chapter 6 notes thermoche mistry. part 1: energy, heat and work thermoche mistry

Chapter 6 Notes

Thermochemistry

Part 1: Energy,

Heat and

Work

Thermochemistry

Energy is the capacity to do work or to produce heat. The law of conservation of energy

states that energy can be converted from one form to another, but can be neither

created nor destroyed.

Kinetic energy is energy due to the motion of the object and depends on

the mass and velocity of the object. KE = ½mv2

Potential energy is energy due to position or composition.

**Mass must be in kilograms and velocity must be in meters/second!!!!! The unit kgm2 = J

s2

Heat and temperature are different.

Temperature is a measure of the kinetic energy of the molecules.

Heat refers to the transfer of energy between two objects due to a

temperature difference.

Heat is not a substance contained by an object, although we often talk of

heat as if this were true.

The pathway is likened to a “path” or “route.” For instance, I can get to the stadium by walking out the front door or the back door.

A state function or state property depends only on the characteristics of the present state – not on the pathway.

The universe is divided into two parts: a. The system is the part of the universe on which we focus.b. The surroundings include everything else in the universe.

For a reaction, the system includes the reactants and products.

The surroundings includes the reaction container, the room, etc.

(i.e. anything else other than reactants and products.)

The study of energy and its interconversions is called

thermodynamics. The law of conservation of energy is often

called the first law of thermodynamics. It states: The

energy of the universe is constant.

Thermodynamic quantities always consist of two

parts: a number, giving the magnitude of the change, and a sign,

indicating the direction of the flow. The sign reflects the system’s point of view.

q = heat

*If energy flows INTO the system via heat (endothermic), then q = +.*If energy flows OUT OF the system via heat (exothermic), then q = .

w = work

*If the surroundings do work on the system (energy flows into the system), then w = +.* If the system does work on the surroundings (energy flows out of the system), then w = .

Work is defined as force acting over a distance. The formula used to solve for work is: W = PΔV

Pressure is measured in atmospheres and volume is measured in liters. Convert to joules using 101.3 J = 1 Latm

The sign changes depending on: compressed gas = +PΔV

expanding gas = PΔV

Memorize!!!

vaporization

melting

sublimation

condensation

freezing

deposition

gas

endothermic liquid exothermic

solid

The internal energy, E, of a system is defined as the sum of the kinetic and potential energies. The formula is ΔE = q + w where q is heat and w is work.

The sign convention is that anything that leaves the system is negative. *q is negative: system releases heat*q is positive: system absorbs heat*w is negative: system does work*w is positive: surroundings do work

Part 2: Properti

es of Enthalpy

Thermochemistry

Enthalpy, H, is defined as:

H = E + PV where E = internal energyP = pressureV = volume

Internal energy, pressure and volume are all state functions (independent of the pathway) therefore enthalpy is also a

state function.

At constant pressure, the change in enthalpy (ΔH) of the system is equal to

the energy flow as heat.

Therefore, ΔH = q at constant P.

At constant pressure, exothermic means that ΔH is negative and

endothermic means that ΔH is positive.

Stoichiometric Calculations

When a mole of methane (CH4) is burned at constant pressure, 890 kJ of energy are released as heat. Calculate H when 5.8 grams of methane are burned at constant pressure.

16.0 g/mol

CH4 + 2 O2 CO2 + 2 H2O H = –890 kJ5.8 g x kJ

– 890. kJ 1 mol CH4 5.8 g CH4 = – 320 kJ1 mol CH4 16.0 g CH4

Part 3: Calorimetry and Heat

Capacity

Thermochemistry

Calorimetry is the study of heat flow

and heat measurement.

Calorimetry

Calorimetry experiments determine the heats (enthalpy

changes) of reactions by making accurate

measurements of temperature changes produced in a

calorimeter.

The formula used incalorimetric calculations is:

q = mcTwhere

q = heat (J)m = mass (g)

c = specific heat capacity (J/g C)T = change in temperature (C)

The heat capacity of an object is the amount of heat needed

to raise the temperature of the object by 1C.

The heat capacity of one gram of a substance is called its

specific heat.

The specific heat is a physical property of the

substance, like its color and melting point. Substances have different specific heat

capacities.

The specific heat capacity of water

= 4.18 J/gC

When calculating T, always subtract the smaller temperature FROM the larger temperature:

T = Tlarger Tsmaller **If the temperature rises,

(ex: from 25C to 30C) then q will be negative and the reaction is exothermic.

**If the temperature drops,(ex: from 40C to 30C) then q will be

positive and the reaction is endothermic.

Example 1: What is the specific heat capacity of iron if the temperature of a 12.3-g

sample of iron is increased by 10.2C when 56.7 J of heat is added?

q = mcT

Example 1: What is the specific heat capacity of iron if the temperature of a 12.3-g sample of iron is increased by

10.2C when 56.7 J of heat is added?

56.7 J = (12.3 g)( c)(10.2C)

56.7 J = (12.3 g) (c) (10.2 C)

0.452 J/gC = c

(12.3 g)(10.2C) (12.3 g)(10.2C)

Example 2: When a 13.7-g sample of solid Pb(NO3)2 dissolves in 85.0 g

of water in a calorimeter, the temperature drops from 23.4C to

19.7C. Calculate H for the solution process.

Pb(NO3)2 (s) Pb+2 (aq) + 2 NO31 (aq) H = ?

q = mcT

Example 2: When a 13.7-g sample of solid Pb(NO3)2 dissolves in 85.0 g of water in a calorimeter, the temperature drops from 23.4C to 19.7C.

Calculate H for the solution process.Pb(NO3)2 (s) Pb+2 (aq) + 2 NO3

1 (aq) H = ?

q= (85.0 g)(4.18 J/gC ) (23.4 C19.7C )

q = 1310 J

**Since the temperature dropped, q will be positive and the reaction is endothermic.

q = mc(Tlarger Tsmaller)

Example 2: When a 13.7-g sample of solid Pb(NO3)2 dissolves in 85.0 g of water in a calorimeter, the temperature drops from 23.4C to

19.7C. Calculate H for the solution process.Pb(NO3)2 (s) Pb+2 (aq) + 2 NO3

1 (aq) H = ?

Calculate the molar mass of Pb(NO3)2:

MM = 331 g/mol

Remember that from the first part, q = +1310 J

Example 2: When a 13.7-g sample of solid Pb(NO3)2 dissolves in 85.0 g of water in a calorimeter, the temperature drops from 23.4C to

19.7C. Calculate H for the solution process.Pb(NO3)2 (s) Pb+2 (aq) + 2 NO3

1 (aq) H = ?

+1310 J

13.7 g Pb(NO3)2

331 g Pb(NO3)2

1 mol Pb(NO3)2

1 mol Pb(NO3)2

= +31,700 J = 31.7 kJ

Energy Calculations

Important Information:

q = mcΔTq = ΔHfusion moles

q = ΔHvaporization moles

specific heat capacity of water = 4.18 J/gC specific heat capacity of ice = 2.1 J/gC

specific heat capacity of steam = 1.8 J/gCΔHfusion of water = 6.0 kJ/mole

ΔHvaporization of water = 40.7 kJ/mole

How much energy does it take to convert 130. grams of ice at 40.0C to steam at

160.C?

Convert grams to moles of water:

1 mol H1 mol H22OO 130. g H130. g H22OO

18.0 g H18.0 g H22OO

= 7.22 mol H= 7.22 mol H22OO

Plan: a. Heat ice from 40.0C to 0.00C.

q = mcTq = (130. g)( 2.1 J/gC)(40.0C)

q = 10,920 Jq 10,900 Jq = 10.9 kJ

b. Add heat to convert ice to liquid water at 0C.

q = Hfusion molesq = (6.0 kJ/mol)(7.22 mol)

q = 43.32 kJq 43.3 kJ

c. Heat liquid water from 0.00C to 100.C.

q = mcTq = (130. g)( 4.18 J/gC)(100.0C)

q = 54,340 Jq 54,300 Jq = 54.3 kJ

d. Add heat to convert liquid water to steam at 100C.

q = Hvaporization molesq = (40.7 kJ/mol)(7.22 mol)

q = 293.854 kJq 294 kJ

e. Heat steam from 100.C to 160.C.

q = mcTq = (130. g)( 1.8 J/gC)(60.0C)

q = 14,040 J q 14.0 kJ

Add the energy values:

Total energy = a + b + c + d + e

Total energy = 10.9 kJ43.3 kJ54.3 kJ294 kJ14.0 kJ

416.5 kJ 417 kJ

Part 4: Hess’ Law

Thermochemistry

The amount of heat that a reaction absorbs or

releases depends on the conditions under which

the reaction is carried out (temperature, pressure,

and physical states of the reactants and products.)

To make comparing enthalpy changes easier, chemists chose a pressure of 1

atmosphere and a temperature of 25C as conditions to carry out reactions.These are called standard states.

An enthalpy change under these conditions is called a standard enthalpy

change. It is denoted with a superscript. It is shown as H.

Conventional Definitions of Standard States

For a compound:*The standard state of a gaseous substance is a pressure of exactly 1 atmosphere.*For a pure substance in a condensed state (liquid or solid), the standard state is the pure liquid or solid.*For a substance present in a solution, the standard state is a concentration of exactly 1 M.

For an element:*The standard state of an element is the form in which the element exists under conditions of 1 atmosphere and 25C. (The standard state for oxygen is O2 (g) at a pressure of 1 atm; the standard state for sodium is Na(s); the standard state for mercury is Hg(ℓ).

In the 19th century, a Swiss chemist named G.H. Hess proposed a way of

finding the enthalpy change for a reaction (even if the reaction could not

be performed directly.)In 1840, Hess demonstrated experimentally that the heat

transferred during a given reaction is the same whether the reaction occurs

in one step or several steps.

His method is now called Hess’ law of heat

summationor simply Hess’ Law.

The method is analogous to solving

simultaneous equations in algebra.

In solving problems using Hess’ Law, there

are some basic rules that must be memorized.

Rules for manipulating reactions:

1. If a reaction is reversed, the sign of H must be reversed.

2. If a reaction is multiplied or divided by a coefficient, H must also be multiplied or divided by that coefficient.

Rules for adding reactions: 1. Identical substances on the

same side of a reaction are added together.

2. Identical substances on opposite sides of a reaction are cancelled.

3. Simply add the H’s of each reaction to get the H of the final reaction.

Use:A + B C + D H = 10 kJ

2 E + C D + 2 F H = 20 kJ

E + A F H = 30 kJ

Determine H for

3 A + B 2C

Example 1:

A + B C + D H = 10 kJ

2E + C D + 2F H = 20 kJ

E + A F H = 30 kJ

Determine H for

3A + B 2C

A + B C + D H = 10 kJ

2 F + D 2 E + C H = + 20 kJ

E + A F H = 30 kJ

2 2 2 2( )

Write the reactions by comparing the substances with the desired

result.

A + B C + D H = 10 kJ

2E + C D + 2F H = 20 kJ

E + A F H = 30 kJ

Determine H for

3A + B 2C

A + B C + D H = 10 kJ

2 F + D 2 E + C H = + 20 kJ

E + A F H = 30 kJ)

2 2 2 2(

Cancel identical items on opposite sides of the arrow.

A + B C + D H = 10 kJ

2E + C D + 2F H = 20 kJ

E + A F H = 30 kJ

Determine H for

3A + B 2C

A + B C + D H = 10 kJ

2 F + D 2 E + C H = + 20 kJ

E + A F H = 30 kJ)

2 2 2 2(

Add identical items on the same side of the arrow.

3 A 2C+ B

A + B C + D H = 10 kJ

2E + C D + 2F H = 20 kJ

E + A F H = 30 kJ

Determine H for

3A + B 2C

A + B C + D H = 10 kJ

2 F + D 2 E + C H = + 20 kJ

E + A F H = 30 kJ)

2 2 2 2(

Verify that the final reaction matches the reaction given in the original problem.

3 A + B 2C

A + B C + D H = 10 kJ

2E + C D + 2F H = 20 kJ

E + A F H = 30 kJ

Determine H for

3A + B 2C

A + B C + D H = 10 kJ

2 F + D 2 E + C H = + 20 kJ

E + A F H = 30 kJ)

2 2 2 2(

Combine the H’s.

3 A 2C+ B H = 50 kJ

When dealing with actual substances in the reactions, the physical state of each substance must be written in parentheses.

For example:

(s) is solid, (ℓ) is liquid, (g) is gas,

and (aq) is aqueous – meaning that it is dissolved in water.

From the following enthalpy changes,

2PbS (s) + 3 O2 (g) 2 PbO (s) + 2 SO2 (g) H = 124 kJ

Pb (s) + CO (g) PbO (s) + C (s) H = 106.8 kJ

calculate the value of H for the following reaction:

2PbS (s) + 3 O2 (g) + 2 C (s) 2 Pb (s) + 2 CO (g) + 2 SO2 (g)

Example 2:


Find H for:


Using:



Leave the first reaction as written:


Find H for:


PbO (s) + C (s) Pb (s) + CO (g) H = + 106.8 kJ

Using:



Reverse the second reaction and change the sign:


Find H for:


PbO (s) + C (s) Pb (s) + CO (g) H = + 106.8 kJ)

Using:



Multiply the second reaction by 2:

2 2 2 2 2(


Find H for:



Using:




2 2 2 2 2(


Find H for:



Using:



2 2 2 2 2(




Find H for:



Using:




2 2 2 2 2(



Find H for:



Using:



2 2 2 2 2(


Combine the H’s.

H = + 89.6 kJ

From the following enthalpy changes,

C (graphite) + O2 (g) CO2 (g) H = 394 kJ

H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (ℓ) H = 890.3 kJ

calculate the value of H for the following reaction:

C (graphite) + 2 H2 (g) CH4 (g)

Example 3:


Find H for: C (graphite) + 2 H2 (g) CH4 (g)Using:


H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (ℓ) H = 890.3 kJ

Leave the first reaction as written:


H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ

Leave the second reaction as written:



H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (ℓ) H = 890.3 kJ


H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ

Reverse the third reaction and change the sign:



H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (ℓ) H = 890.3 kJ

CO2 (g) + 2 H2O (ℓ) CH4 (g) + 2 O2 (g) H = + 890.3 kJ


H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ)

Multiply the second reaction by 2:

22

2 2(



H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (ℓ) H = 890.3 kJ

CO2 (g) + 2 H2O (ℓ) CH4 (g) + 2 O2 (g) H = + 890.3 kJ


H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ)


22

2 2(



H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (ℓ) H = 890.3 kJ

CO2 (g) + 2 H2O (ℓ) CH4 (g) + 2 O2 (g) H = + 890.3 kJ


H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ)2 2

2 2(



H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (ℓ) H = 890.3 kJ

CO2 (g) + 2 H2O (ℓ) CH4 (g) + 2 O2 (g) H = + 890.3 kJ




H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ)

Combine the H’s.

22

2 2(



H2 (g) + ½O2 (g) H2O (ℓ) H = 286 kJ

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (ℓ) H = 890.3 kJ

CO2 (g) + 2 H2O (ℓ) CH4 (g) + 2 O2 (g) H = + 890.3 kJ

C (graphite) + 2 H2 (g) CH4 (g) H = 75.7 kJ

Part 5: Standard Enthalpies

of Formation

Thermochemistry

The standard enthalpy of formation (ΔHf) of a compound is defined as the change in enthalpy that accompanies the formation

of one mole of a compound from its elements with all substances in their

standard states.The ΔHf values for some common substances are shown in Table 6.2.

More values are found in Appendix 4. **ΔHf for an element in its standard state is zero.

The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products:

ΔH˚rxn = Σ npΔHf˚products Σ nrΔHf˚reactants

Use the standard enthalpies of formation in Appendix 4 to calculate the standard enthalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water. 4 NH3 (g) + 7 O2 (g) → 4 NO2 (g) + 6 H2O (ℓ)ΔH˚rxn = Σ npΔHf˚products Σ nrΔHf˚reactants

ΔH˚rxn = [(4 mol NO2) (34 kJ/mol) + (6 mol H2O) (286 kJ/mol)]

[(4 mol NH3) (46 kJ/mol) + (7 mol O2) (0 kJ/mol)]

ΔH˚rxn = [(136 kJ) + (1716 kJ)] [(184 kJ) + (0 kJ)] ΔH˚rxn = [1580 kJ] [184 kJ] ΔH˚rxn = 1396 kJ

It is to your benefit to work every assigned homework problem because

you won’t be able to memorize how to work all of the problems.

The test over this chapter will consist of a multiple choice section (75%) and an essay section (25%). The test will be given over a period of two days.

This format is similar to the AP Chemistry exam. On the AP exam,

each section is worth 50%.

chapter 6 notes thermoche mistry. part 1: energy, heat and work thermoche mistry

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