical, or chemical. Next, we’ll study kinetic energy, potential energy, and internal energy 4

Kinetic Energy, E K The energy associated with an object by virtue of its motion. m = mass (kg) v = velocity (m/s) 6 | 5

Energy & it’s units The SI unit of energy is the joule, J, pronounced “jewel.” The calorie is a non-SI unit of energy commonly used by chemists. It was originally defined as the amount of energy required to raise the temperature of one gram of water by one degree Celsius. The exact definition is given by the equation: 6 | 6

Kinetic Energy, E K Example 1 A person weighing 75.0 kg (165 lbs) runs a course at 1.78 m/s (4.00 mph). What is the person’s kinetic energy? 6 | 7 m = 75.0 kg v = 1.78 m/s E K = ½ mv 2 = 119kg.m 2 /s 2

Kinetic Energy KE Example 2 What is the kinetic energy of a 2100-lb car traveling at 48 miles per hour? (1 lb = kg, 1 mi = km) a)3.3 × 10 –8 J b)2.2 × 10 5 J c)3.7 × J d)1.1 × 10 6 J 8

Potential Energy, E P The energy an object has by virtue of its position in a field of force, such as gravitational, electric or magnetic field. Gravitational potential energy is given by the equation m = mass (kg) g = gravitational constant (9.80 m/s 2 ) h = height (m) 6 | 9

Internal Energy Internal Energy, U The sum of the kinetic and potential energies of the particles making up a substance. Total Energy E tot = E K + E P + U In the laboratory, assuming flasks and test tubes are at rest, both E k and E p are equal to zero. E total = U (internal energy due to molecular motion) 6 | 10

Law of Conservation of Energy Energy may be converted from one form to another, but the total quantity of energy remains constant. 11

Thermodynamic System and Surroundings Thermodynamic system: the substance or mixture of substances under study in which a change occurs surroundings. Thermodynamic surroundings: Everything outside the system is surroundings A system is separated from its surroundings by a boundary across which matter and/or energy is transferred. 12

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Heat, q The energy that flows into or out of a system because of a difference in temperature between the thermodynamic system and its surroundings. Heat flows spontaneously from a region of higher temperature to a region of lower temperature. q is defined as positive if heat is absorbed by the system (heat is added to the system) q is defined as negative if heat is evolved by a system (heat is subtracted from the system) 6 | 14

6 | 15 Heat of Reaction The value of q required to return a system to the given temperature at the completion of the reaction (at a given temperature).

Heat of Reaction Endothermic Process A chemical reaction or process in which heat is absorbed by the system (q is positive). The reaction vessel will feel cool. Exothermic Process A chemical reaction or process in which heat is evolved by the system (q is negative). The reaction vessel will feel warm. 16

Heat of Reaction In an endothermic reaction: The reaction vessel cools. Heat is absorbed. Energy is added to the system. q is positive. In an exothermic reaction: The reaction vessel warms. Heat is evolved. Energy is subtracted from the system. q is negative. 17

Enthalpy of Reaction The change in enthalpy for a reaction at a given temperature and pressure:  H = H(products) – H(reactants) Note:  means “change in.” Enthalpy change is equal to the heat of reaction at constant pressure:  H = q P 6 | 18

Enthalpy of Reaction Enthalpy, H An extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction. Extensive Property A property that depends on the amount of substance. Mass and volume are extensive properties. 6 | 19

Enthalpy change ∆H Example 3 The phrase “the heat absorbed or released by a system undergoing a physical or chemical change at constant pressure” is a)the definition of a state function. b)the change in enthalpy of the system. c)a statement of Hess’s law. d)the change in internal energy of the system. 20

Enthalpy change ∆H Example 4 If ∆ H = –31 kJ for a certain process, that process a)occurs rapidly. b)is exothermic. c)is endothermic. d)cannot occur. 21

6 | 22 The diagram illustrates the enthalpy change for the reaction 2Na(s) + 2H 2 O(l)  2NaOH(aq) + H 2 (g) The reactants are at the top. The products are at the bottom. The products have less enthalpy than the reactants, so enthalpy is evolved as heat. The signs of both q and  H are negative.

Thermochemical Equations The thermochemical equation is the chemical equation for a reaction (including phase labels) in which the equation is given a molar interpretation, and the enthalpy of reaction for these molar amounts is written directly after the equation. For the reaction of sodium metal with water, the thermochemical equation is: 2Na(s) + 2H 2 O(l)  2NaOH(aq) + H 2 (g);  H = –368.6 kJ 6 | 23

Thermochemical Equations A thermochemical equation expresses the enthalpy of the reaction next to a balanced chemical equation. 1.N 2 (g) + 3H 2 (g) → 2NH 3 (g); ΔH = –91.8kJ 2.C 6 H 12 O 6 (s) + 6O 2 (g) →CO 2 (g) + H 2 O(l); ΔH = –2803kJ 3.H 2 (g) + O 2 (g) → H 2 O (g); ΔH = –242kJ 24

Thermochemical Equation Manipulating a Thermochemical Equation When the equation is multiplied by a factor, the value of  H must be multiplied by the same factor. When a chemical equation is reversed, the sign of  H is reversed. 6 | 25

Manipulating Thermochemical Equations Example 5 When sulfur burns in air, the following reaction occurs: S 8 (s) + 8O 2 (g)  8SO 2 (g);  H = – 2.39 x 10 3 kJ 6 | 26 Write the thermochemical equation for the dissociation of one mole of sulfur dioxide into its elements.

6 | 27 S 8 (s) + 8O 2 (g)  8SO 2 (g);  H = –2.39 × 10 3 kJ We want SO 2 as a reactant, so we reverse the given reaction, changing the sign of  H: 8SO 2 (g)  S 8 (g) + 8O 2 (g) ;  H = × 10 3 kJ We want only one mole SO 2, so now we divide every coefficient and  H by 8: SO 2 (g)  1 / 8 S 8 (g) + O 2 (g) ;  H = +299 kJ

Manipulating Thermochemical Equations Example 6 You burn 15.0 g sulfur in air. How much heat evolves from this amount of sulfur? The thermochemical equation is S 8 (s) + 8O 2 (g)  8SO 2 (g);  H = × 10 3 kJ 28

Manipulating Thermochemical Equations S 8 (s) + 8O 2 (g)  8SO 2 (g);  H =  10 3 kJ Molar mass of S 8 = g 6 | 29 q = – 1.40  10 2 kJ

Manipulating Thermochemical Equations Example 7 Given: 4AlCl 3 (s) + 3O 2 (g) → 2Al 2 O 3 (s) + 6Cl 2 (g); ∆H = –529.0 kJ determine ∆H for the following thermochemical equation. Cl 2 (g) + ⅓Al 2 O 3 (s) → ⅔AlCl 3 (s) + ½O 2 (g) a) kJ b)+88.2 kJ c) kJ d) kJ 30

Applying Stoichiometry to Heats of Reaction Example 8 How much heat is evolved upon the complete oxidation of 9.41 g of aluminum at 25°C and 1 atm pressure? (∆H for Al 2 O 3 is –1676 kJ/mol.) 4Al(s) + 3O 2 (g) → 2Al 2 O 3 (s) a)146 kJ b)1169 kJ c)292 kJ d)585 kJ 31

6 | 32 Measuring Heats of Reaction We will first look at the heat needed to raise the temperature of a substance because this is the basis of our measurements of heats of reaction.

Measuring Heats of Reaction Heat Capacity, C, of a Sample of Substance The quantity of heat needed to raise the temperature of the sample of substance by one degree Celsius (or one Kelvin). Molar Heat Capacity The heat capacity for one mole of substance. Specific Heat Capacity, s (or specific heat) The quantity of heat needed to raise the temperature of one gram of substance by one degree Celsius (or one Kelvin) at constant pressure. 6 | 33

Measuring Heats of Reaction The heat required can be found by using the following equations. Using specific heat capacity: q = s x m x  t s = specific heat of the substance m = mass in grams of substance ∆ t= t final – t initial (change in temperature) 6 | 34

Specific Heat Example 9 The units for specific heat are a)J/(g · °C). b)(J · °C). c)(J · g). d)J/°C. 35

6 | 36 Example10 A piece of zinc weighing 35.8 g was heated from 20.00°C to 28.00°C. How much heat was required? The specific heat of zinc is J/(g°C). m = 35.8 g s = J/(g°C)  t = 28.00° q = m  s   t q = 111 J

Specific Heat Example 11 How much heat is gained by nickel when 29.2 g of nickel is warmed from 18.3°C to 69.6°C? The specific heat of nickel is J/(g · °C). a)2.37 × 10 2 J b)9.00 × 10 2 J c)22.7 J d)6.64 × 10 2 J 37

Hess’s Law of Heat Summation For a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps ∆H reaction = ∆H 1 + ∆H 2 + ∆H 3 38

Hess’s Law of Heat Summation Suppose we want  H for the reaction 2C(graphite) + O 2 (g)  2CO(g) It is difficult to measure directly. However, two other reactions are known: C(graphite) + O 2 (g)  CO 2 (g);  H = kJ 2CO 2 (g)  2CO(g) + O 2 (g);  H = – kJ In order for these to add to give the reaction we want, we must multiply the first reaction by 2. Note that we also multiply  H by 2. 39

Hess’s Law of Heat Summation we can add the reactions and the  H values. Cancel the species that appear on both sides. 2C(graphite) + 2O 2 (g)  2CO 2 (g);  H = kJ 2CO 2 (g)  2CO(g) + O 2 (g);  H = – kJ 2 C(graphite) + O 2 (g)  2 CO(g);  H = – kJ 2C(graphite) + O 2 (g)  2CO(g) 40

Hess’s Law of Heat Summation 41

Hess’s law Example 12 Consider the following changes: H 2 O(s) → H 2 O(l); ∆H 1 H 2 O(l) → H 2 O(g); ∆H 2 H 2 O(g) → H 2 O(s); ∆H 3 Using Hess’s law, the sum ∆H 1 + ∆H 2 + ∆H 3 is a)greater than zero. b)equal to zero. c)less than zero. d)sometimes greater than zero and sometimes less than zero. 42

Hess’s law Example 13 Given: Pb(s) + PbO 2 (s) + 2H 2 SO 4 (l) → 2PbSO 4 (s) + 2H 2 O(l); ∆H° = –509.2 kJ SO 3 (g) + H 2 O(l) → H 2 SO 4 (l); ∆H° = –130. kJ determine ∆H° for the following equation. Pb(s) + PbO 2 (s) + 2SO 3 (g) → 2PbSO 4 (s) a) –3.77 × 10 3 kJ b)3.77 × 10 3 kJ c)–639 kJ d)–521 kJ e)–769 kJ 43

Hess’s law Example 14 Given: Fe 2 O 3 (s) + 3CO(g) → 2Fe(s) + 3CO 2 (g); ∆H = –26.8 kJ FeO(s) + CO(g) → Fe(s) + CO 2 (g); ∆H = –16.5 kJ determine ∆H for the following thermochemical equation. Fe 2 O 3 (s) + CO(g) → 2FeO(s) + CO 2 (g) a)–43.3 kJ b)–10.3 kJ c)6.2 kJ d)10.3 kJ e)22.7 kJ 44

Standard Enthalpies of Formation The term standard state refers to the standard thermodynamic conditions chosen for substances when listing or comparing thermodynamic data: 1 atm pressure and the specified temperature (usually 25°C). These standard conditions are indicated with a degree sign (°). When reactants in their standard states yield products in their standard states, the enthalpy of reaction is called the standard enthalpy of reaction,  H°. (  H° is read “delta H zero.”) 6 | 45

Allotropes Elements can exist in more than one physical state Some elements exist in more than one distinct form in the same physical state. For example, carbon can exist as graphite or as diamond; oxygen can exist as O 2 or as O 3 (ozone). These different forms of an element in the same physical state are called allotropes. The reference form is the most stable form of the element (both physical state and allotrope). 6 | 46

Standard enthalpy of formation,  H f ° is the enthalpy change for the formation of one mole of the substance from its elements in their reference forms and in their standard states.  H f ° for an element in its reference and standard state is zero. For example, the standard enthalpy of formation for liquid water is the enthalpy change for the reaction H 2 (g) + 1 / 2 O 2 (g)  H 2 O(l)  H f ° = –285.8 kJ Other  H f ° values are given in Table 6.2 and Appendix C. 6 | 47

Finding heat of reaction from Standard enthalpy of formation,  H f ° Example 15 What is the heat of vaporization of methanol, CH 3 OH, at 25°C and 1 atm? Use standard enthalpies of formation (Appendix C). 6 | 48

6 | 49 We want  H° for the reaction: CH 3 OH(l)  CH 3 OH(g)  H vap = kJ

Finding Δ H from Standard Heats of Formation Example 16 All of the following have a standard enthalpy of formation value of zero at 25°C except a)CO(g). b)Fe(s). c)C(s). d)F 2 (g). e)Ne(g). 50

Finding Δ H from Standard Heats of Formation Example 17 What is ∆H° for the following phase change? NaI(s) → NaI(l) ∆H° f (kJ/mol) NaI(s) – NaI(l) – a) kJ b)– kJ c)0 kJ d)–21.35 kJ e)21.35 kJ 51

Finding Δ H from Standard Heats of Formation Example 18 What is ∆H° for the following reaction? 2C 2 H 2 (g) + 5O 2 (g) → 4CO 2 (g) + 2H 2 O(l) ∆H° f (kJ/mol) C 2 H 2 (g) CO 2 (g)–393 H 2 O(l)–285.8 a) kJ b)– kJ c) kJ d)– kJ e)–452.6 kJ 52