Chapter 17 - Thermochemistry Heat and Chemical Change

Energy and Heat Thermochemistry - concerned with heat changes that occur during chemical reactions.

Energy and Heat Energy - capacity for doing work or supplying heat weightless, odorless, tasteless if within the chemical substances - called chemical potential energy

Energy and Heat Gasoline contains a significant amount of chemical potential energy

Energy and Heat Heat - represented by “q”, is energy that transfers from one object to another, because of a temperature difference between them. only changes can be detected! flows from warmer  cooler object

Exothermic and Endothermic Processes Essentially all chemical reactions, and changes in physical state, involve either: release of heat, or absorption of heat

Exothermic and Endothermic In studying heat changes, think of defining these two parts: the system - the part of the universe on which you focus your attention the surroundings - includes everything else in the universe

Exothermic and Endothermic Together, the system and it’s surroundings constitute the universe Thermochemistry is concerned with the flow of heat from the system to it’s surroundings, and vice-versa. Figure 17.2, page 506

Law of Conservation of Energy The Law of Conservation of Energy states that in any chemical or physical process, energy is neither created nor destroyed. All the energy is accounted for as work, stored energy, or heat.

Endothermic Fig. 17.2a, p.506 - heat flowing into a system from it’s surroundings: defined as positive q has a positive value called endothermic system gains heat as the surroundings cool down

Endothermic

Exothermic Fig. 17.2b, p.506 - heat flowing out of a system into it’s surroundings: defined as negative q has a negative value called exothermic system loses heat as the surroundings heat up

Exothermic

Exothermic and Endothermic Every reaction has an energy change associated with it Exothermic reactions release energy, usually in the form of heat. Endothermic reactions absorb energy Energy is stored in bonds between atoms

Heat Capacity and Specific Heat A calorie is defined as the quantity of heat needed to raise the temperature of 1 g of pure water 1 oC.

Heat Capacity and Specific Heat Used except when referring to food a Calorie, written with a capital C, always refers to the energy in food 1 Calorie = 1 kilocalorie = 1000 cal.

Heat Capacity and Specific Heat The calorie is also related to the joule, the SI unit of heat and energy named after James Prescott Joule 4.184 J = 1 cal

Heat Capacity and Specific Heat Heat Capacity - the amount of heat needed to increase the temperature of an object exactly 1 oC

Heat Capacity and Specific Heat Specific Heat Capacity - the amount of heat it takes to raise the temperature of 1 gram of the substance by 1 oC (abbreviated “C”) often called simply “Specific Heat” Note Table 17.1, page 508

Heat Capacity and Specific Heat Water has a HUGE value, compared to other chemicals

Heat Capacity and Specific Heat For water, C = 4.18 J/(g oC), and also C = 1.00 cal/(g oC) Thus, for water: it takes a long time to heat up, and it takes a long time to cool off! Water is used as a coolant!

Heat Capacity and Specific Heat q = mass (g) x T x C heat abbreviated as “q” T = change in temperature C = Specific Heat Units are either J/(g oC) or cal/(g oC)

q = m x T x C C = q m x T Example When 435J of heat is added to 3.4 g of olive oil at 21 °C, the temperature increases to 85 °C. What is the specific heat of olive oil? q = m x T x C C = q m x T

C = q m x T C = 435J 3.4g x (85°C - 21°C) C = 2 J/g°C

Calorimetry Calorimetry - the accurate and precise measurement of heat change for chemical and physical processes. The device used to measure the absorption or release of heat in chemical or physical processes is called a Calorimeter

Calorimetry Foam cups are excellent heat insulators, and are commonly used as simple calorimeters For systems at constant pressure, the heat content is the same as a property called Enthalpy (H) of the system.

Calorimeter

Calorimetry Changes in enthalpy = H q = H These terms will be used interchangeably in this textbook.

q = H = m x C x T Calorimetry Thus, H is negative for an exothermic reaction. H is positive for an endothermic reaction.

Calorimetry Calorimetry experiments can be performed at a constant volume using a device called a “bomb calorimeter” - a closed system

Bomb Calorimeter

Calorimeter Problem 50ml of HCl (in H2O) Ti= 22.5°C

Calorimeter Problem Add 50ml of NaOH (in H2O) Ti= 22.5°C

Calorimeter Problem Chemical Reaction

Calorimeter Problem 100ml of HCl and NaOH (in H2O) Tf= 26°C

Calorimeter Problem 1ml of H2O = 1g 100ml of H2O = 100g m = 100g Ti= 22.5°C Tf= 26°C CH20= 4.18J/g°C

H = 1460J = 1.46kJ Calorimeter Problem H = m x C x T H = m x C x (Tf –Ti) = 100g x 4.18J/g°C x (26°C - 22.5°C) H = 1460J = 1.46kJ

C + O2 ® CO2 + 395 kJ Energy Reactants Products ® C + O2 395kJ C O2

In terms of bonds Breaking this bond will require energy. C O O C O Breaking this bond will require energy. C O O C Making these bonds gives you energy. In this case making the bonds gives you more energy than breaking them.

Exothermic The products are lower in energy than the reactants Releases energy

CaCO3 + 176 kJ ® CaO + CO2 CaCO3 ® CaO + CO2 Energy Reactants Products

The products are higher in energy than the reactants Absorbs energy Endothermic The products are higher in energy than the reactants Absorbs energy

An equation that includes energy is called a thermochemical equation Chemistry Happens in MOLES An equation that includes energy is called a thermochemical equation CH4+2O2®CO2+2H2O + 802.2 kJ

CH4+2O2®CO2+2H2O + 802.2kJ 1 mole of CH4 releases 802.2kJ of energy. When you make 802.2 kJ you also make 2 moles of water.

Thermochemical Equations A heat of reaction is the heat change for the equation, exactly as written The physical state of reactants and products must also be given. Standard conditions for the reaction is 101.3 kPa (1 atm.) and 25 oC

ΔH = 514 kJ CH4+2O2 ® CO2 + 2 H2O + 802.2 kJ 16.05g CH4 1 mol CH4 If 10. 3 grams of CH4 are burned completely, how much heat will be produced? 16.05g CH4 1 mol CH4 1mol CH4 802.2 kJ 10.3g CH4 ΔH = 514 kJ

CH4+2O2 ® CO2+2H2O + 802.2 kJ How many liters of O2 at STP would be required to produce 23 kJ of heat? 22.4L x 2 = 44.8L produces 802.2kJ 23kJ 802.2kJ = 0.029 0.029 x 44.8L = VO2 = 1.30 L

CH4+2O2 ® CO2+2H2O+802.2 kJ How many grams of water would be produced with 506 kJ of heat? 2 mol of H2O produced with 802.2kJ 2 mol of H2O is 36g 506kJ 802.2kJ x 36g = 22.71g of H2O

Heat of Combustion The heat from the reaction that completely burns 1 mole of a substance Note Table 17.2 page 517

Summary, so far...

Enthalpy The heat content a substance has at a given temperature and pressure Can’t be measured directly because there is no set starting point The reactants start with a heat content The products end up with a heat content So we can measure how much enthalpy changes

Enthalpy Symbol is H Change in enthalpy is DH (delta H) If heat is released, the heat content of the products is lower DH is negative (exothermic) If heat is absorbed, the heat content of the products is higher DH is positive (endothermic)

Energy Change is down DH is <0 Reactants ® Products

Energy Change is up DH is > 0 Reactants ® Products

Heat of Reaction The heat that is released or absorbed in a chemical reaction Equivalent to DH C + O2(g) ® CO2(g) + 393.5 kJ C + O2(g) ® CO2(g) DH = -393.5 kJ

Heat of Reaction In thermochemical equation, it is important to indicate the physical state H2(g) + 1/2O2 (g)® H2O(g) DH = -241.8 kJ H2(g) + 1/2O2 (g)® H2O(l) DH = -285.8 kJ

Section 17.3 Heat in Changes of State

Heats of Fusion and Solidification Molar Heat of Fusion (Hfus) - the heat absorbed by one mole of a substance in melting from a solid to a liquid

Heats of Fusion and Solidification Molar Heat of Solidification (Hsolid) - heat lost when one mole of liquid solidifies

Heats of Fusion and Solidification Heat absorbed by a melting solid is equal to heat lost when a liquid solidifies Thus, Hfus = -Hsolid Note Table 17.3, page 522 Sample Problem 17-4, page 421

It takes 6.01kJ to melt 1 mol of ice. Sample Problem How many kJs of heat are required to melt a 10.0g popsicle at 0°C? Assume the popsicle has the same molar mass and heat of fusion as water. It takes 6.01kJ to melt 1 mol of ice. There are 18g in 1 mol of ice. 10g 18g/mol x 6.01kJ/mol = 3.34 kJ

Heats of Vaporization and Condensation When liquids absorb heat at their boiling points, they become vapors. Molar Heat of Vaporization (Hvap) - the amount of heat necessary to vaporize one mole of a given liquid. Table 17.3, page 521

Heats of Vaporization and Condensation Condensation is the opposite of vaporization. Molar Heat of Condensation (Hcond) - amount of heat released when one mole of vapor condenses Hvap = - Hcond

ΔHvaporization ΔHfusion

Heats of Vaporization and Condensation The large values for Hvap and Hcond are the reason hot vapors such as steam is very dangerous You can receive a scalding burn from steam when the heat of condensation is released!

Heats of Vaporization and Condensation H20(l)  H20(g) m = 63.7g of liquid Hvap = ? 63.7g 18g/mol x 40.7kJ/mol = 144kJ

Heat of Solution Heat changes can also occur when a solute dissolves in a solvent. Molar Heat of Solution (Hsoln) - heat change caused by dissolution of one mole of substance Sodium hydroxide provides a good example of an exothermic molar heat of solution:

H2O(l) NaOH(s)  Na1+(aq) + OH1-(aq) Hsoln = - 445.1 kJ/mol The heat is released as the ions separate and interact with water, releasing 445.1 kJ of heat as Hsoln thus becoming so hot it steams! Sample Problem 17-7, page 526

Section 17.4 Calculating Heat Changes

Hess’s Law If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction. Called Hess’s law of heat summation Example shown on page 529 for graphite and diamonds

 Diamond to Graphite C(s,diamond) C(s,graphite) C(s,diamond) + O2  CO2 ΔH=-393.5kJ C(s,graphite) + O2  CO2 ΔH=-395.4kJ CO2  C(s,graphite) + O2 ΔH=395.45kJ

+ C(s,diamond) + O2  CO2 ΔH=-393.5kJ CO2  C(s,graphite) + O2 ΔH=395.45kJ + C(s,diam)  C(s,graph) ΔH=-1.9kJ

If you turn an equation around, you change the sign: Why Does It Work? If you turn an equation around, you change the sign: H2(g)+1/2 O2(g)®H2O(g) DH=-285.5 kJ H2O(g)®H2(g)+1/2 O2(g)DH =+285.5 kJ If you multiply the equation by a number, you multiply the heat by that number: 2H2O(g)®2H2(g)+O2(g) DH =+571.0 kJ

Why does it work? You make the products, so you need their heats of formation You “unmake” the products so you have to subtract their heats. How do you get good at this?

Standard Heats of Formation The DH for a reaction that produces 1 mol of a compound from its elements at standard conditions Standard conditions: 25°C and 1 atm. Symbol is ΔHf

Standard Heats of Formation The standard heat of formation of an element = 0 This includes the diatomics O2 N2 Cl2 etc……

What good are they? Table 17.4, page 530 has standard heats of formation The heat of a reaction can be calculated by: subtracting the heats of formation of the reactants from the products DHo = ( Products) - ( Reactants)

CH4 (g) = - 74.86 kJ/mol O2(g) = 0 kJ/mol CO2(g) = - 393.5 kJ/mol CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g) CH4 (g) = - 74.86 kJ/mol O2(g) = 0 kJ/mol CO2(g) = - 393.5 kJ/mol H2O(g) = - 241.8 kJ/mol DH= [-393.5+2(-241.8)]-[-74.68+2 (0)] DH= - 802.4 kJ

Examples Sample Problem 17-8, page 531