Thermo Chemistry I
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Readings for this section:

Petrucci: Chapter 7

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 heat

Work

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

 

One of the most common definitions for work is when a force f displaces an object by a distance Dx does work w' = f*Dx. This is a mechanical work.

Kinetic and Potential Energy

both of these can be forms of work and of heat energy

KE = 1/2 mv2         PE = mgh    lifting mass m by height h against             acceleration due to gravity g               = f·d     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 electrical work
(We include this type of work in w').  In first-year chemistry, we'll not often be interested 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 pressure P_atm and moves the piston a distance Dx. The force needed to push back the atmosphere is actually P_atm×A where A is the area of the piston. Hence, we have w = P_atm× A ×Dx = P_atm× DV.

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

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 N·m = kg·m²·s^-2 = Joules (J).

Heat

Energy transfer as a result of a temperature differential.

First Law of Thermodynamics:

Often called the Law of conservation of energy: 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.

Surroundings

the 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:

1. work w can be done on the system by the surroundings (or vice versa). It can take the form of mechanical work or of electrical energy transfer.
2. Heat q flows from the system to the surroundings (or vice 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 convention      
How 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 chemical experience.

We define a term called enthalpy H is the energy transferred between a system and the surroundings under constant pressure.  We cannot measure the absolute enthalpy of a system but we can measure the change in enthalpy for a process.  If the Initial and final states have enthalpies H_i and H_f, respectively then the change in enthalpy for the process can be defined as 

This definition is only good if H is a state function, i.e., the change in enthalpy depends only on the initial and final states, not on the process itself.  

If we simply measure the heat evolved (or absorbed) during a process, we will not have a measure 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 q_p we 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

at constant pressure  

For chemical reactions we can write  

 DH = SH_products - SH_reactants

If DH is negative, then net flow of heat from chemical system to surroundings (heat released). Exothermic (EXO sounds like exit, from the Latin: to leave)

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

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

H2(g) + I2(g) ----> 2 HI(g)              DH = 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 :

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:

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 K^–1mol^–1). The heat capacity per gram is called the specific heat capacity (or just specific heat) (units J K^–1g^–1). 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 of q since q is not a state function. We are better off using the state functions DH rather than q. 

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

• At constant pressure, we measure heat as DH = q_p, and hence the heat capacity we need to use is Cp, where we define

Thus, we can write (at constant P)
DH = C_pDT 

The units of C_p are J/K or J/ºC (remember that DT is the same whether its measured in ºC or in K).  Sometimes, we tabulate heat capacities in per mole or per gram values.  If this is the case, 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

DH = nC_pDT (for molar heat capacity values)

OR

DH = mC_pDT .(for specific heat values)

Obviously, these equations hold C_p 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)

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

Calorimetry

Study 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 1°C (1K).

Molar Heat Capacity

Amount of heat to raise one mole of a substance by 1°C (1K).
(water has molar heat capacity of 75.4 J K^-1·mol^-1 .

Specific Heat

Amount of heat required to raise one gram by 1°C (1K).
(water has specific heat of 1 cal K^-1·g^-1 or 4.18 J K^-1g^-1

Let's try an example:

A sample of 50. mL of a 0.20 M solution of HCl was mixed with 50. mL of 0.20 M NaOH in a coffee cup calorimeter. The initial temperature of both solutions was 22.2°C. After mixing, the temperature rose to 23.5°C. What is the enthalpy change for the neutralization reaction which occurred?

To Start, we need to define the system. Since this case involves a reaction in a liquid solution where the chemicals are intimately involved with the solvent (water), we cannot really separate the water from the chemicals. We use the chemicals and water as the system. The insulated cup is the boundary between the system and the surroundings and we will assume that the cup itself absorbs no heat and that 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 evolved to the surroundings (coffee cup insulation prevented it) the value of q and of DH are zero since w is negligibly small (no volume change in a liquid). For that matter, DU is zero too (See Internal Energy at the end of this section).
2. The system must be returned to the starting temperature by extracting heat from it or adding heat to it. Thus, the DT we calculate is of the step where the system is returned to its original state.

H₃O⁺(aq) + OH^–(aq) ——¾®  2 H₂O(l)

Total volume = 100. mL, dilute aqueous so we assume density = 1.00 g/mL
mass of solution: m_sol = 100. mL × 1.00 g/mL = 100. g

For the cooling process:

DT = T_f – T_i = 22.2°C – 23.5°C = –1.3°C (= –1.3 K)

The temperature was lowered from it's high value back down to its starting point. Thus, we expect a negative temperature change. Another way of viewing this: We removed heat from the system to return it to its starting point. Hence, q and DH are negative (the reaction is exothermic).

This temperature change is happening to the system.

q_p = C_pDT = mC_m DT (C_m is specific heat, i.e., heat capacity per gram)

NOTE: to figure out which equation you need simply look at the units of the quantities you have 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

# mol H3O+ = # mol HCl = 0.50 L × 0.20 M = 0.010 mol.
DH' =  qp =  -540 J

Note that the DH' shown here is for the whole reaction amount. It is not a molar amount; hence, the prime on the DH'. I will use the prime to distinguish the 'extrinsic' quantity and no prime to show an 'intrinsic' quantity (molar).

      -540 J
DH = ---------- = -54. kJ/mol H3O+
      0.010 mol

equation has 1 mol H3O+      so we can write    DH =  -54. kJ/mol reaction.
             1 mol reaction

or it might also look like:

½ H3O+(aq) + ½ OH-(aq) --->  H2O(l)       DH = -27. kJ/mol reaction

Our test calorimeter is not that accurate (significant heat loss to the surroundings). The true DH for 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 22.46°C and is warmed up to 25.52°C as a result of the reaction. What is the DH for one mole of ethanol?

In this case, we can more easily separate the gas reaction mixture from the water in the calorimeter which absorbs (all) the heat from 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 of the surroundings (the water) so we can measure its temperature change.

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

DTwater = 25.52°C - 22.46°C = 3.06°C = 3.06 K

we have grams of water so use a C which has grams to cancel out.

This is the total heat absorbed by the water and is the negative of the heat evolved by the system. Hence,

DH' = qp = -15.3 kJ 

0.510 g ethanol  = 0.0111 mol ethanol
46.05 g/mol 

DH = qp =    -15.3 kJ        = -1380 kJ/mol ethanol 
          0.0111 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 DH is called a heat of combustion. We could simply tabulate DH_comb 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 be able 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 DH and DU do not depend on the path, only the initial and final state. Hence this is a useful thing to do.

Thermodynamic Standard state should not be confused with the standard state for Ideal Gases. 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" (See NOTES 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 DH°.

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

NOTE: all the reactants and products are in their standard state.  

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 the amount 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 DH of reaction. Hence, we either need to include the complete balance chemical reaction with the tabulated values of DH or define a 'standard' way of writing the equations. The latter is the way 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) + ½ O2(g) --->  H2O(l)       DH°f = -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)    DH°comb = -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.

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 H2.
            D
      C2H6 ---> C2H4 + H2                   DH°1 =  +136.2 kJ

[2] Ethene is burned in excess O2.

      C2H4 + 3 O2 ---> 2 CO2 + 2 H2O        DH°2 = -1410.8 kJ

[3] H2 is burned in excess O2.

      H2 + 1/2 O2 ---> H2O                  DH°3 =  -285.8 kJ

[T] overall reaction is sum [1]+[2]+[3]   ____________________

     C2H6 + 3.5 O2 --->  2 CO2 + 3 H2O      DH°T = -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)         DH = ?  

1) DH°comb(C)   = -393.5 kJ
2) DH°comb(H2)  = -285.8 kJ
3) DH°comb(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 the three reactions.

1)  C(s) + O2(g) ---> CO2(g)                DH° = -393.5 kJ/mol
2)  H2(g) + ½ O2(g) ---> H2O(g)             DH° = -285.8 kJ/mol
3)  CH4(g) + 2 O2(g) ---> CO2(g) + 2H2O(l)  DH° = -890.4 kJ/mol
Now lets add up the three reactions to give the desired reaction

1)   C(s) + O2(g) ---> CO2(g)                DH° = -393.5 kJ/mol
2)2×[H2(g) + ½ O2(g) ---> H2O(g)             DH° = -285.8 kJ/mol]
3)  CO2(g) + 2H2O(l) ---> CH4(g) + 2 O2(g)   DH° = +890.4 kJ/mol
________________________________________________________________
      C(s) + 2 H2(g)  ---> CH4(g)            DH° = -74.7 kJ/mol

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Standard Enthalpies of Formation

Þ Enthalpy change for the reaction in which one mole of a compound is formed from its elements in their standard states.

Ex.

C(graphite) + O2(g) ---> CO2                DHf° = -393.5 kJ/mol
H2(g) + ½ O2 ---> H2O(l)                    DHf° = -285.8 kJ/mol
C(graphite) + 2H2(g) ---> CH4(g)            DHf° = -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)       DH1°
(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)    DH2°
C(s) + 2H2(g) + 2O2(g) ---> CO2(g) + 2H2O(l)     DH3°
From Hess' law, we have,    DH1° = DH2° + DH3°

Notice that the following is true:

DH2° = - sum of Std enthalpies of formation of reactants
DH3° = + 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

DH₁° = [DH_f°(CO₂) + 2 DH_f°(H₂O)] – [DH_f°(CH₄) + 2 DH_f°(O₂)]

We can look up the values for Heat of formation in any standard reference book.  

DH₁° = [(–393.5 kJ) + 2 (–285.8 kJ)] – [(–74.7kJ) + 2 (0)]

DH₁° = –890.4 kJ

In general, we can write:

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) + 6 H2O(l) ?

DH° = åDHf°(prod) - åDHf°(reactants)
DH° = [4 DH°(NO) + 6 DH°(H2O)] - [4 DH°(NH3) + 5 DH°(O2)] 
DH° = [4 (90.25 kJ/mol) + 6 (-285.83 kJ/mol)] - [4 (-46.11 kJ/mol) + 5 (0)] 
DH° = -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 of 1 mol of B₅H₉(g)?

balance this     B5H9(g) +    O2(g) --->   B2O3(s) +   H2O(l)
               2 B5H9(g) + 12 O2(g) ---> 5 B2O3(s) + 9 H2O(l)
for 1mol       1 B5H9(g) +  6 O2(g) ---> 5/2 B2O3(s) + 9/2 H2O(l)

  DH° = åDHf°(prod) - åDHf°(reactants)
  DH° = [5/2 DH°(B2O3) + 9/2 DH°(H2O)] - [DH°(B5H9) + 6 DH°(O2)]
  DH° = [5/2 (-1272.77) + 9/2 (-285.83)] - [(73.2) + 6 (0)]
  DH° = -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) ---> 6 CO2(g) + 3 H2O(l)

      DH°comb = -3267 kJ/mol
      DH°comb =  åDHf°(prod) - åDHf°(reactants)
-3267 kJ/mol = [6 DHf°(CO2,g)  + 3 DHf°(H2O,l)] - 
               [DHf°(C6H6,l) + 7.5 DHf°(O2,g)]

Substitute in values from data table.  

-3267 kJ/mol = [6 (-393.5)  + 3 (-285.83)] - 
               [DHf°(C6H6,l) + 7.5 (0)]

Solve for the only unknown:  

DHf°(C6H6,l) = 49 kJ/mol

One could look at the DH_f° 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 25ºC. This data may not be exactly the information we need. What if a reaction doesn't actually happen at standard 25ºC. 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 25ºC we should take the latter option.

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

We could calculate the DH for the process 2 using standard thermodynamic data. Steps 1 and 2 involve temperature changes and we can calculate the DH for 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.

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 processes involved 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 state function 

 DU = U_final - U_initial.

Since we can never actually measure either U_initial nor U_final we measure, instead, the energy transferred into or out of the system in the form of heat and work to obtain.

DU = 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 enthalpy calculations.  It turns out that while neither q nor w are state functions, their sum is.

Example:

If 15 kJ of work is done by the surroundings on the system and the system looses 10 kJ heat to the surroundings then

DU = -10 kJ + 15 kJ
   = +5 kJ
    
    

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

C3H8(g) + 5 O2(g) ---> 3 CO2(g) + 4 H2O(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(º pressure * area) * displacement = work.

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

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

DU = q + w = q - PsurrDV

or qp = DU + PsurrDV 

in this case, qp = DU + DngRT   (the only volume increase came from the fact 
                                 that the number of moles of gas is different 
                                 in product than in reactants).

In a closed container (no volume change) we have

 DV = 0      so:

subscript v refers to constant volume conditions.
  

Enthalpy is now better defined as

H = U + PV

Changes in enthalpy are

DH = DU + DPV  since we normally work in a fixed pressure
                (atmosphere) situation, we can rewrite this as

DH = DU + PDV   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.

• At constant pressure, we measure heat as DH = q_p, and hence the heat capacity we need to use is C_p, where we define

• At constant volume, the heat change we measure is DU = q_v and the heat capacity C_V is defined as

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

DH – DU = D(PV) = RDT

.

Therefore

Solids and liquids often have values of C_P and C_V that don't differ much. On the other hand there is no consistent rule. Take for example water and benzene; for water, C_P – C_V = 0.075R, while for benzene we have C_P – C_V = 5.1R.

let's do an example where we look at the PV work 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 25°C and constant pressure of 1.00 atm, 198 kJ of energy is released as heat. Calculate DU and DH for this reaction.

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

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

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

DH = –198 kJ/mol (trivial in this case since the molar quantities given exactly match the numbers in the equation)

DU' = q_p + w = q_p – PDV (total PV work)

or (since we have the molar values)

DU = DH - PDV (remember, this is the PV work per mole which just happens to equal the total PV work in this case)

Since we have constant P and T conditions, the only change in PDV will be due to change in numbers of moles of gas molecules. We can use the stoichiometry from the equation as written to determine PDV :

Dn = #moles prod gas – #moles reactant gas = 2 – 3 = –1 mol.

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

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Prof. Michael J. Mombourquette.
Copyright © 1997
Revised: November 25, 2008.

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