HeatHeatHeatHeat

Solid Liquid Gas

Heat = AMOUNT of internal energy

Temperature = a MEASURE of the average molecular kinetic energy

10 g PbT = 40 °C

100 g PbT = 40 °C

Both blocks are at the same temperature.

Do they both contain the same amount of heat?

Which substance requires more heat to increase the temperature by 5 °C?

Specific heat capacity (Cp): amount of heat(q)

required to raise 1 g of substance by 1 °C

Cp(Pb) = 0.126 J/g°CCp(paraffin) = 2.1 J/g°C

Pb100 g

How much heat is required by the 100 g candle to increase the temperature by 5 °C?

Cp(paraffin) = 2.1 J/g°C q = Cp(mass)(T)

q = (2.1 J/g°C)(100 g)(5 °C)

q = 1050 J

q = Cp(mass)(T)

1050 J = (4.184 J/g°C)(100g)(T)

T = 2.5 °C

If the same amount of heat was used to heat 100 g of water [Cp(liquid water) = 4.184 J/g°C], what would be the T of the water?

For the same amount of heat and mass, T decreasesdecreases as the

specific heat of the substance increasesincreases

For the same amount of heat and mass, T decreasesdecreases as the

specific heat of the substance increasesincreases

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200 ft

100 kg Pb

If the temperature of the lead is 327°C before it hits the water, what is the final temperature of the lead after hitting the water?

If the temperature of the lead is 327°C before it hits the water, what is the final temperature of the lead after hitting the water?

Cp(Pb) = 0.13 J/g°C

Cp (H2O) = 4.18 J/g°C

q = mCpT T = Tf - Ti

Tf = 93 °C

-qPb = qH2O

-(1x105 g)(0.13 J/g°C)(Tf – 327°C) =

(1x104 g)(4.18 J/g°C)(Tf – 20°C)

10 kg H2O

Ti = 20 °C

Melting one 14-gram Al soda can requires 5.55 kJ of energy. What is its molar heat of fusion?

Melting one 14-gram Al soda can requires 5.55 kJ of energy. What is its molar heat of fusion?

105,000 cans are recycled in the US every minute.How many kJ/s are being used in recycling Al cans?105,000 cans are recycled in the US every minute.How many kJ/s are being used in recycling Al cans?

Almol 0.519g/mol 26.982 Alg 14

kJ/mol 10.7 Almol 0.519

can 1x

can 1kJ 5.55

(Al)Hfus Δ

kJ/s 9.72x10 Almole 1kJ 10.7

x can 1

Almol 0.519 x

s 60cans 1.05x10 3

5

That’s equivalent to burning 2300 food Calories/s!

18 g H2O = 1 mole H2O

- 10 °C 90 °C

Experiment: Heat two beakers containing 18 g of water at the same rate, and monitor their temperatures.

Question: Will their temperatures increase at the same rate?

0 °C 100 °C

Experiment: Heat two beakers containing 18 g of ice and water at the same rate, and monitor their temperatures.

Question: Will their temperatures increase at the same rate?

Answer: It takes twice as long to increase the temperature of the liquid water by 10 °C than it does to increase the temperature of the ice by the same amount.

Tem

pera

ture

(°

C)

0

100

Heating curve of water

solid warming

solid + liquid present

liquid warming liquid + gas present

Gas warming

Heat (kJ/s)

Tem

pera

ture

(°

C)

0

100

Heating curve of water

melting/freezing point

boiling/condensation point

Temperature is constant during phase transitions!!

All heat energy goes to changing the state of matter.

Heat (kJ/s)

Heat (kJ/s)

Tem

pera

ture

(°

C)

0

100

Heating curve of water

Hfus = the amount of heat needed to covert a solid into its liquid phase

Hfus

Hvap

(heat of fusion)

(heat of vaporization)

Hvap = the amount of heat needed to convert a liquid into its gaseous phase

Tem

pera

ture

(°

C)

0

100

Heating curve of water

Heat (kJ/s)

H2O: Hfus = 6.01 kJ/mol Hvap = 40.7 kJ/mol

Hfus = 20.2 kJ/mol Hvap = 10.3 kJ/molH2PEw:

A greater Hfus = more time to melt

And vice versa

Heating Curve Wrap Up:

•The specific heat capacity (Cp)of a substance determines the temperature change observed when heat is added or withdrawn from the substance.

•Temperature is INVARIANT during phase transitions.

•The amount of heat required to convert one mole of the substance from one phase to another is its molar enthalpy of transition (Hfus, Hvap, Hsub).

•The amount of heat given off for one mole of a substance during a phase transition while cooling is its molar enthalpy of transition (Hcond, Hsol, Hdep).

•The shape of a heating curve depends upon the heating rate, specific heat capacities of the phases involved, and the enthalpies of transition.

What is the sign for all three?

+H

What is the sign for all three?

-H

1.0 gram of solid sodium metal is added to 100 g of water. The reaction produces sodium hydroxide and hydrogen gas. Calculate the molar heat of reaction if the water’s temperature increased by 2C.

Step 1: Write out the chemical equation and balance it. Na (s)+ H2O (l) NaOH-1+1 + H2 (g)(aq)22 2

Step 2: Determine if there’s a limiting reagent.*

2H mol 1.26Na 22H 1

x g/mol 22.98977Na g 1.0

2H mol 2.77O2H 22H 1

x g/mol 18.015

O2H g 100.0

Na is the limiting reagent. Only 1.26 mol of H2 will be formed.

Na is the limiting reagent. Only 1.26 mol of H2 will be formed.

*Choose a product that has a coefficient of 1 for best results.

Step 3: Determine the amount of heat involved in the reaction.

q = mCpT

q = ? m = 100 g Cp = 4.184 g/JC T = 2 C

J 837C)C)(2J/g O)(4.1842H g (100 q

Step 4: Calculate your molar heat of reaction.

If a reaction that produced 1.26 moles of H2 also released 837 J of heat, then the molar enthalpy (heat) change for this reaction would be:

J/mol 664mol 1.26J 837

rxnΔH

A simpler problem:

How much heat is given off when 1.6 g of CH4 are burned in an excess of oxygen if Hcomb = -802 kJ/mol?

Step 1: Write the reaction equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Step 2: Calculate molar amount involved

combusted 4CH moles 0.100g/mol 16.042

4CH g 1.6

Step 3: Calculate amount of heat given off

Hrxn = (-802 kJ/mol)(0.100 mol CH4) = -80.2 kJ

Q: Is this an exothermic or endothermic reaction?

EX 3: What is the molar heat of combustion of propene (C3H6) if burning 3.2 g releases 156 kJ of heat?

Step 1: Write the reaction equation.

2 C3H6 (g) + 9 O2 (g) 6 CO2 (g) + 6 H2O (g)

• This reaction equation involves the combustion of 2 moles of C3H6 and we want to find out what it is for one mole. • Save yourself a headache and simplify future calculations by dividing the reaction equation through by 2.

C3H6 (g) + 4.5 O2 (g) 3 CO2 (g) + 3 H2O (g)

Step 2: Convert grams of propene to moles.

6363 HC mol 0.076

g/mol 42.081HC g 3.2

Step 3: Divide the heat released by moles of propene.

kJ/mol 2053HC mol 0.076

kJ 156combHΔ

63

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What have we learned?

Sometimes heat is given off during a chemical reaction.This makes it feel hotter.

Sometimes heat is absorbed during a chemical reaction.This makes it feel colder.

What causes it to be different?

• Chemical bonds contain energy.

• Add the energy of all of the bonds in the reactants together to find their total energy.

• Add the energy of all of the bonds in the products together to find their total energy.

• If the two numbers aren’t the same (and they almost never are), then there will be heat energy given off or taken in.

2 H2 + O2 2 H2O

Energy

2 H2 + O2

2 H2O

•If the products contain less energy, energy must have been given off during the reaction.

Energy barrier

2 H2 + O22 H2O

Energy

2 H2 + O2

2 H2O

•If the products contain more energy, energy must have been absorbed during the reaction.

Energy barrier

If heat energy is given off during a reaction, it is called an EXOTHERMIC REACTION.

Heat exits = exothermic

Exothermic reactions can be recognized by a temperature INCREASE.

If heat energy is absorbed during a reaction, it is called an ENDOTHERMIC REACTION.

Heat enters = endothermic

Endothermic reactions can be recognized by a temperature DECREASE.

2 AlBr3 + 3 Cl2 2 AlCl3 + 3 Br2

Energy

2 AlBr3 + 3 Cl2

2 AlCl3 + 3 Br2

Hrxn = Heat content of products – heat content reactants

Hrxn < 0 Reaction is exothermic

But how do we determine the heat content in the first

place?

Heat of formation, Hf

• The Hf of all elements in their standard state equals zero.

• The Hf of all compounds is the molar heat of reaction for synthesis of the compound from its elementsHf (AlBr3):2 Al + 3 Br2 2 AlBr3

Hrxn = 2Hf(AlBr3)

Hrxn

2Hf(AlBr3) =

• Since the Hrxn can be used to find Hf, this means that Hf can be used to find Hrxn WITHOUT having to do all of the calorimetric measurements ourselves!!

The Law of Conservation of Energy strikes again!!

Hess’s Law: Hrxn = Hf(products) – Hf(reactants)

6 CO2 (g) + 6 H2O (l) C6H12O6 (s) + 6 O2 (g)

Hrxn = [Hf(C6H12O6) + 6 Hf(O2)] – [6 Hf(CO2) + 6 Hf(H2O)]

From Hf tables: Hf(C6H12O6) = -1250 kJ/mol

Hf(CO2) = -393.5 kJ/mol Hf(H2O) = -285.8 kJ/mol

Hrxn = [-1250 kJ/mol] – [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)]

Hrxn = +2825.8 kJ/molHrxn = +2825.8 kJ/mol

Using Hess’ Law with Hrxn

What is the Hcomb for ethane?

2 C6H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2O (g)

C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g)Hrxn = -1323 kJ/mol

C2H4 (g) + H2 (g) → C2H6 (g) Hrxn = -137 kJ/mol

H2O (l) H2O (g)

Energy

H2O (l)

H2O (g)

Hvap = +40.7 kJ/mol

• Water will spontaneously evaporate at room temperature even though this process is endothermic.

• What is providing the uphill driving force?

a measure of the disorder or randomness of the

particles that make up a system

• Water will spontaneously evaporate at room temperature because it allows the disorder of the water molecules to increase.

• The entropy, S, of gases is >> than liquids or solids.

• If Sproducts > Sreactants, S is > 0

Predict the sign of S:

ClF (g) + F2 (g) ClF3 (g) S < 0

CH3OH (l) CH3OH (aq) S > 0

Are all +S reactions spontaneous?

2 H2O (l) 2 H2 (g) + O2 (g)

S is large and positive…

…but H is large and positive as well.

• Gibb’s Free Energy, G, allows us to predict the spontaneity of a reaction using H AND S.

If –G spontaneous reaction

2 H2O (l) 2 H2 (g) + O2 (g)

What is G for this reaction at 25C?

Hrxn = Hf(products) –Hf(reactants)

Hrxn = [2(0) + 0] - 2(-285.83 kJ/mol) = 571.66 kJ/mol

Srxn = [2(130.58 J/molK) + 205.0 J/molK] - 2(69.91 J/molK)

Srxn = Sf(products) –Sf(reactants)

Srxn = 326.34 J/molK = 0.32634 kJ/molK

Grxn = Hrxn – TSrxn = 571.66 kJ/mol - 298K(0.32634 kJ/molK)

Grxn = +474.41 kJ/mol

2 H2O (l) 2 H2 (g) + O2 (g)

Grxn = Hrxn – TSrxn = 571.66 kJ/mol - T(0.32634 kJ/molK)

What is the minimum temperature needed to make this reaction spontaneous?

Set Grxn = 0 to find minimum temperature

0 = 571.66 kJ/mol - T(0.32634 kJ/molK)

T = (571.66 kJ/mol)/(0.32634 kJ/molK) = 1751 K

T > 1479 C