Topics 5 and 15.pdf

Thermochemistry

Topics 5 and 15

Energetics “Thermodynamics”

Hess’s Law

Calorimetry

Enthalpy

Energetics

•

Relationship between chemical reactions and

heat

•

What causes chemical reactions to occur

Two Trends in Nature

• Order



Disorder

 

• High energy



Low energy

System v. Surroundings

•

System

–

The actual

chemical

reaction taking

place

•

Surroundings

Heat Flows

•

Heat always flows from hot to cold

Exothermic reaction one that gives off heat – transfers thermal energy from the system to the surroundings.

2H₂ (g) + O₂ (g) 2H₂O (l) + energy

H₂O (g) H₂O (l) + energy

6.2

Reactants had more energy than the

Endothermic reaction one in which heat has to be supplied to the system from the surroundings.

energy + 2HgO (s) 2Hg (l) + O₂ (g)

energy + H₂O (s) H₂O (l)

Products end with more potential energy the

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

D

H

=

H

(products) –

H

(reactants)

DH = heat given off or absorbed during a reaction at constant pressure

H_products < H_reactants

DH < 0

H_products > H_reactants

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 00_{C and 1 atm.}

Thermochemical Equations

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 250_{C and 1 atm.}

Heat v. Temperature

•

Heat

–

form of energy measured

in J (SI)

•

Temperature

–

average kinetic energy of

molecules in a system

–

Measured in K (SI) 0

o

_C=

4 ways to find H

•

Using calorimeter

•

Using Hess’s Law

•

Enthalpies of formation

Calorimetry

•

Calorimetry

–

The measurement of heat flow

•

Calorimeter

–

Device used to measure heat flow

Specific heat (c or s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the

substance by one degree Celsius (or K).

Heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.

C = ms

Heat (q) absorbed or released:

q = msDt

q = CDt

Dt = t_final - t_initial

How much heat is given off when an 869 g iron bar cools from 940_{C to 5}0_C?

s of Fe = 0.444 J/g • 0C

Dt = t_final – t_initial = 50C – 940_{C = -89}0_C

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

Constant-Pressure Calorimetry

No heat enters or leaves!

Calorimetry

•

When measuring heat lost or gained by the

reaction, we measure it using water as the

basis for the calculations.

Q

_w

=

m x s x Dt

m of water Q of water

s of water

Calorimetry

•

The heat gained or lost by the reaction is

equal to, but opposite in sign, to the heat

gained or lost by the water

Sample problem

• 0.075 mol HCl and 0.075 mol NaOH are added to 150.0 ml of water in a calorimeter. The temperature of the solution increased from 23.00_{C to 29.9}0_{C. Calculate the enthalpy}

change for the formation of 1.0 mol of water in this rxn. Assume no heat is lost the surroundings and the density of the solution is the same as the water.

H

+

(aq)

+ OH

-(aq)

H

2

O

(l)

Sample problem

0.075 mol HCl and 0.075 mol NaOH are added to 150.0 ml of water in a calorimeter. The temperature of the solution increased from 23.00_{C to 29.9}0_{C. Calculate the enthalpy}

change for the formation of 1.0 mol of water in this rxn.

Assume no heat is lost the surroundings and the density of the solution is the same as the water.

H

+

(aq)

+ OH

-(aq)

H

2

O

(l)

Step 1: find q

_water

q = m s

D

t

Sample problem

0.075 mol HCl and 0.075 mol NaOH are added to 150.0 ml of water in a calorimeter. The temperature of the solution increased from 23.00_C

to 29.90_{C. Calculate the enthalpy change for the formation of 1.0}

mol of water in this rxn. Assume no heat is lost the surroundings and the density of the solution is the same as the water.

H

+_(aq)

+ OH

-_(aq)

H

₂

O

_(l)

Step 1: find

D

q

_water

q = m s

D

t

•

2

nd

Step: Find

D

q

_rxn

•

Water absorbed heat : heat gained by water is

lost in the reaction

•

2

nd

Step: Find

D

q

_rxn

•

Water absorbed heat : heat gained by water is

lost in the reaction

•

2

nd

Step: Find

D

q

_rxn

•

Water absorbed heat : heat gained by

water is lost in the reaction

-q

_water

= q

_rxn

-(+4330 J) = q

_rxn

-4330 J = q

_rxn

•

Step 3: find the

D

H of the reaction

H

+_(aq)

+ OH

-_(aq)

H

₂

O

_(l)

D

H

_rxn

=

D

q

_rxn

=

-4330 J = -5.8x10

4

_J

n

_H2O

0.075 mol H

₂

O

D

H

_rxn

vs.

D

q

_rxn

•

D

H

_rxn

= heat lost or gained in the balanced

chemical equation

4 ways to find H

•

Using calorimeter

•

Using Hess’s Law

•

Enthalpies of formation

Standard Enthalpy of Formation

•

Standard enthalpy of formation

(

D

H

0

) is the

heat change that results when

one mole

of a

compound is formed from its

elements

in

their standard state .

standard state (101.3kPa and 298K or 1atm, 250_C)

D

H

_f0

Standard Enthalpy of Formation

For Methanol

C

_(s)

+ 2H

_2(g)

+ ½ O

_2(g)

CH

₃

OH

D

H

0

_{= -201 KJ/mol}

Heat of formation reactions are written so that the reactants exist as they would under standard

Standard enthalpy of formation

The standard enthalpy of formation of any

element

in its most stable form

is zero.

DH0 _(O

2) = 0

f DH

0 _(O

3) = 142 kJ/mol

f

DH0 _{(C, graphite) = 0}

f DH

0 _{(C, diamond) = 1.90 kJ/mol}

f

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

Thermochemical Equations

• 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

H₂O (s) H₂O (l) DH = 6.01 kJ

• The physical states of all reactants and products must be specified in thermochemical equations.

Thermochemical Equations

6.4

H₂O (l) H₂O (g) DH = 44.0 kJ

How much heat is evolved when 266 g of white phosphorus (P₄) burn in air?

P₄ (s) + 5O₂ (g) P₄O₁₀ (s) DH_reaction = -3013 kJ

266 g P₄ 1 mol P4 123.9 g P₄

x 3013 kJ

1 mol P₄

aA + bB cC + dD

DH0

rxn = [cDH0f (C) + dDH0f (D)] - [aDH0f (A) + bDH0f (B)]

f

= n DH0 (products)

f

DH0

rxn S - S n DH0 (reactants)

6.6

Enthalpy of a Reaction

Sum of coefficients

Benzene (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)

DH0

rxn = S DH0_f (products) - S DH0_f (reactants)

DH0

rxn = [12DH0_f (CO2)+ 6DH0_f (H2O)] - [ 2DH_f0 (C6H6)+ 15(O2)]

DH0

rxn = [ 12(-393.5) + 6(-285.8) ] – [ 2(49.04) + 15(0) ] = -6535 kJ

-6535 kJ

2 mol = - 3267 kJ/mol C6H6

Example 2 Find the

D

H

0

_rxn

) ( 2 ) ( 2 ) ( 2 ) (

2

3

2

2

2

H

S

_g



O

_g



SO

_g



H

O

_g

= n DH0 (products)

f

DH0

rxn S - S n DH0 (reactants)

))]

(

(

3

))

(

(

2

[

)]

(

(

2

(

))

(

(

2

[

) ( 2 0 ) ( 2 0 ) ( 2 0 ) ( 2 0 0 g f g f g f g f rxn

O

H

S

H

H

O

H

H

SO

H

H

D



D

S



D



D

S



D

)]

0

(

3

)

2

.

20

(

2

[

)]

8

.

241

(

2

(

)

8

.

296

(

2

[

0















D

H

_rxn

mol

KJ

H

_rxn0





1036

.

8

/

4 ways to find H

•

Using calorimeter

•

Using Hess’s Law

•

Enthalpies of formation

State Functions

•

A property that depends on the present state

of the system, not the path taken to arrive at

that state.

The energy contained within 1.0 mole of water

at 25

0

C is the same whether or not the ice was

Hess’s Law

If a reaction is carried out in a series of steps,

the overall change in enthalpy will be equal to

the sum of the enthalpy changes for the

individual steps

Hess’s Law

Calculate the

D

H for the reaction

3C(s) + 4H

_2(g)

C

₃

H

_8(g)

Use the following information:

Rxn Chem equation Enthalpy Change

1

_2H

_2(g)

_{+ O}

_2(g)

_2H

₂

_O

_(l) DH= -571.7 kJ

2 C₃H_8(g) + 5O_2(g) 3CO_2(g) + 4H₂O_(l) DH= -2220.1 kJ

Hess’s Law Cont…

1

2 H

_2(g)

+ O

_2(g)

2 H

₂

O

_(l)

DH= -571.7 kJ

2 C₃H_8(g) + 5O_2(g) 3CO_2(g) + 4H₂O_(l)

DH= -2220.1 kJ

3 C(s) + O₂ (g) CO₂(g)

DH= -393.5 kJ

3C(s) + 4H

_2(g)

C

₃

H

_8(g)

x2

4 H

_2(g)

+ 2O

_2(g)

4 H

₂

O

_(l)

DH= -1143.4 kJ

X -1 DH= +2220.1 kJ

x3

3C(s) + 3O₂ (g) 3CO₂(g) _D

H= -1180.5 kJ

Calculate the standard enthalpy of formation of CS₂ (l) given that:

C(graphite) + O₂ (g) CO₂ (g) DH0_rxn = -393.5 kJ

S(rhombic) + O₂ (g) SO₂ (g) DH0_rxn = -296.1 kJ

CS₂(l) + 3O₂ (g) CO₂ (g) + 2SO₂ (g) DH0_rxn = -1072 kJ

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.

rxn

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

2S(rhombic) + 2O₂ (g) 2SO₂ (g) DH0_rxn= -296.1x2 kJ

CO₂(g) + 2SO₂ (g) CS₂ (l) + 3O₂ (g) DH_rxn0 = +1072 kJ

+

C(graphite) + 2S(rhombic) CS₂ (l)

DH0 = [1072] – [-393.5 + (2x-296.1)] = 2057 kJ

rxn

4 ways to find H

•

Using calorimeter

•

Using Hess’s Law

•

Enthalpies of formation

Bond Enthalpies

•

Enthalpy changes

of reaction are

the result of

breaking and

making bonds

•

Using average

Example

•

3F

_{2 (g)}

+ NH

_{3 (g)}



3 HF

_(g)

+ NF

_{3 (g)}

Energy in Energy out

Chemistry in Action:

Fuel Values of Foods and Other Substances

C₆H₁₂O₆ (s) + 6O₂ (g) 6CO₂ (g) + 6H₂O (l) DH = -2801 kJ/mol

1 cal = 4.184 J

The enthalpy of solution (DH_soln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.

DH_soln = H_soln - H_components

6.7

Which substance(s) could be used for melting ice?

The Solution Process for NaCl

DH_soln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol

Energy Diagrams

50 kJ/mol 300 kJ/mol

150 kJ/mol 100 kJ/mol

-100 kJ/mol +200 kJ/mol

Exothermic Endothermic

(a) Activation energy (Ea) for the forward reaction

(b) Activation energy (Ea) for the reverse reaction

Entropy (S) is a measure of the randomness or disorder of a system.

order S disorder S

If the change from initial to final results in an increase in randomness DS > 0

For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state

S_solid < S_liquid << S_gas

H₂O (s) H₂O (l) DS > 0

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.

DS_univ = DS_sys + DS_surr > 0 Spontaneous process:

DS_univ = DS_sys + DS_surr = 0 Equilibrium process:

Entropy Changes in the System (D

S

_sys

)

aA + bB cC + dD

DS0

rxn = [ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B)]

DS0

rxn = S S0(products) - S S0(reactants)

The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250_C.

rxn

18.4

What is the standard entropy change for the following reaction at 250_{C? 2CO (g) + O}

2 (g) 2CO2 (g)

S0_{(CO) = 197.9 J/K•mol}

S0_(O

2) = 205.0 J/K•mol

S0_(CO

2) = 213.6 J/K•mol

DS0

rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]

DS0

Entropy Changes in the System (D

S

_sys

)

18.4

When gases are produced (or consumed)

• If a reaction produces more gas molecules than it consumes, DS0 _{> 0.}

• If the total number of gas molecules diminishes,

DS0 _{< 0.}

• If there is no net change in the total number of gas molecules, then DS0 _{may be positive or negative}

BUT DS0 will be a small number.

What is the sign of the entropy change for the following reaction? 2Zn (s) + O₂ (g) 2ZnO (s)

Spontaneous

Physical and Chemical Processes

• A waterfall runs downhill

• A lump of sugar dissolves in a cup of coffee

• At 1 atm, water freezes below 0 0_{C and ice melts above 0} 0_C

• Heat flows from a hotter object to a colder object

• A gas expands in an evacuated bulb

• Iron exposed to oxygen and water forms rust

spontaneous

nonspontaneous

DS_univ = DS_sys + DS_surr > 0 Spontaneous process:

DS_univ = DS_sys + DS_surr = 0 Equilibrium process:

Gibbs Free Energy

For a constant-temperature process:

DG = DH_sys -TDS_sys Gibbs free

energy (G)

DG < 0 The reaction is spontaneous in the forward direction.

DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.

DG = 0 The reaction is at equilibrium.

D

G

= D

H

-

T

D

S

18.5

aA + bB cC + dD

DG0

rxn = [cDG0f (C) + dDG0f (D)] - [aDG0f (A) + bDG0f (B)]

DG0

rxn = S DG0f(products) - S DG0 f(reactants)

The standard free-energy of reaction (DG0 _{) is the}

free-energy change for a reaction when it occurs under standard-state conditions.

rxn

Standard free energy of

formation (DG0_{) is the free-energy}

change that occurs when 1 mole of the compound is formed from its elements in their standard states.

f

DG0 _{of any element in its stable}

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

DG_rxn0 DG0 _(products)

f

= S - S DG0 _f(reactants)

What is the standard free-energy change for the following reaction at 25 0_C?

DG0

rxn = [12DG0_f (CO2)+6DG0_f (H2O)] - [ 2DG0_f (C6H6)]

DG0

rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ

Is the reaction spontaneous at 25 0C?

DG0 _{= -6405 kJ < 0}

spontaneous

Recap: Signs of Thermodynamic Values

Negative

Positive

Enthalpy (ΔH) Exothermic

Endothermic

Entropy (ΔS)

Less disorder More disorder

Gibbs Free

Energy (ΔG)

Spontaneous

Not

Gibbs Free Energy and Chemical Equilibrium

DG = DG0 _{+ RT lnQ}

R is the gas constant (8.314 J/K•mol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

DG = 0 Q = K

0 = DG0 _{+ RT lnK}

DG0 ₌  _{RT lnK}

DG0 ₌  _{RT lnK}

The boiling point is the temperature at which the

(equilibrium) vapor pressure of a liquid is equal to the external pressure.

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm.

11.8

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

The critical pressure (P_c) 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)?

Melting

11.8

Fr

eezing

H₂O (s) H₂O (l)

Su bli mat io n 11.8 De posi tion

H₂O (s) H₂O (g)

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

DH_{sub =} DH_fus + DH_vap

Molar heat of fusion (DH_fus) is the energy required to melt 1 mole of a solid substance.

Sample Problem

• How much heat is required to change 36 g of

H

₂

O 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

Sample Problem

• How much heat is required to change 36 g of

H

₂

O 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