Zumdahl Ch 06 Thermochemistry powerpoint

THERMOCHEMISTRY

Zumdahl

Chapter 6

6.1 Energy

 Energy – the capacity to do work or produce heat.  Law of Conservation of Energy (1st law of

thermodynamics)– Energy is neither created nor destroyed. (Energy of universe is constant)

 Potential Energy _{– energy stored due to position}

or composition.

 Kinetic Energy – energy due to motion of an

object = ½ mv2

 Energy is a state function (or property) – a

People at LAX:

People at LAX:

Same

Same::

latitude/longitude/altitude, regardless of how they got there

latitude/longitude/altitude, regardless of how they got there

(state functions)

(state functions)

Different

Different:: Miles flown

Miles flown

Tiredness

Tiredness

Frequent flier miles earned

Frequent flier miles earned

(non-state functions)

6.1 System vs Surroundings

 System – a defined environment  Surroundings – everything else

 The text is written from the system’s perspective

6.1 Components of Energy

 Heat (q) – transfer of energy due to a temperature

difference.

 q is positive, E_sys increases  q is negative, E_sys decreases

 Work (w) – force acting over a distance.

 w is positive, E_sys increases  w is negative, E_sys decreases

 Unlike energy, heat and work are not state

functions.



E = q + w



E

= -



E



E

=



E



E

= -



E

6.1 Exothermic vs Endothermic

Reactions

 Exothermic reactions release energy.

 Endothermic reactions absorb energy.

 E_sys = E_surround  PE (bonds)  KE

(heat)

 _{New bonds are stronger}  E_sys = -E_surround

 KE (heat)  PE

(bonds)

6.1 Gas expansion work (p. 233-4)

w = -P



V

 Gases do work via volume change according to the formula below. The system is often a gas-filled piston.

 When a gas expands (positive 

V

 When a gas compresses (negative V), work is positive.

6.1 Fluid Work (p. 233-4)



E = q + w

w =

-P



V

What work is done when a gas is compressed

from 45 L to 20 L at a constant external

pressure of 5 atm?

w = -P



V

w = -

(5 atm)(-25 L)

The Thermite Reaction

 This is one of the most exothermic chemical reactions known (-850. kJ).

 Thermite is used in industrial welding.

 Iron (III) oxide is reduced at such a high temp that it produces molten iron.

 The mp of iron is 1538 

◦

thermite

Endothermic Processes

Ba(OH)₂.8H

2O(s) + 2NH4SCN(s)  Ba(SCN)2(s) + 10H2O(l) + 2NH3(g)

 The temperature of this reaction after completion is about -20 _◦C.

 _{An example of an endothermic process is when ammonium nitrate}

(found in chemical cold packs) dissolves:

NH₄NO_3(s)  NH₄+

(aq) + NO3-(aq) ammonium

Endothermic Processes

 The temperature of this reaction after completion is about -20 °C.

 _{An example of an endothermic process is when ammonium nitrate}

(found in chemical cold packs) dissolves:

NH₄NO_3(s)  NH₄+

(aq) + NO3-(aq) ammonium

End of lesson

6.2 Enthalpy and calorimetry

Objective:

• _{Students will be able to carry out calculations}

involving specific heat capacity.

• _{Students will understand various terms relating to}

6.2 Enthalpy and calorimetry

 Enthalpy (H) is the heat involved in a chemical

reaction at constant pressure. It is defined as:

H = E +

PV

E = system energy

P = system

pressure

V = system volume

 Since E, P and V are all state functions, so is H.

 H represents only heat exchanged; E

6.2 Enthalpy



H =

q

 At constant pressure:

 For a reaction at constant pressure, H is given by:

 positive H = endothermic reaction.  negativeH = exothermic reaction.



H

= H

- H

6.2 Calorimetry

 Calorimetry is the science of measuring heat.

 Specific heat capacity (C_p; s in Zumdahl) – the heat

required to raise 1 g of a substance by 1 ºC (or 1 K).

 Molar heat capacity – the energy required to raise 1

“Cold” Fire (No need to write this

down.)

 The specific heat of

water is so large (4.18 J/ºC · g) that it can

absorb large amounts of energy with very little change in temperature.

C

H

OH

+ 3O



3H

O

+

2CO

6.2 Constant Pressure Calorimetry

 Constant-pressure

calorimeters are run at

atmospheric pressure and are used to determine H for

reactions in solution.

 Any observed temperature

increase is due to release of energy, calculated by:

s = specific heat capacity

m = mass of solution

T = increase in temperature

Specific heat capacity problem

12.00 g of a metal at 100.0 ºC is poured into a styrofoam cup calorimeter

containing 75.0 g of water at 20.0 ºC. The final temperature of the metal + water is 22.1 ºC. What is the specific

heat capacity of the metal? For H₂O_(l) , s

= 4.18 J/g ·ºC

H = s · m · T

H (water) = 4.18 J 75.0 g 2.1ºC = 660 J

g °C

s (metal) = H = 660 J = 0.71 J/g ºC

Specific heat capacity problem

Ex. 2 3Ba2+ + 2PO

43-  Ba3(PO4)2(s)

In a reaction, 1.03 g of barium phosphate is formed, and the temp of the 250. g of solution in the reaction vessel increases by 1.5 ºC. Find the molar heat of reaction for the above reaction.H = s · m · T

H (water) = 4.18 J 250. g 1.5ºC = 1600 J

g ºC

molar heat of reaction

mol = 1.03 g mol = 1.71 x 10-3 _mol

601.84 g

= - 9.4 x 105 _J/mol

meaning “per mole”

= H

mol

= -1600 J

End of section

6.2 Constant Volume Calorimetry

 _{Constant-volume calorimeters}

are called bomb calorimeters. The reaction occurs in an

internal steel chamber.

 _{Since the volume of the}

chamber is constant, no work is done. All the energy,

therefore, is released as heat.



E = q + w



E = q

q_V = heat at constant volume



E = q

=

C

·



T

C

= heat capacity of

calorimeter



T

= temperature

6.2 Constant Volume Calorimetry

Silver(II) fluoride is an extremely rare example of silver in the +2

oxidation state. Using a bomb calorimeter with a heat capacity of 10.9

kJ/ºC, decomposition of 249.4 g AgF₂ reacts violently in the presence of

xenon gas. If the temperature of the bomb increased from 22.12 to

29.62 ºC, calculate the energy released per mole of AgF₂.

AgF_2(s) + Xe_(g)  Ag_(s) + XeF_2(s)

q

=



T

· (heat capacity of calorimeter)

q

=

(29.62 – 22.12 ºC)

· (10.9 kJ/ºC)

q

=

(7.50 ºC)

· (10.9 kJ/ºC) =

81.8 kJ

249.4 g AgF₂ 1 mol AgF₂ = 1.71 mol AgF₂

145.865 g AgF₂

81.8 kJ

1.71 mol AgF₂

= 47.8 kJ/mol AgF

=

-

47.8 kJ/mol AgF

First solve the problem mathematically

keeping all quantities positive and

then

6.3 Hess’s Law



Objective: Students will be able to

apply Hess’s Law to solve

6.3 Hess’s Law



Hess’s law states that if a reaction can be

6.3 Calculating



H

 There are three quantitative relationships between a chemical equation and H:

1) If a chemical equation is multiplied by some integer, H is also multiplied by the same integer.

2) If a chemical equation is reversed, then H

changes sign.

3A + B  2C H

2 · (3A + B  2C) 2 · (H)

6A + 2B  4C 2H

A + B  C + D H

6.3 Calculating



H

 Since enthalpy is a state function, H is dependent only on initial and final states.

Hess’ Law

3) If a chemical reaction can be expressed as the sum of a series of steps, then H for the overall equation is the sum of the H values for each step.

3A + B  C H₁

C  2D H₂

6.3 Calculating



H

Ex. 1 Find H for the formation of propane using the

reactions and corresponding H values below.

3C

+ 4H



C

H

C_(s) + O_2(g)  CO_2(g) H = -393.5 kJ

O_2(s) + 2H_2(g)  2H₂O_(g) H =

-483.6 kJ

C₃H_8(s) + 5O_2(g)  3CO_2(g) + 4H₂O_(g) H = -2043 kJ

3CO_2(g) + 4H₂O_(g)  C₃H_8(s) + 5O_2(g) H =

+2043 kJ

3 · [C_(s) + O_2(g)  CO_2(g)] 3 · H = -393.5 kJ

3C_(s) + 3O_2(g)  3CO_2(g) H = -1180.5 kJ

2 2O·[O2(s)_2(s) + 4H + 2H_2(g)2(g)   4H 2H₂O_(g)] 2O(g) 2 · H = -483.6 kJH =

-967.2 kJ

6.3 Calculating



H

Ex. 2 Find H for the formation of sulfur dioxide using

the reactions and corresponding H values below.

S_(s) + O_2(g)  SO_2(g)

2SO_2(g) O_2(g) 2SO_3(g) H = -198.2 kJ

S_(s) + 3/2O_2(g)  SO_3(g) H = -395.2 kJ

S(s) + O2(g)  SO2(g) H = -296.1 kJ

A

B

Hint: Keep “A” as is. Reverse “B” and divide by

2. Add.

End of section

6.3 Hess’s law quiz (2019)

Given the following equations and ΔHo values,

determine the heat of reaction (in kJ) at 298 K for the reaction:

B₂H₆(g) + 6Cl₂(g)  2BCl₃(g) + 6HCl(g)

BCl₃(g) + 3H₂O(l)  H₃BO₃(g) + 3HCl(g) ΔHo =

-112.5 kJ

B₂H₆(g) + 6H₂O(l)  2H₃BO₃(s) + 6H₂(g) ΔHo = -493.4 kJ

6.4 Standard enthalpy of formation



Objective: Students will be able to

6.4 Standard enthalpy of formation

 Standard conditions are needed …

 standard states - arbitrary sets of conditions

for calculating thermodynamic state properties (H, E, G).

 For compounds:

 Gases - 1 atm of pressure

 Liquids and solids – pure, 1 atm

 Solutions - exactly 1 M

 For elements:

 1 atm and 25 oC

6.4 Standard Enthalpy of Formation

 Standard Enthalpy of Formation ( ) –

the change in enthalpy that accompanies

the formation of 1 mol of a compound from

its elements in their standard states.

 For a pure element in its standard state, = 0.

6.4 Standard Enthalpy of Formation

Write equations for the formation of MgCO_3(s) and C₆H₁₂O_6(s) from their elements in their

standard states.

Mg

+ C

+ O



MgCO

Mg

+ C

+ O



MgCO

Mg

+ C

+ O



MgCO

Mg

+ C

+ O



MgCO

6C

+ 6H

+ 3O



C

H

O

6C

+ 6H

+ 3O



C

H

O

6C

+ 6H

+ 3O



C

H

O

6C

+ 6H

+ 3O



C

H

O

6.4 Standard Enthalpy of Reaction,

1111111

 _{The enthalpy change for a given reaction is calculated by subtracting the enthalpies of formation}

of the reactants from the enthalpies of formation of the products:

 _{Remember elements in their standard states have a}

= zero.

6.4 Standard Enthalpy of Reaction,

1112111

Calculate for the combustion of octane

(C

H

).

C

C

H

H

+ O

+ O

 CO

 8CO

+ H

+

O

9H

O

Substance

(kJ/mol)

C₈H_18(l) -250.1

O_2(g) 0

CO_2(g) -393.5

H₂O_(g) -242

This data was obtained from Appendix 4, p.

A19.

=



[products] -



[reactants]

 Note: This answer is slightly different from the one in

your text on pg. 252, even though both problems solve for the H_rxn of combustion of octane. Two differences: 1)This problem solves for gaseous water. Your text solves

for liquid water.

2)This problems solves kJ/mol. The problem in your text is off by roughly a factor of two since it solves for 2 mols of octane.

 Don’t worry. Test questions will not be ambiguous. You

End of section

 See homework on board.

Ch. 6 Review Questions

1. Using the appendix, calculate the enthalpy change

for the reaction: 2SO₂ + O₂  2 SO₃. Is it endo- or exothermic?

2. Using your answer to number 1, how much heat is

absorbed or produced when 150. g of SO3 is produced?

3. How much heat is needed to raise the temperature

of 168 g of nickel (C_p = 0.445 J/mol ºC) from -15.2 ºC to +23.6 ºC?

4. If a substance absorbs 200. kJ of heat energy while

doing 150. kJ of work on its surroundings, what is the total ΔE?

5. Write an equation for which ΔHº = ΔHºf for