u P 4 + N 2 O  Warm up P 4 O 6 + N 2 u Balance the equation.

u What is the Limiting/Excess reactant for 12 mol P 4 and 14 mole N 2 O

Warm up u True or false: all chemical reactions release energy.

Thermochemistry Heat and Chemical Change

Section 11.1

The Flow of Energy - Heat u

OBJECTIVES:

• Explain the relationship between

energy

and

heat

.

4

Section 11.1

The Flow of Energy - Heat u

OBJECTIVES:

• Distinguish between heat capacity and specific heat .

5

Energy and Heat u Thermochemistry - concerned with heat changes that occur during chemical reactions u Energy - capacity for doing work or supplying heat 6

Energy and Heat u Heat represented by “q”, is energy that transfers from one object to another, because of a temperature difference between them.

7

Exothermic and Endothermic Processes u In studying heat changes, think of defining these two parts: • the system • the surroundings

Exothermic and Endothermic Processes u 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.

Exothermic and Endothermic Processes u Heat flowing into surroundings: a system from it’s • defined as positive • q has a positive value • called endothermic – system

gains heat

as the surroundings cool down

Exothermic and Endothermic Processes u 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

u GUMMY BEAR SACRIFICE

Exothemic and Endothermic u Every reaction has an energy change associated with it u Exothermic reactions release energy, usually in the form of heat.

u Endothermic energy reactions absorb u Energy is stored in bonds between atoms 14

Warm up u Define exothermic and give one CHEMICAL reaction example.

u Define endothermic and give one CHEMICAL reaction example.

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

• 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.

16

Heat Capacity and Specific Heat u Joule- • the SI unit of heat and energy 4.184 J = 1 cal u

Specific Heat Capacity

- the amount of heat it takes to raise the temperature of

1 gram

of the substance by 1 o C (abbreviated “C”) 17

Heat Capacity and Specific Heat u For water, C = 4.18 J/(g o C), and also C = 1.00 cal/(g o C) u Thus, for water: • it takes a long time to heat up, and • it takes a long time to cool off!

u Water is used as a coolant!

18

Heat Capacity and Specific Heat u To calculate, use the formula: u q = mass (g) x  T x C u heat abbreviated as “q” u  T = change in temperature u C = Specific Heat u Units are either J/(g o C) or cal/(g o C) 20

q = mass (g) x  T x C u You have 250ml of water at 25 o C . In order to start your experiment you need the water to be exactly the same as a human core temperature, 37 o C . The specific heat of water is 1cal/(g o C) .

u How much heat energy(in calories) do you need to add to your sample?

u q = mass (g) x  T x C You are the scientist assigned to explore an asteroid, caught by NASA, to look for exotic materials. You have decided to use specific heat to identify your sample.

u Mass = 100g u q = 100 cal u  T = 0.00 to 32.46 o C u What is the rock of the asteroid made of?

q = mass (g) x  T x C u A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from 25 °C to 175 °C . Calculate the specific heat capacity of iron.

Pg. T18

Warm up u Imagine how the graph of an endothermic and an exothermic reaction would look.

u In your lab journal, make a line graph for each with energy on the y-axis and time on the x-axis.

Section 11.2

Measuring and Expressing Heat Changes u

OBJECTIVES:

• Construct equations that show the heat changes for chemical and physical processes.

25

Section 11.2

Measuring and Expressing Heat Changes u

OBJECTIVES:

• Calculate heat changes in chemical and physical processes.

26

Calorimetry u Calorimetry - the accurate and precise measurement of heat change for chemical and physical processes.

Calorimetry u For systems at constant pressure, the heat content is the same as a property called system Enthalpy (H) of the

Calorimetry u Changes in enthalpy =  H u q =  H These terms will be used interchangeably u Thus, q =  H = m x C x  T u  H is negative for an exothermic reaction u  H is positive for an endothermic reaction

C + O 2 C + O 2  CO 2 + 395 kJ Reactants  C O 2 Products 395kJ 30

C In terms of bonds O O C O O Breaking this bond will require energy.

O C O C O O Making these bonds gives you energy.

In this case making the bonds gives you more energy than breaking them.

31

Exothermic u The products are lower in energy than the reactants u Releases energy 32

CaCO 3   2 2 CaO + CO 2 176 kJ CaCO 3 Reactants  Products 33

Graph these diagrams!

2C 4 H 10 + 13O 2 → 8CO 2 + 10H 2 O ΔH comb = -5320 kJ N 2(g) 2H 2(g) + O 2(g) ---> 2H 2 O (g) + heat + 2 O 2(g) ----> 2 NO 2(g) ΔH = +67.6 kJ NH 3 (g) + HCl(g) ---> NH 4 Cl(s) ΔH = -176 kJ E a = 50kJ 4 NO(g) + 6 H 2 O(l) ---> 4 NH 3 (g) + 5 O 2 (g) E a = 456kJ ΔH = +1170 kJ

Graph these diagrams!

Chemistry Happens in

MOLES

u An equation that includes energy is u called a thermochemical equation CH 4 + 2O 2  CO 2 + 2H 2 O + 802.2 kJ u 1 mole of CH 4 energy.

releases 802.2 kJ of u When you make 802.2 kJ you also make 2 moles of water 37

Thermochemical Equations u 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 o C 38

CH 4 + 2 O 2



CO 2 + 2 H 2 O + 802.2 kJ

u If 10. 3 grams of CH 4 are burned completely, how much heat will be produced?

10. 3 g CH 4 1 mol CH 4 16.05 g CH 4 802.2 kJ 1 mol CH 4 = 514 kJ

39

CH 4 + 2 O 2



CO 2 + 2 H 2 O + 802.2 kJ

u How many liters of O 2 at STP would be required to produce 23 kJ of heat?

u How many grams of water would be produced with 506 kJ of heat?

40

Summary, so far...

u Enthalpy The heat content a substance has at a given temperature and pressure u Can’t be measured directly because there is no set starting point u The reactants

start

with a heat content u The products

end up

content with a heat u So we can measure how much enthalpy changes 42

Enthalpy u Symbol is H u Change in enthalpy is  H (delta H) u If heat is

released

, the heat content of the products is

lower

 H is negative (exothermic) u If heat is

absorbed

, the heat content of the products is

higher

 H is positive (endothermic) 43

Change is down  H is <0 Reactants  Products 44

Change is up  H is > 0 Reactants  Products 45

Heat of Reaction u The heat that is released or absorbed in a chemical reaction u u u Equivalent to  H C + O 2 (g)  C + O 2 (g)  CO CO 2 2 (g) + 393.5 kJ (g)  H = -393.5 kJ In thermochemical equation, it is important u u u to indicate the physical state H 2 (g) + 1/2O 2 (g)  H 2 (g) + 1/2O 2 (g)  H H 2 2 O(g) O(l)   H = -241.8 kJ H = -285.8 kJ 46

Heat of Combustion u The heat from the reaction that completely burns 1 mole of a substance u Note Table 11.4, page 305 47

Section 11.3

Heat in Changes of State u

OBJECTIVES:

• Classify , by type, the heat changes that occur during melting, freezing, boiling, and condensing.

48

Section 11.3

Heat in Changes of State u

OBJECTIVES:

• Calculate heat changes that occur during melting, freezing, boiling, and condensing.

49

Heats of Fusion and Solidification u Molar Heat of Fusion (  H fus ) - the heat absorbed by one mole of a substance in

melting

from a solid to a liquid u Molar Heat of Solidification (  H solid ) - heat lost when one mole of liquid solidifies 50

Heats of Vaporization and Condensation u Molar Heat of Vaporization (  H vap ) the amount of heat necessary to vaporize one mole of a given liquid.

u Table 11.5, page 308 51

Heats of Vaporization and Condensation u Molar Heat of Condensation (  H cond ) - amount of heat released when one u mole of vapor condenses  H vap =  H cond 52

Heats of Vaporization and Condensation u Note Figure 11.5, page 310 u The large values for  H vap  H cond and 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!

53

Heats of Vaporization and Condensation u H 2 0 (g)  H 2 0 (l)  H cond = - 40.7kJ/mol u Sample Problem 11-5, page 311 54

Heat of Solution u Heat changes can also occur when a solute

dissolves

in a solvent.

u Molar Heat of Solution (  H soln ) heat change caused by dissolution of one mole of substance 55

Section 11.4

Calculating Heat Changes u

OBJECTIVES:

• Apply Hess’s law of heat summation to find heat changes for chemical and physical processes.

56

Section 11.4

Calculating Heat Changes u

OBJECTIVES:

• Calculate heat changes using standard heats of formation.

57

Hess’s Law u 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 u Example shown on page 314 for graphite and diamonds 58

Why Does It Work?

u If you turn an equation around, you change u the sign: If H 2 (g) + 1/2 O 2 (g)  H 2 O(g)  H=-285.5 kJ u then, H 2 O(g)  H 2 (g) + 1/2 O 2 (g)  H =+285.5 kJ u also, u If you multiply the equation by a number, u you multiply the heat by that number: 2 H 2 O(g)  H 2 (g) + O 2 (g)  H =+571.0 kJ 59

Standard Heats of Formation u The  H for a reaction that produces 1 mol of a compound from its elements at standard conditions u u Standard conditions: 25 °C and 1 atm.

Symbol is  H f 0 u

The standard heat of formation of an element = 0

u

This includes the diatomics

60

What good are they?

u Table 11.6, page 316 has standard heats of formation u The heat of a reaction can be calculated by: • subtracting the heats of formation of  H o the reactants from the products = (  H f 0 Products) - (  H f 0 Reactants) 61

Examples u CH 4 (g) + 2 O 2 (g)  H f 0  CO 2 (g) + 2 H 2 O(g)

CH 4 (g) = - 74.86 kJ/mol

 H f 0

O 2 (g) = 0 kJ/mol

u  H f 0  H f 0

CO 2 (g) = - 393.5 kJ/mol H 2 O(g) = - 241.8 kJ/mol



H= [-393.5 + 2(-241.8)] - [-74.68 +2 (0)]



H= - 802.4 kJ

62