energy, enthalpy and thermochemistry

8/7/2019 Energy, Enthalpy and Thermochemistry

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Energy, Enthalpy and Thermochemistry

Energy and the First Law of Thermodynamics

To begin the study of the transformation of energy in chemical (or other) processes, we need

to first develop a few terms that we will use. These definitions must be very tightly specified

and understood in order to properly follow some of the more complex logic in later pages.

Definitions:

Thermodynamics

The science of transformation of energy.Energy

The capacity to do work or to transfer heatWork

Work has several aspects to it. There is mechanical work, electrical work, light-energy work and work of expansion or contraction of a gas (PVwork).

One of the most common definitions for work is when a force fdisplaces an object by

a distance (x does workw' =f*(x. This is a mechanical work.

Kinetic and Potential Energy

both of these can be

forms of work and ofheat energy

KE= 1/2 mv2

PE= mgh lifting mass m by height h against

acceleration due to gravityg

= fd force times distance.

Work is organized energy, lifting a book, etc.

Heat (defined later) is random (disorganized) energy heat released, sound dissipates

through air, when book drops and hits the floor.

One can also transfer energy into or out of a sample via light-energy and by electricalwork

(We include this type of work in w'). In first-year chemistry, we'll not often beinterested in these terms directly. For the most part, therefore, we'll set w'to be zero.

Work can also be done by pushing back the atmosphere (piston in a cylinder). In this

case, one pushes back against the atmospheric pressurePatmand moves the piston a

distance(x. The force needed to push back the atmosphere is actuallyPatmA where

A is the area of the piston. Hence, we have w =Patm A (x = Patm (V.


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We call this atmospheric work.(Note that this has the same units as the previous definition ofwork).

In chemical situations, we are normally interested only in the atmospheric work done

when gases are evolved or used up in the reaction.

SI units:w has units of Nm = kgm

2s

-2= Joules (J).

HeatEnergy transfer as a result of a temperature differential.

First Law of Thermodynamics:

Often called the Law ofconservation ofenergy: Energy can neither be created nor

destroyed but only changed from one form to another.

OR

The energy of the universe is constant

OR

The energy of a system which is isolated from its surroundings is constant.

More definitions

System

whatever we are interested in:

Chemicals in a reaction

chemical, reaction mixture, container

etc.

Surroundingsthe rest of the universe

Closed System

Heat may transfer but not matter.

Isolated system

No heat or matter can transfer between the system and its surroundings.

Energy transfer can be done in one of two ways:


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1. workw can be done on the system by the surroundings (orvice versa). It can take theform of mechanical work or of electrical energy transfer.

2. Heat q flows from the system to the surroundings (orvice versa)There is a general sign convention that chemists use when describing q and w. When energy

flows into the system as a result of heat or work, the sign is positive (the system gains

energy). When energy flows out of a system, the sign is negative (the system looses energy.Both q and w refer, not to an amount of energy, but to an amount of energy transferred as a

result of the process.

Note the sign conventionHow do we measure the actual amount of heat contained in a material. Can we even measure

it.

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Enthalpy

For the initial part of this chapter, we will be looking at a specific type of cases where the

system undergoing the change is held at constant pressure. This, in fact, represents the way

the majority of reactions are done in our chemicalexperience.

We define a term called enthalpy His the energytransferred between a system and the surroundings

under constant pressure. We cannot measure theabsolute enthalpy of a system but we can measure the

change in enthalpy for a process. If the Initial and final states have enthalpies Hiand Hf,respectively then the change in enthalpy for the process can be defined as

(H= Hf- Hi.

At this point, we're looking at ablack box. We can't really

understand what's inside, we justdefine terms to try to understand it

We're familiar with state functions. For instance, the

distance between Kingston and Toronto can bemeasured by a "straight" line on the map. It never

changes. It is only a function of the position of the two

cities. However, the distance driven to get from

Kingston to Toronto depends on the route we take.

Obviously, some routes are shorter than others. So, wecannot use a non-state function (distance driven) to

measure a state function (distance separating two cities)


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This definition is only good ifHisa state function, i.e., the change in

enthalpy depends only on the initialand final states, not on the process

itself.

If we simply measure the heat evolved (or absorbed) during a process, we will not have ameasure of the change in enthalpy since, in general,q is not a state function. It depends on

how the process occurs.

We need to restrict our process measurements to specific conditions in order to be able to

measure enthalpy directly. In this case, if we maintain constant pressure and then simply

measure the heat transferred as a result of the process qpwe will have a measure of the change

in enthalpy. The subscript p refers to the fact that this heat was measured at constant pressure

conditions.

Thus, we have simply

(H= qp

at constant pressure

For chemical reactions we can write

(H= 7Hproducts - 7Hreactants

If(His negative, then net flow of heat from chemical system to surroundings (heat released).

Exothermic (EXO sounds like exit, from the Latin: to leave)

If(His positive, then net flow of heat from surroundings to chemical system (heatabsorbed). Endothermic (ENDO sounds like enter)

CH4(g) + 2O2(g) ----> CO2(g) + 2H2O(l) (H= -890 kJ/mol(that's per mole of Rx n as written)(EXOTHERMIC)

H2(g) + I2(g) ----> 2 HI(g) (H= 52.2 kJ/mol(ENDOTHERMIC)

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Heat Capacity

One of the easiest ways to measure a heat transfer is to measure a change in temperature and

then calculate from that, the heat transferred. We use the concept of Heat capacity to do this.Heat capacity of a substance describes the amount of heat that can be absorbed by the

substance for a given unit rise in temperature (unit = 1 K). It can be expressed in general as :

without some specific restrictions on the measurements.

In this case, we would need to drive our vehicle in astraight line over the surface of the earth, ignoring roads

lakes and hills (a "crow" could do it hence the term "asthe crow flies" when referring to distances).


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Heat capacity of a substance describes the amount of heat that can be absorbed by thesubstance for a given unit rise in temperature (unit = 1 K). It can be expressed in general as:

The Heat capacity can be expressed as a function of the amount of material. The heat

capacity per mole is called the molar heat capacity (units J K1

mol1

). The heat capacity per

gram is called thespecific heat capacity (or justspecific heat) (units J K1

g1

). It can simply

be the capacity of the unit object. For example, a calorimeter (made up of or containing

several materials, container, insulation, water, etc.) will have a heat capacity unique to it.

It is not completely correct to talk of heat capacity in terms ofq since q is not a state function.

We are better off using the state functions (Hrather than q.

We assume that no electrical or mechanical workw'is being done.

y At constant pressure, we measure heat as (H= qp,and hence the heat capacity weneed to use is Cp, where we define

Thus, we can write (at constantP)

(H= Cp(T

The units ofCp are J/K or J/C (remember that (Tis the same whether its measured in C or

in K). Sometimes, we tabulate heat capacities in per mole or per gram values. If this is thecase, we need to multiply the molar heat capacity by the number of moles or the gram heat

capacity (specific heat) by the number of grams to get the total heat capacity. In these cases,we can think of modifying our equation for enthalpy to be

(H= nCp(T (for molar heat capacity values)

OR

(H= mCp(T .(for specific heat values)

Obviously, these equations hold Cp to be independent of temperature. In reality, that's a

reasonable estimate for our purposes but is not strictly correct.

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Thermochemistry (calorimetry)


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Now, Let's quickly review a few definitions we will need to use in this section.

CalorimetryStudy of heat absorbed or evolved in chemical reactions

Calorimeter

Device used in Calorimetry to measure heat processes (normally thermally insulated

from the surroundings)Heat Capacity

Amount of heat to raise an object or a given amount of substance by 1C (1K).

Molar Heat Capacity

Amount of heat to raise one mole of a substance by 1C (1K).

(water has molar heat capacity of 75.4 J K1

mol1

.

Specific Heat

Amount of heat required to raise one gram by 1C (1K).

(water has specific heat of1 cal K1

g1

or 4.18 J K1

g1

Let's try an example:

A sample of 50. mL of a 0.20 Msolution of HCl was mixed with 50. mL of 0.20 MNaOH in

a coffee cup calorimeter. The initial temperature of both solutions was 22.2C. After mixing,

the temperature rose to 23.5C. What is the enthalpy change for the neutralization reaction

which occurred?

To Start, we needto define the system. Since this case involves a reaction in a liquidsolution

where the chemicals are intimately involvedwith the solvent (water), we cannot really

separate the waterfrom the chemicals. We use the chemicals andwateras the system. The

insulatedcup is the boundary between the system andthe surroundings andwe willassume

that the cup itselfabsorbs no heatandthat no heat passes through the cup to the

surroundings.

We have thus, a two step process to deal with.

1. The reaction occurs, warming the system. Since no heat was evolvedto thesurroundings (coffee cup insulation preventedit) the value ofqandof(Hare zero

since w is negligibly small (no volume change in a liquid). For that matter, (U is zerotoo (See InternalEnergy at the endofthis section).

2. The system must be returnedto the starting temperature by extracting heatfrom it oradding heat to it. Thus, the (T we calculate is ofthe step where the system is returned

to its original state.

H3O+(aq) + OH(aq) p 2 H2O(l)

Total volume = 100. mL, dilute aqueous so we assume density = 1.00 g/mLmass of solution: msol = 100. mL 1.00 g/mL = 100. g

For the cooling process:

(T= Tf Ti = 22.2C 23.5C = 1.3C (= 1.3 K)


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The temperature was loweredfrom it's high value backdown to its starting point. Thus, weexpecta negative temperature change. Another way ofviewing this: We removedheatfrom

the system to return it to its starting point. Hence, qand(Hare negative (the reaction isexothermic).

This temperature change is happening to the system.

qp = Cp(T =mCm(T(Cmis specific heat, i.e., heat capacity per gram)

NOTE: to figure out which equation you need simply look at the units of the quantities youhave and figure out how to cancel out the undesired units by the correct combination of C

with the values you have. In this case, we have

qp = 4.18 J K-1g-1 100. g -1.3 K= -540. J

# molH3O+ = # molHCl = 0.50 L 0.20 M = 0.010 mol.

(H' = qp = -540 J

Note that the (H' shown here is for the whole reaction amount. It is nota molaramount;hence, the prime on the (H'. I will use the prime to distinguish the 'extrinsic'quantity andno

prime to show an 'intrinsic'quantity (molar).

-540 J

(H = ---------- = -54. kJ/molH 3O+

0.010 mol

equation has 1 molH3O+ so we can write (H= -54. kJ/mol reaction.

1 mol reactionor it might also look like:

H3O+(aq) + OH-(aq) ---> H2O(l) (H= -27. kJ/mol reaction

Our test calorimeter is not that accurate (significant heat loss to the surroundings). The true

(Hfor the first reaction is -56.02 kJ/mol and is valid for the heat of neutralization of any

strong acid with a strong base in water.

Here's another example:

0.510 g of ethanol is burned in a flame calorimeter containing 1200 g of water. The water is

initially at 22.46C and is warmed up to 25.52C as a result of the reaction. What is the (H

for one mole of ethanol?

In this case, we can more easily separate the gas reaction mixture from the water in thecalorimeter which absorbs (all) the heatfrom the reaction. Therefore, we can use the

chemicals themselves as the system. In this case, the insulation in the calorimeter serves to

isolate a small portion ofthe surroundings (the water) so we can measure its temperature

change.

C2H5OH(l) + 3 O2(g) ---> 2CO2(g) + 3 H2O(l)

(Twater = 25.52C - 22.46C = 3.06C = 3.06 K


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we have grams ofwaterso usea Cwhich has gramsto cancel out.

This is the total heatabsorbedby the waterandis the negative ofthe heat evolvedby thesystem. Hence,

(H' = qp = -15.3 kJ0.510 g ethanol = 0.0111 mol ethanol 46.05 g/mol

(H = qp= -15.3 kJ = -1380 kJ/mol ethanol0.0111 mol ethanol

(1 mol ethanol in 1 mole reaction)qp = -1380 kJ/mol ethanol or simply -1380 kJ/mol.

Note this reaction is written such that one mole of a reactant is burned completely in oxygen.

This type of reaction is called a combustion reaction and the (His called a heat of

combustion. We could simply tabulate (Hcomb and know that the reaction is as written above.

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Standard Enthalpy Changes

Reaction energies depend on the conditions under which they were measured. In order to beable to record energies and tabulate them in a way that others can make sense of them, we

define a standard state. This standard state is then used to define the state of both the reactants

(before reaction) and products (after reaction).Recall that since (Hand(Udo notdependon

the path, only the initialandfinal state. Hence this is a useful thing to do.

Thermodynamic Standard state should not be confused with the standard state for IdealGases. They aren't the same.

Standard State:

Pressure = 100 kPa (1 bar)*

Concentration = 1 mol/L, (solutions)

[Temperature = Normally, 25 C, ]

We'll see later that a simpler definition for standard state is "activity = 1" (SeeNOTES 12.b)

Actually, Temperature is not part of the definition. It is merely the experimental temperature

for which most of the results are tabulated.

We indicate Thermodynamic quantities measured at standard conditions by using the

superscript zero (sometimes pronounced "naught") as in (H.

eg. CH4(g) + 2 O2(g) ---> CO2(g) + 2 H2O(l) (H = -890.4 kJ/mol


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NOTE:allthereactantsandproductsare intheirstandardstate .

Reactions may well occur at temperatures far different from Standard State but we always

start with the reactants at Std State and then perform the reaction at whatever conditions

desired, then return the products to their standard state, all the while, recording carefully theamount of energy used in each of the steps. The overall effect is the standard enthalpies (or

internal energies) of reaction. This is valid, of course, since H is a state function.

There are several ways of recording standard enthalpies of reaction. Recall that the enthalpy

of a reaction depends on how it is written, (if all coefficients are doubled, so is the (Hof

reaction. Hence, we either need to include the complete balance chemical reaction with the

tabulated values of(Hor define a 'standard' way of writing the equations. The latter is theway we do it.

We've already seen two such reactions and their enthalpy changes. Formation reactions are

written by default such that the compound of interest is formed from its constituent elements

in their standard states. (see the formation of water reaction:

H2(g) + O 2(g) ---> H2O(l) (Hf = -286. kJ/mol

We could also specify a reaction where the compound of interest is a reactant consumed

completely, for example in a combustion with excess oxygen.

C2H6(g) + 7/2 O2(g) ---> 2 CO2(g) + 3 H2O(l) (Hcomb = -1560.4 kJ/mol

Since both of these types of reactions are now well defined, we can re-create the chemical

equation exactly knowing only the compound of interest and the type of reaction. Hence, the

values can be tabulated without writing the complete reaction.

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Hess' Law

Since enthalpy is a state function, we can go to products via several routes and the enthalpy

change will still be the same.

or

The enthalpy change in a reaction does not depend on the reaction pathway.

Let's look at the example of ethane burning in excess oxygen to produce carbon dioxide and

water.


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We can consider the reaction to occur in one complete process [T] or in three distinct steps

[1], [2] and [3]. In either case, the total enthalpy change should be the same.

[1] Ethane is heated to drive off H 2.

(C2H6 ---> C2H4 + H2(H1 = +136.2 kJ

[2] Ethene is burned in excess O 2.

C2H4 + 3 O2 ---> 2 CO2 + 2 H2O (H2 = -1410.8 kJ[3] H2 is burned in excess O 2.

H2 + 1/2 O2 ---> H2O (H3 = -285.8 kJ[T] overall reaction is sum [1]+[2]+[3] ____________________

C2H6 + 3.5 O2 ---> 2 CO2 + 3 H2O (HT = -1560.4 kJ

We can use this idea to calculate enthalpies of reactions for reactions where measurement is

impractical.

Example: Use the following standard enthalpies of combustion to calculate the enthalpy

change for the following reaction

C(s) + 2 H2(g)---> CH4(g) (H= ?

1) (Hcomb(C) = -393.5 kJ

2) (Hcomb(H2) = -285.8 kJ

3) (Hcomb(CH4) = -890.4 kJ

Knowing the standard format for the combustion reaction (1 mol of reactant burns to

completion in oxygen to produce carbon dioxide and water) we can easily write down thethree reactions.

1) C(s) + O 2(g) ---> CO2(g) (H = -393.5 kJ/mol

2) H2(g) + O 2(g) ---> H2O(g) (H = -285.8 kJ/mol

3) CH4(g) + 2 O2(g) ---> CO2(g) + 2H2O(l) (H = -890.4 kJ/mol


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Now lets add up the three reactions to give the desired reaction

1) C(s) + O 2(g) ---> CO2(g) (H = -393.5 kJ/mol

2)2[H2(g) + O2(g) ---> H2O(g) (H = -285.8 kJ/mol]

3) CO2(g) + 2H2O(l) ---> CH4(g) + 2 O2(g) (H = +890.4 kJ/mol ________________________________________________________________

C(s) + 2 H2(g) ---> CH4(g) (H = -74.7 kJ/mol

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Standard Enthalpies of Formation

Enthalpy change for the reaction in which one mole of a compound is formed from its

elements in their standard states.

Ex.

C(graphite) + O 2(g) ---> CO2(Hf = -393.5 kJ/mol

H2(g) + O 2 ---> H2O(l) (Hf = -285.8 kJ/mol

C(graphite) + 2H 2(g) ---> CH4(g) (Hf = -74.7 kJ/mol

Question: What will be the standard enthalpy of formation of an element in its standard state

[like O2(g)]?

Answer: zero.

Now let's use these with Hess' law to determine the reaction enthalpy for the following

CH4(g) + 2 O2(g) ---> CO2(g) + 2H2O(l) (H1(combustion of methane)

We could chose to break down the reactants to their constituent elements in their standard

state and then reconstruct them into the products.

CH4(g) + 2 O2(g) ---> C(s) + 2H2(g) + 2O2(g) (H2

C(s) + 2H2(g) + 2O2(g) ---> CO2(g) + 2H2O(l) (H3

From Hess' law, we have, (H1 =(H2 + (H3

Notice that the following is true:

(H2 = - sum of Std enthalpies of formation of reactants

(H3 = + sum of Std enthalpies of formation of products

So now, we can write:

For the combustion of methane we're dealing with here, we can write

(H1 = [(Hf(CO2) + 2 (Hf(H2O)][(Hf(CH4) + 2 (Hf(O2)]


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We can look up the values forHeat offormation in any standardreference book.

(H1 = [(393.5 kJ) + 2 (285.8 kJ)] [(74.7kJ) + 2 (0)]

(H1 = 890.4 kJ

In general, we can write:

(H =(Hf(prod)(Hf(reactants)

A more completely mathematically correct equation is written as follows

This (these) reaction is merely a special application of Hess' Law whereby we are using Heats

of formation as the heats to add up.

What are elements in their standard states? How do we decide which ones to use?

Take hydrogen:

Most hydrogen exists at standard conditions as H2(g) [minute amounts may exist as

H(g)] so we choose the most common form at normal (standard) conditions.

Now try Carbon:

Obviously, carbon is solid at Std Conditions but do we choose graphite, diamond,

bucky balls (Buckminster Fullerenes)?

Graphite is the defined standard, it is more common and more stable than diamond.

Let's try a few examples to get some practice with these calculations:

What is the standard enthalpy change for

4 NH3(g) + 5 O2(g) ---> 4NO(g) + 6H 2O(l) ?

(H = (Hf(prod) - (Hf(reactants)(H = [4(H(NO) + 6(H(H2O)] - [4(H(NH3) + 5(H(O2)]

(H = [4(90.25 kJ/mol) + 6( -285.83 kJ/mol)] - [4(-46.11 kJ/mol) + 5 (0)]

(H = -1169.54 kJ/mol

Now here's another one.

B5H9(g) reacts exothermically with O2 to form B2O3(s) and water. What is the standard

enthalpy change for the reaction of1 mol of B5H9(g)?

balance this B 5H9(g) + O2(g) ---> B2O3(s) + H2O(l)2 B5H9(g) + 12 O2(g) ---> 5 B2O3(s) + 9 H2O(l)


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for 1mol 1 B 5H9(g) + 6 O2(g) ---> 5/2 B2O3(s) + 9/2 H2O(l)

(H = (Hf(prod) - (Hf(reactants)(H = [5/2 (H(B2O3) + 9/2 (H(H2O)] - [(H(B5H9) + 6(H(O2)]

(H = [5/2 (-1272.77) + 9/2 ( -285.83)] - [(73.2) + 6(0)]

(H = -4541.4 kJ/mol

1 mol of benzene burns in air at standard state conditions and gives off 3267 kJ of heat. What

is the enthalpy of formation of C6H6?

C6H6(l) + 7.5O2(g) ---> 6CO2(g) + 3 H2O(l)

(Hcomb = -3267 kJ/mol

(Hcomb = (Hf(prod) - (Hf(reactants)-3267 kJ/mol = [6(Hf(CO2,g) + 3 (Hf(H2O,l)] -

[(Hf(C6H6,l) + 7.5 (Hf(O2,g)]Substitute in values from data table.-3267 kJ/mol = [6( -393.5) + 3 (-285.83)] -

[(Hf(C6H6,l) + 7.5 (0)]Solve for the only unknown:(Hf(C6H6,l) = 49 kJ/mol

One could look at the (Hf for a compound and tell if there is much chemical energy stored

in it or not. If the heat of formation is positive then energy was stored (less stable) in it's

production, if negative, it lost energy (more stable) when produced.

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Temperature change and Enthalpy

The standard enthalpy data we find tabulated in thermodynamic tables is all for data at 25C.

This data may not be exactly the information we need. What if a reaction doesn't actually

happen at standard 25C. We have two options: 1, we can assume that there is no significant

difference in enthalpy change or, 2, we can calculate the effect.

In cases where there is only small temperature changes the former option may not be too bad.

However, in some cases, especially where the temperature of reaction is far from 25C we

should take the latter option.

Consider a reaction between chemicals A and B to produce C and D at some temperature T.


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We could calculate the (Hfor the process 2 using standard thermodynamic data. Steps 1 and

2 involve temperature changes and we can calculate the (Hfor these steps if we know the

heat capacities of the compounds involved. According to Hess' law, the overall enthalpy

change for the reaction at temperature T is the sum of the steps 1, 2 and 3.

Step 1: This is simply a temperature change and we can calculate the enthalpy change using the heat

capacities of A and B.

(H1 = {aCP(A) + bCP(B)}(T1 (298 T)

Step 2: (H2 = (H298 (calculate from whatever means possible, for example you could use (Hf

values.)

Step 3: Like step 1, this is simply a temperature change. Calculate the enthalpy change using the

heat capacities of C and D. Note that the process is the reverse direction of step 1 and the

sign of the temperature change is opposite.

(H3 = {cCP(C) + dCP(D)}

(T3 (T 29

8) NOTE:

(T1

= (

T3

Step T: (HT =(H1 + (H298 + (H3

= {aCP(A) + bCP(B)}(T1 + (H298 + {cCP(C) + dCP(D)}(T3

= {cCP(C) + dCP(D) [aCP(A) + bCP(B)]}(T3 + (H298

= {(CP}(T3 + (H298

The change in heat capacity (capacity of products capacity of reactants) is a actually

slightly dependent on temperature and hence this formulation is not valid if the temperature T

is very different from standard temp. (298 K). In most cases, the temperature effect is not

extremely large but that is not general.

Like the general concepts involved in Hess' Law, we simply add up the heats of the processesinvolved in order to determine the overall heat. This is merely another special application of

Hess' Law.

Internal Energy

Internal energy of the system is represented by the symbol U. It is a more general description

of the energy of the system than enthalpy. We can never measure the absolute energy of a

system but we can measure changes in the internal energy of the system using the statefunction

(U = Ufinal - Uinitial.


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Since we can never actually measure eitherUinitialnorUfinal we measure, instead, the energytransferred into or out of the system in the form of heat and work to obtain.

(U = q + w ( = heat transferred + work done)

This is, in fact, the mathematical expression of the First Law of Thermodynamics

and is completely general, not relying on any of the restrictions we placed on enthalpycalculations. It turns out that while neitherq norw are state functions, their sum is.

Example:

If15 kJ of work is done by the surroundings on the system and the system looses 10 kJ heat

to the surroundings then

(U = -10 kJ + 15 kJ= +5 kJ

Chemical reactions often involve changes in Volume (exp. gases)

C3H8(g) + 5 O2(g) ---> 3 CO2(g) + 4H2O(g)6 mol gas 7 mol gas ====> increase in volume

at const. P

If this reaction is to happen, it needs to push back the rest of universe (atmosphere) to make

room for itself. That's a force(| pressure * area) * displacement = work.

work done =>p(V(since the system did the work, it's negative)

w = - Psurr(V (gas expanding against an external pressure Psurr.)

(U = q + w = q - Psurr(V

or qp =(U + Psurr(V

in this case, qp = (U + (ngRT (the only volume increase came from thefact

that the number of moles of gas isdifferent

in product than in reactants).

In a closed container (no volume change) we have

(V = 0 so:

(U = qv.

subscript v refers to constant volume conditions.

Enthalpy is now better defined as


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H= U + PV

Changes inenthalpy are

(H= (U + (PV since we normally work in a fixed pressure(atmosphere) situation, we can rewrite this as

(H= (U + P(V or we can write:

To do these measurements under general conditions, we need to expand a few previous

definitions. First, lets look at heat capacity.

y At constant pressure, we measure heat as (H= qp,and hence the heat capacity weneed to use is Cp, where we define

y At constant volume, the heat change we measure is (U = qv and the heat capacity CVis defined as

In many substances, the two heat capacities differ considerably. Consider ideal gases. We

have

(H(U = ((PV) = R(T

.

Therefore

CP CV = R.

Solids and liquids often have values ofCPand CVthat don't differ much. On the other hand

there is no consistent rule. Take for example water and benzene; for water, CP CV= 0.075R,

while for benzene we have CP CV= 5.1R.


17/17

let's do an example where we look at the PVwork done.

When 2.00 mol of SO2(g) react completely with 1.00 mol O2(g) to form 2.00 mol of SO3(g) at

25C and constant pressure of1.00 atm, 198 kJ of energy is released as heat. Calculate (U

and (Hfor this reaction.

2 SO2(g) + 1 O2(g) ---> 2 SO3(g)

The heat released is |qp| since it was measured under constant pressure conditions. so we can

quickly say

(H'= 198 kJ (negative sign since the heat is released, i.e., exothermic)

(H= 198 kJ/mol (trivial in this case since the molar quantities given exactly match the

numbers in the equation)

(U' = qp + w = qp P(V(total PVwork)

or (since we have the molar values)

(U = (H- P(V(remember, this is the PVwork per mole which just happens to equal thetotal PVwork in this case)

Since we have constant Pand Tconditions, the only change in P(Vwill be due to change in

numbers of moles of gas molecules. We can use the stoichiometry from the equation as

written to determine P(V:

(U = (H- (nRT or (H= (U + (nRT

Note: Either of these two equations above are valid for any situation wherethe TandPare constant but where the number of moles of gas moleculeschanges because of stoichiometry. Pick one and learn it.

(n = #moles prod gas #moles reactant gas = 2 3 = 1 mol.

(U = -198 kJ/mol + -(-1 * 8.31451 J K -1mol-1 * 298 K)= -198 kJ/mol + 2.48 kJ/mol = -196 kJ/mol

energy, enthalpy and thermochemistry

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