Chapter 5 Chemistry Lecture

8/10/2019 Chapter 5 Chemistry Lecture

http://slidepdf.com/reader/full/chapter-5-chemistry-lecture 1/65



Energy

Types of energy

– Kinetic

– Potential

• Gravitational

• Electrostatic

• Chemical Measured in joules (J), or calories (cal)

1 cal = 4.184 J



Systems and Surroundings

The system is what we are studying

The surroundings are everything else

Universe = system + surroundings

Types of systems

– Closed - common in chemistry• Exchanges energy but not mass with

surroundings



What Energy Can Do

Energy can be gained by a system if:

– Work is done on the system

– Heat is gained by the system Energy can be lost by a system if:

– The system does work

– The system loses heat So, what is work?

For that matter, what is heat?



Who’s Doing The Work?



Question

If a student compresses a gas by pushingdown on a piston:

a. The student has done work on the system;energy of the system decreases

b. The system has done work on the student;energy of the system decreases

c. The student has done work on the system;

energy of the system increasesd. The system has done work on the student;

energy of the system increases



First Law of Thermodynamics

Energy can neither be created nor

destroyed

An easier way to say this:

– Energy is conserved

An even easier way to say this:

– YOU CAN’T WIN!



What Does This Mean?

Internal energy is the energy of a system (E)

If a system changes its energy, the new

energy is: – ∆E = Efinal - Einitial

∆E is negative if system has lost energy…to? – The surroundings!

∆E is positive if system has gainedenergy…from? – The surroundings!



Put Another Way…



Chemical Reactions and

Energy



E, Heat, and Work

Systems and surroundings may exchangeenergy as – Heat

– Work

∆E = q + w – q = heat

• +q is heat GAINED by system, -q is heat LOST bysystem

– w = work• +w is work done ON system, -w is work done BY system



Example

What is the ∆E for a system for a

process in which the system absorbs

140 J of heat from the surroundings,and does 85 J of work on the

surroundings?



Heat

Endothermic (heat in) processes

– The system absorbs heat

Exothermic (heat out) processes

– The system loses heat

Temperature is proportional to the

internal energy of a system



Exothermic and Endothermic

Processes



State Functions

The total energy (E) for a system is hard

to know

– We generally talk about change in energy

Factors influencing E

– Temperature

– Pressure

– Total quantity of matter



State Functions

Properties of systems that are defined

by the system’s condition (state) are

state functions – State functions depend upon conditions,

NOT pathway

– Internal energy is a state function



Not State Functions

Heat (q) is not a state function – A system can get from one state to another

through loss/gain of various amounts of heat

Work (w) is not a state function – A system can get from one state to another

through doing/being subjected to various amountsof work

For a given energy change, q and w can vary,but E does not



Reactions and Work

Most reactions in chemistry are open to

the atmosphere

– Pressure is constant

Work is associated with a change in

volume



Question

Which of the following represents the

relationship between system and

surroundings energy?a. ∆Esystem = ∆Esurroundings

b. ∆Esystem = -∆Esurroundings

c. ∆Esystem = 2(∆Esurroundings)d. ∆Esystem = -2(∆Esurroundings)



Question

Which of the following is a state

function?

a. Internal energy (of a system)

b. Heat

c. Work



PV Work

PV work

– Change in gas volume

– W = -P∆V • Expansion: ∆V is positive, work BY system

• Compression: ∆V is negative, work ON system



PV Work



An Important Note

Temperature is proportional to the internal

energy of the system

– All other factors held constant, temperatureincreases if heat (energy) is absorbed

– All other factors held constant, temperature

increases if work is done ON system

– All other factors held constant, temperaturedecreases if work is done BY system



Question

For gas in a piston canister, which of

the following would NOT constitute an

increase in E of the system?a. Pushing down on the piston

b. Lighting a candle under the canister

c. Pulling up on the piston



Question

What does it mean to say temperature of asystem is proportional to E?

a. The temperature of the system can be used tocalculate E

b. Comparing the temperatures of differentsystems allows you to compare their Es

c. If ∆E for a system is positive, the temperature

will decreased. If ∆E for a system is positive, the temperature

will increase



Enthalpy (H)

Provides a quantification of heat flow if

– Constant pressure

– No work other than PV work

– H = E + PV

• Internal energy plus product of

pressure/volume

• H is a state function



∆H

∆H is change in enthalpy

∆H = ∆(E + PV), or ∆E + P∆V

But…easier…

– In systems at constant pressure, ∆H = q

• If ∆H is positive, ENDOTHERMIC reaction

• If ∆H is negative, EXOTHERMIC reaction



Enthalpies of Reaction

For a reaction, ∆Hrxn = ∆Hprod - ∆Hreact



Other Enthalpy Issues

What happens if we reverse the

reaction?

– Magnitude of ∆Hrxn is the same, signchanges

What happens if we burn two moles of

methane? – Twice as much heat is evolved



Definition

Squir-rel-y adj. Slang

– 1.Eccentric.

– 2.Cunningly unforthcoming or reticent.

– The American Heritage Dictionary of the

English Language, Fourth Edition.



Question

A squirrel is sitting on an electrical

wire. What kind of energy does he

have?a. Gravitational potential

b. Electrical potential

c. Kinetic



Question

The squirrel leaps for a neighboring

wire; as he moves through the air

(quite some distance above theground), what kind of energy does he

have?

a. Gravitational potentialb. Kinetic

c. Gravitational potential and kinetic



Calorimetry

How can we measure ∆Hrxn experimentally?

Need to know a few facts… – As substances gain heat, they get hotter (and vice

versa)

– Substances increase/decrease in temperature

predictably• Can quantify heat gained/lost based upon temperaturechange



Heat Capacity (C)

The amount of heat necessary to raisetemperature of an object by 1 ºC (1 K)

– Larger heat capacity means more heatrequired to raise temperature

Molar heat capacity (Cmolar ) is heatcapacity of one mole of a substance

Specific heat (s) is heat capacity of onegram of a substance



Specific Heat (s)

s = q

m x ∆T

Quantity of heat

transferred

Change in

temperatureGrams of

substanceCan also write in terms of heat:

q = ms∆T



Example

What is the specific heat of water, given

that it takes 418 J of heat to raise the

temperature of 50.0 g by 2.00 K?



Question

By what amount will the temperature of

50.0 g of water change if it absorbs 815

J of heat, given that swater = 4.18 J/gK?



The “Coffee Cup” Calorimeter

Not sealed• Reaction occurs at

constant (atm) pressure

• Coffee cup is insulating,has low heat capacity, so

low heat absorption

• qsoln = (ssoln)(gsoln)(∆T) = -qrxn



Example

50.0 mL 0.100 M AgNO3 and 50.0 mL

0.100 M HCl are mixed in a coffee cup

calorimeter. The temperature of thesolution increases from 22.20ºC to

23.11ºC; calculate enthalpy of reaction

for the reaction: AgNO3 (aq) + HCl (aq) AgCl (s) + HNO3 (aq)



“Bomb” Calorimetry

Combustion in

sealed “bomb”

Heat absorbed bycontents of

calorimeter

qrxn = -Ccal∆T

Constant volume



Example

A 1.800 g sample of phenol (MW =

94.11 g/mol) was burned in a bomb

calorimeter with a heat capacity of11.66 kJ/ºC. The temperature of the

calorimeter and contents increased from

21.36ºC to 26.37ºC. What is the heat ofcombustion per mole of phenol?



Question

If a system changes temperature a

LOT for the amount of heat it absorbs,

what kind of specific heat does it have:a. High

b. Low



Question

In a coffee-cup calorimeter, what do

we assume comprises the system to

SIMPLIFY our calculations?a. Everything in the universe

b. The cup plus the water in the cup

c. The water in the cup ONLY



Hess’s Law

Very formal definition

– If a reaction is carried out in a series of

steps, ∆H for the overall reaction will equalthe sum of the ∆Hs for individual steps

What this really means

– We can use known ∆H values to determineunknown ∆H values



Example

Calculate ∆Hrxn for the formation of CO,

given the reactions below:

– C(s) + 1/2 O2 (g) CO(g) ∆Hrxn = ? – C(s) + O2 (g) CO2(g) ∆Hrxn = -393.5 kJ

– CO(g) + 1/2 O2 (g) CO2(g) ∆Hrxn = -283.0 kJ



Enthalpies of Formation

∆Hf is enthalpy change for formation of

compound from constituent elements

– Depends upon conditions of reactants andproducts

• Temp, pressure, state

– For simplicity, “standard state” is defined



Standard State

Set of conditions (the ones found in

most laboratories) for which enthalpies

are recorded – Pure form of a substance

– Pressure = 1 atm

– Temperature = 298K (25ºC)



∆Hº = Enthalpy at Standard

StateStandard enthalpy change

– Indicates that all reactants and products

are in standard state conditions – Can be for formation or reaction



∆Hºf

Standard enthalpy of formation

–Change in enthalpy for reaction that forms

one mole of compound from elements – All substances in standard state

2 C(graphite) + 3 H2 (g) + 1/2 O2 (g) C2H5OH(l) ∆Hºf = -277.7 kJ

Note that all substances are in their most stable form



Table of ∆Hºf Values

QuickTime™ and a

TIFF (Uncompressed) decompressor are needed t o see this picture.



Question

True or False: to use Hess’s Law to

find the enthalpy of reaction, the

reaction MUST have taken place via aseries of steps, for which the

enthalpies of reaction are known.

a. TRUEb. FALSE

Careful! Tricky question!



Question

For which of the following is the

enthalpy of formation ZERO?

a. O2(g)

b. O2(l)

c. O(g)

d. All of these



Question

True or False: Because ethane (C2H6)

is most stable as a gas, the enthalpy

of formation of gaseous ethane isZERO

a. TRUE

b. FALSE



Example

What is the equation for formation of

glucose in terms of standard enthalpy of

formation?



Enthalpy of Reaction

Enthalpies of formation (∆Hºf ) can be

used to calculate standard enthalpy of

reaction (∆Hºrxn) – ∆Hºrxn = (sum of ∆Hºf of products) - (sum of

∆Hºf of reactants)



Finding ∆Hºf

Find ∆Hºf for C3H6O(l), given the

equation:

C3H6O(l) + 4 O2(g) → 3 CO2(g) + 3 H2O(l), ∆H°

rxn = -1790 kJ



Food and Fuel

Combustion is the burning of a chemicalto release heat and produce simpler

substancesFoods are burned in the presence ofoxygen to produce various products

– CO2

– H2O

– Chemical energy or heat



Food Calories

A food Calorie is actually a kcal (1000

cal)

Three types of nutrients – Carbohydrates, 4 Cal/g

– Proteins, 4 Cal/g

– Fats, 9 Cal/g



Heat IS Energy

The Calories in a meal can be used to

fuel biological processes



Example

A meal consisting of a Double Quarter-

Pounder with Cheese, large fries, and large

Triple-Thick Chocolate Shake contains 2470

food calories. If the energy from this meal

were used to heat 10 gallons (approximately

38L or 38kg) of water initially at room

temperature (25°

C), how hot would thewater get?

Chapter 5 Chemistry Lecture

Documents

Transcript of Chapter 5 Chemistry Lecture