chapter 17: thermodynamics thermochemistry chapter 17: thermodynamics energy is the capacity to do...

Chapter 17: Thermodynamics

Thermochemistry


Energy is the capacity to do work.• Radiant energy comes from the sun and is

earth’s primary energy source

• Thermal energy is the energy associated with the random motion of atoms and molecules

• Chemical energy is the energy stored within the bonds of chemical substances

• Nuclear energy is the energy stored within the collection of neutrons and protons in the atom

• Potential energy is the energy available by virtue of an object’s position

• Kinetic energy is the energy of motion


Heat is the transfer of thermal energy between two bodies that are at different temperatures.

Energy Changes in Chemical Reactions

Temperature is a measure of the thermal energy.

Temperature = Thermal Energy


Thermochemistry is the study of heat change in chemical reactions.

The system is the specific part of the universe that is of interest in the study.

open

mass & energyExchange:

closed

energy

isolated

nothing

Surrounding: the rest of the universe.


Collision Theory

• To explain factors affecting rxn rate

1. For two molecules to react, they must come in contact or collide.

2. When molecules collide,1. Ineffective collision occurs: molecules may

not orient in right position. No Rxn.

2. Effective collision occurs: molecules possess enough kinetic energy to position.


What Happen Just Before Rxn

• Molecules approaches each other.• Repulsive force increases as they

approach each other.• There is a minimum energy to push the

molecules to collide Activation Energy (Ea)

• Supplying enough kinetic energy will “kick-start” the reaction.


Graphing Rxn: Spontaneous


Gibbs Free Energy (ΔG)

• If ΔG is positive, or ΔG > 0Indication: Rxn Nonspontaneous

• If ΔG is negative, or ΔG < 0Indication: System Spontaneous

• The previous example is spontaneous.


Rate Influencing Factor

1. Surface Area

2. Temperature

3. Concentration

4. Presence of Catalysts


Factor 1: Surface Area

• Similar to solubility• More surface area enhances collision

frequency• Homogeneous Rxn

– The reactants are in the same physical state.For example, gas reacting with gas

• Heterogeneous Rxn– The reactants are in DIFFERENT physical state

For example: Solid reacting with liquid/aq. soln


Factor 2: Temperature

• Increase in temperature provide more kinetic energy for reaction to proceed

• Fast-moving molecules have better chance of “effective collision.”

• It also enhances frequency of collision.• Refrigerator is designed to SLOW reaction

rate.


Factor 3: Concentration

• Higher chemical concentration generally promotes reaction rate

• Concentration is directly proportional to the frequency of collision, in general.

• Just like medicine, we need higher dosage for severely-ill patient.


Factor 4: Catalyst

• In human body, we have enzymes acting as catalyst.

• It does not undergo permanent change.• It provides an alternative reaction pathway

of Lower Activation Energy.

• Inhibitor: reduces a reaction rate by preventing the reaction occurring in the usual way.– Example: Food preservatives


Effect of Catalyst on Rxn


Relating DE to Heat(q) and Work(w)• Energy cannot be created or destroyed.• Energy of (system + surroundings) is constant.• Any energy transferred from a system must be

transferred to the surroundings (and vice versa).

• From the first law of thermodynamics:

The First Law of Thermodynamics

wqE


First Law of Thermodynamics

Energy can be converted from one form to another but energy cannot be created or destroyed.

Second Law of Thermodynamics

The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.

DSuniv = DSsys + DSsurr > 0Spontaneous process:

DSuniv = DSsys + DSsurr = 0Equilibrium process:


Enthalpy Changes

• The difference between the potential energy of the reactants and the products during a physical or chemical change is the Enthalpy change or ∆H.

• AKA: Heat of Reaction at constant pressure


Thermochemical Equations

H2O (s) H2O (l) DH = 6.01 kJ

Is DH negative or positive?

System absorbs heat

Endothermic

DH > 0

6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.



CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH = -890.4 kJ

Is DH negative or positive?

System gives off heat

Exothermic

DH < 0

890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.


Graphing Rxn: EndothermicThe temperature goes down


Graphing Rxn: ExothermicThe temperature goes up


How does ΔH and ΔS affect spontaneity?

-ΔH +ΔH

+ΔS

-ΔS

Alwaysspontaneous

Spontaneity dependson temp

Spontaneity depends on

temp

neverspontaneous


Cold and hot packs

• How do instant hot and cold packs work?


Hot pack• Pressing the bottom , the diaphragm breaks.• Calcium chloride dissolves in water and warms it.• The beverage gets warm.


EnergySurroundings

Exothermic process

• Heat flows into the surroundings from the system in an exothermic process.

Temperature rises

Hot pack


Cold pack• Water and ammonium nitrate are kept in separate

compartments.• Pressing the wrapper, the ammonium nitrate dissolves in

water and absorbs heat.• The pack becomes cold.

It is used to treat sports injuries.


Cold pack

Energy

Surroundings

Endothermic process

• Heat flows into the system from the surroundings in an endothermic process.

Temperature falls


ExplosionsThis reaction is exothermic!


PhotosyntesisThis reaction is endothermic!


Combustions

These reactions are exothermic!

Chapter 17: ThermodynamicsEnthalpy Changes in Reactions

• All chemical reactions require bond breaking in reactants followed by bond making to form products

• Bond breaking requires energy (endothermic) while bond formation releases energy (exothermic)

Chapter 17: ThermodynamicsEnthalpy Changes in Reactions

endothermic reaction - the energy required to break bonds is greater than the energy released when bonds form.

exothermic reaction - the energy required to break bonds is less than the energy released when bonds form.


Enthalpy Changes in Reactions

∆H can represent the enthalpy change for a number of processes

1. Chemical reactions

∆Hrxn – enthalpy of reaction

∆Hcomb – enthalpy of combustion


2. Formation of compounds from elements

∆Hof – standard enthalpy of formation

The standard molar enthalpy of formation is the energy released or absorbed when one mole of a compound is formed directly from the elements in their standard states.


3. Phase Changes

∆Hvap – enthalpy of vaporization

∆Hfus – enthalpy of melting

∆Hcond – enthalpy of condensation

∆Hfre – enthalpy of freezing

4. Solution Formation

∆Hsoln – enthalpy of solution


Time

Temp. (°C )

-40

0

100

140

q = mc∆T

q = n∆H

q = mc∆T

q = mc∆T

q = n∆H


Data:

cice = 2.01 J/g.°C

cwater = 4.184 J/g.°C

csteam = 2.01 J/g.°C

ΔHfus = +6.02 kJ/mol

ΔHvap = +40.7 kJ/mol


warming ice:

q = mc∆T

= (40.0)(2.01)(0 - -40)

= 3216 J

warming water:

q = mc∆T

= (40.0)(4.184)(100 – 0)

= 16736 J

warming steam:

q = mc∆T

= (40.0)(2.01)(140 -100)

= 3216 J


melting ice:q = n∆H

= (2.22 mol)(6.02 kJ/mol)

= 13.364 kJ

boiling water:q = n∆H

= (2.22 mol)(40.7 kJ/mol)

= 90.354 kJ

n = 40.0 g 18.02 g/mol

= 2.22 mol

moles of water:


Total Energy

90.354 kJ

13.364 kJ

3216 J

3216 J

16736 J

127 kJ


Sample Problem• How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C?

Step 1: Heat the ice Q=mcΔT

Q = 36 g x 2.06 J/g deg C x 8 deg C = 593.28 J = 0.59 kJ

Step 2: Convert the solid to liquid ΔH fusion

Q = 2.0 mol x 6.01 kJ/mol = 12 kJ

Step 3: Heat the liquid Q=mcΔT

Q = 36g x 4.184 J/g deg C x 100 deg C = 15063 J = 15 kJ


Sample Problem• How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C?

Step 4: Convert the liquid to gas ΔH vaporization

Q = 2.0 mol x 44.01 kJ/mol = 88 kJ

Step 5: Heat the gas Q=mcΔT

Q = 36 g x 2.02 J/g deg C x 20 deg C = 1454.4 J = 1.5 kJ

Now, add all the steps together

0.59 kJ + 12 kJ + 15 kJ + 88 kJ + 1.5 kJ = 118 kJ


specific heat capacity - the amount of energy , in Joules (J,SI unit), needed to change the temperature of one gram (g) of a substance by one degree Celsius (°C).

• c: does not depend on mass, so intensive property.

• The symbol for specific heat capacity is a lowercase c

• The unit is J/ g°C or kJ/ g°C

Heat/Enthalpy Calculations


• A substance with a large value of c can absorb or release more energy than a substance with a small value of c.

• with the larger c will undergo a smaller temperature change with the same amount of heat applied.


FORMULA

q = mc∆T

q = heat (J)

m = mass (g)

c = specific heat capacity

∆T = temperature change

= T2 – T1

= Tf – Ti


eg. How much heat is needed to raise the temperature of 500.0 g of water from 20.0 °C to 45.0 °C?

Solve q = m c ∆T

for c, m, ∆T, T2 & T1


heat capacity - the quantity of energy , in Joules (J, SI unit), needed to change the temperature of a substance by one degree Celsius (°C)

• The symbol for heat capacity is uppercase C

• C: depends on mass, so extensive property.

• The unit is J/ °C or kJ/ °C


FORMULA

C = mc

q = C ∆T

C = heat capacity

c = specific heat capacity

m = mass

∆T = T2 – T1


Heat Capacity and Specific Heat• Calorimetry = measurement of heat flow (follow Law of

conservation of energy).• Heat lost by one substance = Heat gained by other

substance• Calorimeter = apparatus that measures heat flow.• Heat capacity = the amount of energy required to raise the

temperature of an object (by one degree). • Molar heat capacity = heat capacity of 1 mol of a

substance.• Specific heat = specific heat capacity = heat capacity of 1

g of a substance.

Calorimetry

Tq substance of gramsheat specific

Chapter 17: ThermodynamicsCalorimetry

Constant Pressure Calorimetry

TSqrxn solution) of mass total(

interest of species of moles**mol

qH rxn

rxn

The heat capacity (C) of a substance

C = ms

Heat (q) absorbed or released:

q = msDt

q = CDt

Dt = tfinal - tinitial


How much heat is given off when an 869 g iron bar cools from 940C to 50C?

s of Fe = 0.444 J/g • 0C

Dt = tfinal – tinitial = 50C – 940C = -890C

q = msDt = 869 g x 0.444 J/g • 0C x –890C = -34,000 J

6.5


More practice1. 5 g of copper was heated from 20C to 80C.

How much energy was used to heat the Cu?

2. If a 3.1 g ring is heated using 10.0 J, it’s temp. rises by 17.9C. Calculate the specific heat capacity of the ring. Is the ring pure gold?

q=cmT = 0.38 J/(gC) x 5 g x 60 C = 114 J

3.1 g

q=cmT10.0 J

x 17.9C= 0.18 J/(g°C)c = q/mT=

The ring is not pure. Gold is 0.13 J/(gC)


H2O (s) H2O (l) DH = 6.01 kJ/mol ΔH = 6.01 kJ

• The stoichiometric coefficients always refer to the number of moles of a substance


• If you reverse a reaction, the sign of DH changes

H2O (l) H2O (s) DH = -6.01 kJ

• If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.

2H2O (s) 2H2O (l) DH = 2 mol x 6.01 kJ/mol = 12.0 kJ


Standard enthalpy (heat) of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.

f

The standard enthalpy of formation of any element in its most stable form is zero.

DH0 (O2) = 0f

DH0 (O3) = 142 kJ/molf

DH0 (C, graphite) = 0f

DH0 (C, diamond) = 1.90 kJ/molf


The standard enthalpy of reaction (DH0 ) (heat of reaction) is the enthalpy of a reaction carried out at 1 atm.

rxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

ΔH for the overall rxn will equal to the sum of the

enthalpy changes for the individual steps.


1. Calculate H forNH3 (g) + HCl (g) NH4Cl (s).

Standard heats of formation:NH3(g) = - 45.9 kJ/mol, HCl(g) = - 92.3 kJ/mol, NH4Cl(s) = - 314.4 kJ/mol

a. 176.2 kJ.b 360.8 kJ.c 176.2 kJ.d 268 kJ .

17.4 Section Quiz.

Chapter 17: ThermodynamicsBenzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn DH0 (products)f= S DH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn = [ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ

-6535 kJ2 mol

= - 3267 kJ/mol C6H6


2. The heat of formation of Cl2(g) at 25°C is

a. the same as that of H2O at 25°C.

b. larger than that of Fe(s) at 25°C.

c. undefined.

d. zero.

17.4 Section Quiz.


Example:

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

2H2O(g) 2H2O(l) H= - 88 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ

Hess’s Law

Given:

C(s) + ½ O2(g) CO(g) DH = -110.5 kJ

CO2(g) CO(g) + ½ O2(g) DH = 283.0 kJ

Calculate DH for: C(s) + O2(g) CO2(g)


Calculate the standard enthalpy of formation of CS2 (l) given that:C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJrxn

S(rhombic) + O2 (g) SO2 (g) DH0 = -296.1 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1. Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)

2. Add the given rxns so that the result is the desired rxn.

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ

2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -296.1x2 kJrxn

CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)

DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJrxn


1. According to Hess’s law, it is possible to calculate an unknown heat of reaction by usinga. heats of fusion for each of the compounds in the

reaction.b. two other reactions with known heats of reaction.

c. specific heat capacities for each compound in the reaction.

d. density for each compound in the reaction.

17.4 Section Quiz.


The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure.

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature.


Where’s Waldo? Can you find…

The Triple Point?

Critical pressure?

Critical temperature?

Where fusion occurs?

Where vaporization occurs?

Melting point (at 1 atm)?

Boiling point(at 6 atm)?Carbon Dioxide


Mel

ting

Fre

ezin

g

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium


Sub

limat

ion

Dep

ositi

on

H2O (s) H2O (g)

Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid.

DHsub = DHfus + DHvap

( Hess’s Law)

chapter 17: thermodynamics thermochemistry chapter 17: thermodynamics energy is the capacity to do...

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