Warmup: (3 min)Warmup: (3 min)

1.1. What is the difference between What is the difference between temperature and heat?temperature and heat?

2.2. Do you think heat flows from cold to Do you think heat flows from cold to hot or from hot to cold? Give an hot or from hot to cold? Give an example.example.

Energy and Energy and EnthalpyEnthalpy

The First Law of The First Law of ThermodynamicsThermodynamics

states that energy:states that energy:

can NOT be created or can NOT be created or destroyeddestroyed

can be converted from can be converted from one type (potential, one type (potential, kinetic, thermal) to the kinetic, thermal) to the otherother

can be exchanged can be exchanged between the system between the system and the surroundingsand the surroundings

EnergyEnergy (E): ability to do work or produce heat (E): ability to do work or produce heat

In chemistry, we talk about work and energy In chemistry, we talk about work and energy in terms of rearranging atoms and in terms of rearranging atoms and molecules by forming and breaking bonds.molecules by forming and breaking bonds.

Heat Heat (q):(q):

thermal energythermal energy

“COOL” IS NOT A

THING. IT IS AN ABSENCE

OF HEAT

HOTHOTtoto

COLDCOLD

Heat flowsHeat flowsfromfrom

Temperature:Temperature:-measures the -measures the amount of thermal amount of thermal energy (heat) in a energy (heat) in a systemsystem-depends on how fast -depends on how fast the molecules are the molecules are moving (KE)moving (KE)

TTKK = T = T°°CC + 273 + 273

T°C = (TF – 32)(5)9

average

KE of a

sample of

molecule

s

are different.

Energy UnitsEnergy Units

1.1. the Joule(J) or kJthe Joule(J) or kJ

2.2. the the calorie (cal) (calorie (cal) (amount amount of energy needed to of energy needed to raise the temperature of raise the temperature of 1 gram of water by 1 °C)1 gram of water by 1 °C)

This is a Calorie

(kcal) NOT a calorie!

One serving of yogurt contains 6.01 x 10One serving of yogurt contains 6.01 x 1044 calories. Convert this to kiloJoules of energy.calories. Convert this to kiloJoules of energy.

1 kJ = 1000 J1 kJ = 1000 J 1 cal = 4.184 J1 cal = 4.184 J1000 cal = 1 Cal1000 cal = 1 Cal

6.01 x 104 cal

(1 calorie)

(4.184 J)= 251 kJ

(1000 J)

(1 kJ)

How many Joules is 26000 Calories?How many Joules is 26000 Calories?

1 kJ = 1000 J1 kJ = 1000 J 1 cal = 4.184 J1 cal = 4.184 J

1000 cal = 1 Cal1000 cal = 1 Cal

26000 Cal

(1 Cal)

(1000 cal)= 1.1 x 108 J(4.184 J)

(1 cal)

Enthalpy (Enthalpy (∆∆H)H)

Potential energyPotential energy is stored in bonds is stored in bonds In chemical reactions, reactant bonds In chemical reactions, reactant bonds

are broken (energy absorbed) and are broken (energy absorbed) and product bonds form (energy released)product bonds form (energy released)

∆∆H: H: netnet amount of heat released or amount of heat released or absorbed; depends on the strength of absorbed; depends on the strength of bonds made/brokenbonds made/broken

Cold packs Cold packs (read only)(read only)

In cold packs, the chemical ammonium In cold packs, the chemical ammonium nitrate is often used because it absorbs a nitrate is often used because it absorbs a

lot of heat when it dissolves in water.  Ammonium lot of heat when it dissolves in water.  Ammonium nitrate dissolves in water “endothermically.”nitrate dissolves in water “endothermically.”

Water and ammonium nitrate are kept in separate Water and ammonium nitrate are kept in separate compartments in the pack until the pack is needed.  compartments in the pack until the pack is needed.  Then the chambers are broken and the ammonium Then the chambers are broken and the ammonium nitrate dissolves in the water, absorbing heat and nitrate dissolves in the water, absorbing heat and making the pack as cold as 0making the pack as cold as 0°°C. C.

19.2 kJ + NH19.2 kJ + NH44NONO33(s) → NH(s) → NH44++(aq) + NO(aq) + NO33

--(aq)(aq)

reactantreactant

Hot packs Hot packs (read only)(read only)CaClCaCl22(s)(s) → Ca → Ca2+2+(aq) (aq) + 2Cl+ 2Cl--(aq)(aq) +13.9kJ +13.9kJ

In hot packs, calcium chloride or In hot packs, calcium chloride or magnesium sulfate frequently are used magnesium sulfate frequently are used because these chemicals dissolve in because these chemicals dissolve in water exothermically. water exothermically. 

In other words, they release a lot of heat In other words, they release a lot of heat when they are dissolved in water.  Hot when they are dissolved in water.  Hot packs can reach temperatures around packs can reach temperatures around 9090°°C.C.

productproduct

Exothermic Rxns

Endothermic Rxns

Heat flows from:

system surroundings

surroundings system

ΔH - +

Surroundings feel:

HOT COLD

because: Absorbed energy

released by reaction

System stole energy to break

bonds

Xe(g) + 2FXe(g) + 2F22(g) + 251 kJ --> XeF(g) + 251 kJ --> XeF44(s) (s)

endo or exo? endo or exo? ∆∆H = H =

XeFXeF44(s) --> Xe(g) + 2F(s) --> Xe(g) + 2F22(g) + 251 kJ (g) + 251 kJ

endo or exo? endo or exo? ∆∆H = H =

- 251 kJ- 251 kJ

WWhen a reaction is reversedhen a reaction is reversed, ,

the the sign of sign of ∆∆HH is reversed too is reversed too

xenon xenon tetrafluoridetetrafluoride

+251 kJ+251 kJ

Identify each as an Identify each as an exothermic or endothermic exothermic or endothermic reaction:reaction:*write just the *write just the answeranswer

1)1) When solid KBr is dissolved in water, When solid KBr is dissolved in water, the liquid gets colderthe liquid gets colder

2)2) 2KClO2KClO33(s) + 84.9 kJ (s) + 84.9 kJ 2KCl(s) + 3O 2KCl(s) + 3O22(g)(g)

3)3) When concentrated sulfuric acid is When concentrated sulfuric acid is added to water, the solution gets very added to water, the solution gets very hothot

4)4) The net enthalpy change for a reaction The net enthalpy change for a reaction is negative 768kJ (is negative 768kJ (ΔΔH = - 768 kJ)H = - 768 kJ)

Enthalpy and StoichiometryEnthalpy and StoichiometryEx. 2CEx. 2C22HH44O(l)+ 2HO(l)+ 2H22O(l)O(l)2C2C22HH66O(l) +OO(l) +O22(g) ∆H = (g) ∆H =

915.7 kJ915.7 kJ

a.a. Considering the reaction above, how much Considering the reaction above, how much energy is required to produce 0.467 moles of energy is required to produce 0.467 moles of oxygen gas?oxygen gas?

0.467 moles O0.467 moles O22

b. If I want to completely burn 3.4 moles liquid b. If I want to completely burn 3.4 moles liquid

CC22HH44O, how much energy must be present?O, how much energy must be present?

3.4 moles 3.4 moles CC22HH44OO

(915.7 kJ)(915.7 kJ)

(1 mole (1 mole OO22 ) )

= 428 kJ= 428 kJ

(915.7 kJ)(915.7 kJ)

(2 mole C(2 mole C22HH44O)O)= 1600 kJ= 1600 kJ

Energy must be added or absorbed to break bonds and Energy must be added or absorbed to break bonds and that energy is released when bonds are formed. You can that energy is released when bonds are formed. You can calculate the total enthalpy of the reaction using the calculate the total enthalpy of the reaction using the following formula:following formula:

ΔΔHH = bonds broken - bonds formed = bonds broken - bonds formed

Calculate the change in energy that accompanies the Calculate the change in energy that accompanies the reaction:reaction:

HH2(g)2(g) + F + F2(g) 2(g) 2HF 2HF(g)(g)

Bond type Bond type Bond energy Bond energy (kJ/mol )(kJ/mol )H—H H—H 432432

F—F F—F 154154H—F H—F 565565

ΔH = {breaking H-H + F-F} – (forming 2 H-F bonds)= [1(432) + 1(154)] - [2(565)]= [1(432) + 1(154)] - [2(565)]

ΔH = = -544 kJ-544 kJMore energy is released by forming the new bonds More energy is released by forming the new bonds

than was required to break the old bondsthan was required to break the old bonds

+ 544 kJ

22HH22O O 22HH22 + + 11OO22

brokenbroken formedformed22(two O-H bonds) – {(two O-H bonds) – {22(one H-H bond) and (one H-H bond) and 11(one O=O (one O=O bond)}bond)}

22(2 x 464kJ) - {(2 x 464kJ) - {22(436kJ)(436kJ)++ 11(498kJ)}(498kJ)}

∆H = 1856 kJ 1856 kJ - - 1370 kJ1370 kJ= + 486 kJ= + 486 kJ

BondBond

Energy(kJ/mol)

BondBond

Energy (kJ/mol)

BondBond

Energy(kJ/mol)

BondBond

Energy(kJ/mol)

C and H 414 O and O 142 Ca and Cl 398 H and H 436

H and Cl

431 O = O 498 Ca and O 452 Cl and Cl 243

H and N 389 O and C 360 Na and Cl 310 N and Cl 425

H and O 464 O = C 736 K and Cl 437 C and C 512

+ 486 kJ+ 486 kJ