Ch 14 Suggested HW: 19, 27, 39, 23, 42, 82

Suggested HW: (Ch 14) 19, 27, 39, 23, 42, 82

Energy

• We depend on energy for our existence, and to maintain our standard of life.

• It is not something that we can touch, but we experience it in many ways

• Commonly, we think of energy as being either kinetic or potential (stored), but there are many variations of these classifications.

Potential Energy At Work

Chemical Energy

• Chemical energy is a kind of potential energy that is stored within chemical bonds.

• For example, combusting fuel (oil, coal) breaks the C-C and C-H bonds, releasing their trapped energy as heat.

• Connecting the ends of a battery allows a redox reaction to occur, forcing electrons to flow from one end to the other (current)

Introduction• In this unit, we will discuss thermochemistry

• Thermochemistry is the study of the energy changes associated with chemical processes, and will lead to a much deeper discussion on thermodynamics, which investigates the flow of energy between systems.

• The first law of thermodynamics, introduced in this lecture, allows us to “keep track” of energy change.

• Two fundamental concepts of thermochemistry and thermodynamics are heat and work.

Energy Basics

• What is Energy? – Energy is defined as the capacity to perform “work”

• How do we define work?• Work is defined as the result of a force acting through

some distance (F x d)– Ex. Pushing an object along a rough surface

» The force can be exerted by a human, steam engine, electric motor, etc.

• In SI units, force is in Newtons (N), distance is in meters. Thus, work (and therefore, energy) is expressed in Newton-meters (N•m), also known as the JOULE (J)

(1 J = 1 N•m)

What is Energy? Work?

• You can not perform work without energy – Ex.: You need at least 100 J of energy to do 100 J of

work• Let’s use a gas cylinder equipped with a piston as an

example. The gas inside of the cylinder is our system. Everything else is our surroundings.

• The application of pressure to the piston causes displacement of the system by a distance Δh

Calculating Work of A Trapped Gas

w=F x (∆ h)

• Remember…. pressure is force divided by area!!

P=FA →𝐏𝐀=𝐅

w=−𝐏𝐀 (∆h)

• Since the pressure acts against the natural expansion of the gas, the pressure term is negative

• Volume = (LW)H = (Area) x height

w=− P∆V

PΔV Work For Trapped Gas

• Compression: Since the compression is caused by the surroundings, we say that work is done on the system by the surroundings.

• Ex. A gas is trapped inside of a cylinder sealed with qa movable piston. The gas is compressed under a constant pressure of 1 atm from 3.5L to 1.2L. Calculate the work of the system.

• Work done on the system is ALWAYS POSITIVE. Energy from the surroundings is transferred into the system, so the system gains that energy.

expansion

w=− P∆V¿− (1 atm )(−2.3 L)¿𝟐 .𝟑𝐋𝐚𝐭𝐦

A Quick Note

• You may have noticed that the units of the previous answer were L•atm, not J. You must convert to joules.

• Now, we can complete the previous problem.

𝟏𝐋𝐚𝐭𝐦=𝟏𝟎𝟏 .𝟑𝟐𝟓 𝐉

2.3 Latm x 101.325 JLatm =𝟐𝟑𝟑 𝐉

PΔV Work For Trapped Gas

• Expansion: If the gas expands against the piston, then work is done by the system on the surroundings.

• Work done by the system is ALWAYS NEGATIVE because the system now has that much less energy.

• Ex. Following a reaction, a gas expands against a constant pressure of 1 atm from 5L to 9L. Calculate the work of the system.

w=− (1atm )(4 L)(101.325 JLatm )

w=−𝟒𝟎𝟓 .𝟑 𝐉

Heat

• Energy that is not used to perform work or stored is lost as heat. Heat is the flow of energy caused by a temperature gradient.

• For example, only 25% of the thermal energy released from the combustion of gasoline is actually used to perform work (move the car). – The rest is lost as heat to the surroundings (heating

up the car and the air around the car). – Without this limitation, cars could easily exceed 100

mpg.

• Heat, denoted with the symbol q, will always flow when there is a temperature difference between the system and the surroundings (from high temp to low temp).

Heat• The sign convention for heat is the same as for work. •

– When heat is lost by the system, it must be transferred to the surroundings. In such a case, qsys is negative, and qsurr is positive (-qsys = qsurr)

– A reaction in which qsys < 0 is called an exothermic reaction

Fuel + O2 CO2 + H2O + heat

CaCl2(s) Ca2+(aq) + 2Cl-(aq) + heat

H2O

Heat• When heat is absorbed by the system, it is taken from the

surroundings, and qsys is positive (qsurr is negative).

• When qsys >0, the reaction is endothermic.

Al(NO3)3(s) + heat ------> Al3+(aq) + 3NO3-(aq)

H2O(s) + heat -----> H2O (L)

First Law of Thermodynamics

• If a system could be perfectly isolated from the world around it so that no energy transfer could occur with the surroundings, then the total energy of the system, or internal energy (U) would simply be the sum of kinetic and potential energies of the molecules.

• Of course, its not possible to isolate a system. In real situations, chemicals are in physical contact, so energy flows readily between systems. We can measure the change in internal energy of a system by:

∆𝐔=𝐪+𝐰

First Law of Thermodynamics

Energy is never created or destroyed, merely converted between forms and transferred from place to place as heat or work. The total energy of the universe is finite (conservation of energy).

Conservation of Energy

• Imagine we have a perfectly insulated box. Inside that box is trapped air, a light bulb, wires and a battery

• When a connection is made between the battery and bulb, chemical energy is converted to electric current which transmits the wiring, causing emission.

• Light interacts with air, generating heat

• The total heat energy in the box is equal to the total internal energy that was in the battery.

• The energy in the battery was in a more usable form.

State Functions• In thermodynamics, we must distinguish between those

terms that are state functions, and those that are not.

• A state function is a property of the system that is based on the system’s condition, but NOT on the path taken to reach that condition. Internal energy is a state function.

• Example: Imagine that our system is 50g of water at 25oC. The value of U is the same regardless of how that state is reached.

Heat and Work are NOT State Functions

• U is a state function, but q and w are not.

• The amount of heat and work transferred will depend on the path, although their sum does not.

• Take the system (battery) below. The value of ΔU doesn’t depend on HOW the energy is used, but q and w do.

Battery is short circuited. Coil heats up. No work is done. All energy is lost as heat.

Fully Charged

Dead

Heat

Heat

Work

Some energy is used to do work (turn fan).

ΔU

Enthalpy• We can represent the total energy change of a system at

constant pressure as:

• Rearranging the equation to solve for q, we obtain:

• If a reaction is carried out in a closed, rigid container, the volume is constant. Thus, no PV work can be done, and the energy change in the system is entirely heat.

• More commonly, a reaction would be carried out in an open flask under constant atmospheric pressure, and the energy change would be due to both heat and work. q would then follow expression (1).

∆𝐔=𝐪+𝐰=𝐪−𝐏 ∆𝐕

(𝟏 )𝐪𝐩=∆𝐔+𝐏 ∆𝐕 (𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐩𝐫𝐞𝐬𝐬𝐮𝐫𝐞)

(𝟐 )𝐪𝐯=∆𝐔 (𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 𝐯𝐨𝐥𝐮𝐦𝐞)

Enthalpy• The heat change of a reaction at constant pressure is

known as ENTHALPY (H), or heat of reaction.

• In most chemical reactions, only ΔH is important. As with ΔU, ΔH is a state function. Typically, the difference between ΔU and ΔH is small.

• For a reaction at a constant pressure of 1 atm and 25oC, we can define the standard enthalpy change (denoted by ‘o’ symbol) of a reaction

• Enthalpy is a molar quantity, and is usually expressed in KJ/mol. As with q, positive values of ΔH indicate endothermic reactions and negative values indicate exothermic reactions.

∆𝐇❑𝐨𝐫𝐱𝐧=𝐇𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬−𝐇𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬

Calculating ΔHo

• The enthalpy change associated with a reaction can be determined by several methods, depending on the information provided. These methods include:

1. Stoichiometric calculation, using standard enthalpies

2. Hess’s Law, using standard reactions of known ΔHo

rxn

3. Heats of formation4. Bond enthalpy calculations

Stoichiometric Enthalpy Calculations

• Lets take the combustion of hydrogen gas as an example of an exothermic reaction. 2H2(g) + O2(g)

2H2O(g)

ΔH < 0exothermic

2H2(g) + O2(g) 2H2O(g) ΔHorxn= -483.6 kJ/mol

1. Enthalpy is an extensive property. The magnitude of ΔH is proportional to the amount of reactant consumed. So, 2 mol of H2 will evolve -483.6 kJ. 4 mol of H2 will evolve double that.

2. Enthalpy of the reverse reaction has the opposite sign. The decomposition of 2 moles of H2O into 2 moles of H2 and 1 mole of O2 is +483.6 kJ.

3. The state of the reactants matters. If liquid water formed instead of steam, ΔH would be different.

Enthalpy Diagram

Example

• Calculate the heat released when 30.0 g of iron(III)oxide is combined with 15.0 g of Al(s) at 1 bar.

• To determine the enthalpy change of the reaction, we must first determine the moles of the limiting reactant.

Fe2O3(s) + 2Al(s) Al2O3(s) + 2 Fe(s) ΔHorxn= -851.5

kJ/mol

15 g Al x mol Al27 g Al=.556mol Al

30 g F e2O3 xmol F e2O3

159.7 g Fe2O 3=.376mol F e2O3

Limiting Reactant

.556mol Al x −851.5 kJ heat2mol Al reacted=−𝟐𝟑𝟕𝐤𝐉

Hess’s Law

• For multi-step processes, enthalpy changes are additive. • Lets take the following reaction:

• This reaction is two steps, and proceeds through an intermediate (short-lived species), as shown below.

Sn (s )+2Cl2 (g )→SnC l4 (L)

Sn (s )+C l2 (g )→SnC l2 (s )∆ H orxn❑ (1)=−325.1 kJ /mol

SnC l2 (s )+C l2 (g )→SnC l4 (L )∆ H orxn❑ (2)=−186.2 kJ /mol

Hess’s Law, continued

• We can add these reactions together, canceling intermediates on opposing sides of the arrow.

• Since we know the enthalpy change in each step, we can determine the overall enthalpy change of the reaction in question.

Sn (s )+C l2 (g )→SnC l2 (s )∆ H orxn❑ (1)=−325.1 kJ /mol

SnC l2 (s )+C l2 (g )→SnC l4 (L )∆ H orxn❑ (2)=−186.2 kJ /mol

𝑆𝑛 (𝑠 )+2𝐶𝑙2 (𝑔 )→𝑆𝑛𝐶 𝑙4(𝐿)

• This additive property is known as Hess’s Law.

Hess’s Law

• The usefulness of Hess’s Law is that it allows you to express the value of ΔHo of an unknown reaction as the sum of the ΔHo values of known standard reactions.

Example

• Given the following ΔHorxn values, calculate ΔHo

rxn for the reaction: 2SO2(g) + O2(g) 2SO3(g)

(1) SO2(g) ---> S(s) + O2(g) ΔHorxn (1) = 296.8 kJ/mol

(2) 2S(s) + 3O2(g) ---> 2SO3(g) ΔHorxn (2) = -791. 4 kJ/mol

(3) 2SO2(g) + O2(g) ---> 2SO3(g) ΔHorxn (3) = -197.8

kJ/mol

It is necessary to multiply reaction (1) by a factor of 2 in order to get 2 moles of SO2(g) and to cancel out S(s) and 2 moles of O2(g).

(1) 2 [SO2(g) ---> S(s) + O2(g) ] ΔHorxn (1) = 2 (296.8

kJ/mol)

(2) 2S(s) + 3O2(g) ---> 2SO3(g) ΔHorxn (2) = -791. 4

kJ/mol

Group Example

• Given the following values of ΔHorxn:

• Calculate the value of ΔHorxn for the following reaction:

(1 )2P ( s )+3C l2(g)→2 PC l3(L)(2 )2 P (s )+5Cl2(g )→2PCl5(s)

= -639.4 kJ/mol= -887.0 kJ/mol

(3)

Enthalpies of Formation

• The enthalpy change associated with a the formation of a molecule is called the enthalpy change of formation, ΔHf (also called heat of formation).

• For a given reaction, if you know the enthalpies of formation of the reactants and products:

∆𝐇❑𝐨𝐫𝐱𝐧=∑𝐧 ∆𝐇❑

𝐨𝐟 (𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬 )−∑ 𝐧∆𝐇❑

𝐨𝐟 (𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬)

Enthalpies of Formation

For pure elements, and for elements in their most stable form, ΔHo

f = 0 kJ/mol

The enthalpies of formation are exactly known for many elements.

Example

• Calculate the enthalpy change for the combustion of benzene, C6H6 to form CO2 and H2O using standard heats of formation.

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

Using the heats of formation of the reactants and the products, we can directly calculate

∆𝐻❑𝑜𝑟𝑥𝑛=∑ 𝑛∆𝐻❑

𝑜𝑓 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 )−∑ 𝑛∆𝐻❑

𝑜𝑓 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)

∆𝐻❑𝑜𝑟𝑥𝑛=−6535𝑘𝐽 /𝑚𝑜𝑙

Molar Bond Enthalpies

• The enthalpy change for the complete decomposition of water:

is = + 925 kJ/mol. • In a water molecule, there are 2 O-H

bonds. The positive value of enthalpy indicates that heat (energy) must be absorbed by the system in order to break these bonds.

• Therefore, we can estimate that the molar bond enthalpy, Hbond, of each O-H bond (energy needed to break the bond) is 463.5 kJ/mol

H 2O (L )→O (g )+2H (g)

Molar Bond Enthalpies

• In general, if we consider a reaction in the most basic sense, then we would imagine that the formation of a compound proceeds as follows:

∆𝑯❑𝒐𝒓𝒙𝒏=¿ energy absorbed to

break reactant bonds + energy released when product bonds form

Table of common bond enthalpies. These are the energies to break the bonds. When you form bonds, the sign of H will be negative.

Example• Using molar bond enthalpies, estimate of the following:

• As you can see, this reaction proceeds by breaking one C-H bond, and a Cl-Cl bond, followed by the formation of a C-Cl bond and an H-Cl bond

C H 4 (g )+Cl2 (g )→CH 3Cl (g )+HCl (g)

C

H

H

H

H Cl Cl

Energy absorbed. Bonds broken

C

H

H

H

Cl H Cl

Energy released. New bonds.

∆𝐇❑𝐨𝐫𝐱𝐧=¿ +(414 kJ/mol + 243

kJ/mol)(-331 kJ/mol - 431 kJ/mol) = -105

kJ/mol

Group Example

• Nitrogen gas reacts with hydrogen gas to produce ammonia, NH3. Use molar bond enthalpies to calculate the enthalpy of formation ammonia. Compare this with the known value, -45.9 kJ/mol.

Molar Heat Capacity

• The molar heat capacity, denoted as Cp, defines the energy needed to raise a mole of a substance one degree K.

• We can relate heat capacity to the total transferred heat (enthalpy) by the following:

𝐪𝐩=𝐧𝐂𝐩∆𝐓

Example

• Calculate the joules of heat required to raise the temperature of 150 kg of water from 18oC to 60oC, assuming no loss to the surroundings.

• Note: A change of x oC is exactly equal to a change of x oK.

𝐪𝐩=∆𝐇=𝐧𝐂𝐩∆𝐓

q p=(150,000 g H 2O xmol H 2O18g H 2O )x (75.3 J

mol K❑o ) x (42 K❑

o )=𝟐 .𝟔𝟒𝐱𝟏𝟎𝟕 𝐉

Heat Transfer

• Heat ALWAYS flows from hotter substances to cooler substances until both substances are the same temperature.

• Note: Heat is energy. Temperature determines the flow of heat.

• In an insulated system, no heat is transferred to the surroundings. Therefore, if we have two materials in contact, the heat lost by one material is exactly equal to the heat gained by the other.

Group Example

• 50g of copper metal at 80oC is placed in 100 mL of water at 10oC within an insulated system. What will the final temperature of the water and copper be? Use the molar heat capacities of Cu and H2O(L) as 24.4 J/moloK and 75.3 J/moloK,

𝐓=𝟏𝟑 .𝟎𝟕 𝐂❑𝐨