Topic 6 Thermochemistry

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Topic 6

Thermochemistry

Nearly all chemical reactions involve either the release or the absorption of heat which is one form of energy. Heat is involved in reactions by either going into or out of the reaction system and is one aspect of a reaction that can be monitored.

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ThermochemistryThermochemistry is the study of the energy effects that accompany chemical reactions.

Why do chemical reactions occur? What is the driving force of rxn?

Answer: stability: reaction (rxn) wants to get to lower E. For a rxn to take place spontaneously, the products of reaction must be more stable (lower E) than the starting reactants. Nonspontaneous reaction means the rxn will never happen by itself; the reaction can happen, but it will need external help.

E

R

P

rxn itself releases E; therefore, rxn is spon

reactants at higher E are less stable; therefore, more reactive

E R

P

rxn must absorb E from external source; therefore, rxn is nonspon

reactants at lower E are more stable; therefore, less reactive

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Energy

Energy is defined as the capacity to move matter (“do work”). The concept of energy gives us an alternative way to describe motion.

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EnergyThere are three broad concepts of energy:

– Kinetic Energy, Ek, is the energy associated with an object by virtue of its motion.

– Potential Energy, Ep is the energy an object has by virtue of its position in a field of energy.

– Internal Energy, U, is the sum of the kinetic and potential energies of the particles making up a substance.

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Energy

The total energy of a system is the sum of its kinetic energy (Ek), potential energy (Ep), and internal energy (U).

UEEE pktot

Typically in a laboratory, the experimental vessel is at rest making the kinetic and potential energies contribution to the total energy zero.

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Energy

The Law of Conservation of Energy: Energy may be converted from one form to another, but the total quantities of energy remain constant.

Energy may be transferred or converted but not destroyed.

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Reaction EnthalpyIn chemical reactions, heat is often transferred from the “system or reaction” to its “surroundings,” or vice versa. system is the substance or mixture of substances under study in which a change occurs.surroundings are everything in the vicinity of the thermodynamic system (room, flask, solvent).

system or rxn

surroundings

+ into system- out system

We tend to measure the change in internal energy of the system. The sign (+ or -) for energy

indicates the direction of flow meaning into or out of the system.

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Heat of ReactionHeat flow is defined as the energy that flows into or out of a system. We follow heat flow by watching the difference in temperature between the system and its surroundings.

Often we follow the surroundings temp (solvent) and must realize that the opposite is happening to the system. If the system is absorbing heat (+) from the surroundings than the temp of the surroundings must be decreasing.

Tsystem (+) Tsurr (-)

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Heat of ReactionHeat flow or heat of reaction is denoted by the symbol q and is the amount of heat required to return a system to the given temperature at the completion of the reaction.

For an endothermic rxn, the sign of q is positive; heat is absorbed by the system from the surroundings.

E

P

system absorbs heat from surroundings; nonspon (endo)

R

Dq > 0

Surroundings

+q

Tsystem

System

Tsurr

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Heat of Reaction

Dq < 0

-q

System

Surroundings

For an exothermic rxn, the sign of q is negative; heat is evolved (released) by the system to the surroundings.

Tsystem

E

R

P

system releases heat to surroundings; spon (exo)

Tsurr

HW 50code: exo

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Enthalpy and Enthalpy ChangeThe heat absorbed or evolved by a reaction depends on the conditions under which it occurs. ex. constant pressure

Usually, a reaction takes place in an open vessel, and therefore under the constant pressure of the atmosphere.

heat of this type of reaction is denoted qp; this heat at constant pressure is named enthalpy and given the symbol H. H is the heat flow at constant pressure.

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Extensive property - depends on the quantity of substance (i.e. mass, volume, heat); contrary to intensive property that does not depend on amount of substance (i.e. color, boiling point).

Enthalpy is a state function. State functions are properties associated with a system in equilibrium; for any given state, a state function is uniquely defined (i.e. T, P, V, E). If a system undergoes a change (from one equilibrium state to another), and X is a state function, then DX (the change in the value of X) is the same regardless of how the changed occurred. It only depends on the final state (Xf) and the initial state (Xi) .

Enthalpy and Enthalpy ChangeEnthalpy, denoted H, is an extensive property of a substance that can be used to obtain the heat absorbed or evolved in a chemical reaction at constant pressure.

ooo

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The reaction enthalpy, DHrxn, for a reaction carried out at constant pressure is the enthalpy change associated with the conversion of reactants to products (at the same temperature).

)reactants((products) HHH D

Enthalpy and Enthalpy Change

)i((final) nitialHHH D

-DH is associated with an exothermic rxn (releases heat)+DH is associated with an endothermic rxn (aborbs heat)

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As we already stated the reaction enthalpy is equal to the heat of reaction at constant pressure, qp. This represents the entire change in internal energy (DU) minus any expansion “work” done by the system; therefore, we can define enthalpy and internal work by the 1st law of thermodynamics:

In any process, the total change in energy of the system, DU, is equal to the sum of the heat

absorbed, q, and the work, w, done by the system.

DU = qp + w = DH + w

Enthalpy and Enthalpy Change

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Changes in E manifest themselves as exchanges of energy between the system and surroundings.

These exchanges of energy are of two kinds: heat and work that must be accounted.Heat is energy that moves into or out of a system because of a temperature difference between system and surroundings.

Work is the energy exchange that results when a force F moves an object through a distance d; work (w) = Fd

In chemical systems, work is defined as a change in volume at a given pressure, that is

VPw D

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The negative sign is present in the equation to keep the sign correct in terms of system. For expansion work, DV will be a positive value; however, expansion involves the system doing work on the surroundings and a decrease in internal energy (negative sign keeps overall work term a negative value). For contraction work, DV will be a negative value; however, contraction involves the surroundings doing work on the system and an increase in internal energy (negative sign keeps overall work term a positive value).

Giving us the 1st law of thermo in its more useful form:

VPHU DDDnote:

system absorbs heat (+)

system releases heat (-)

surroundings doing work on system (+)

system doing work on surroundings (-)

VPw D

HW 51code: energy

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Thermochemical EquationsA thermochemical equation is the chemical equation for a reaction (including physical states {important}) in which the equation is given a molar interpretation, and the reaction enthalpy for these molar amounts is written directly after the equation.

kJ -91.8H );g(NH2)g(H3)g(N 322 D

Note: physical states are important because the DH for the formation of H2O (l) from H2 (g) and O2 (g) is not the same as for the formation of H2O (g).

Enthalpy is an extensive property. This means that the value of reaction enthalpy is directly proportional to the amount of reactants and products involved in the reaction. This equation basically gives a recipe of what and how much to mix (1 mole N2 and 3 mols H2) to produce 2 mols NH3 and release 91.8 kJ of heat. Note that a negative value of DH indicates that the reaction is exothermic; it releases heat. Since H is an extensive property, if the recipe is varied, the amount of heat generated will also vary (i.e. only used ½ mole of N2 gas, only ½ of -91.8 kJ of heat would be released).

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Since enthalpy is an extensive property, we can set up conversion factors based on the coefficients and DH values in a thermochemical equation and perform calculations similar to stoichiometric calculations.

Thermochemical Equations

kJ -91.8H );g(NH2)g(H3)g(N 322 D

−𝟗𝟏 .𝟖𝒌𝑱𝟏𝒎𝒐𝒍𝑵𝟐

−𝟗𝟏 .𝟖𝒌𝑱𝟑𝒎𝒐𝒍𝑯𝟐

−𝟗𝟏 .𝟖𝒌𝑱𝟐𝒎𝒐𝒍 𝑵𝑯𝟑

Note: the minus sign is not a “negative sign”; it indicates heat is being released by the system.

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If the DH for a reaction is x, then the DH for the reverse reaction is –x. If a reaction is exothermic, its reverse reaction is endothermic and vice versa; magnitude of DH is the same, but the algebraic sign is reversed.

When a thermochemical equation is multiplied by any factor, the value of DH for the new equation is obtained by multiplying the DH in the original equation by that same factor.

Thermochemical Equations

kJ 967.4 H ; )(4)(2)(4

kJ 483.7- H ; )(2)()(2o

222

o222

D

D

gOHgOgH

gOHgOgH

kJ 483.7 H ; )()(2)(2

kJ 483.7- H ; )(2)()(2o

222

o222

D

D

gOgHgOH

gOHgOgH exo

endo

Note: if DH has a degree symbol (o) like it does in the examples above (DHo), it indicates thermo standard conditions at 25oC (298K) and 1 atm.

equation multiplied by factor of 2

equation reversed

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Applying Stoichiometry and Heats of ReactionsConsider the reaction of methane, CH4, burning in the presence of oxygen at constant pressure. Given the following equation, how much heat could be obtained by the combustion of 10.0 g CH4? 10.0 g O2? 10.0 g of each?

);l(OH2)g(CO)g(O2)g(CH 2224 kJ -890.3Ho D

HW 52

We will determine the amount of heat by performing a stoichiometry calculation similar to what we did in earlier sections except we will use a conversion factor involving heat. We will convert mass to mols and then convert mols to kJ.

= -556 kJ

= -139 kJ

The second part of this problem will be conducted the same way but using O2 factors.

The third part involves a limiting reagent problem. We must determine which reactant will be exhausted first because it will dictate the overall heat involved in the reaction. Since O2 is generating less kJ of heat, it is the limiting reagent meaning -139 kJ is also the answer to part 3.

code: stoich

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To determine how heats of reactions are measured, we must examine the heat required to raise the temperature of a substance because a thermochemical measurement is based on the relationship between heat and temperature change.

Measuring Heats of Reaction

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– The heat capacity, C, of a sample of substance is the quantity of heat required to raise the temperature of the sample of substance one degree Celsius (kJ/oC, cal/oC).

– Changing the temperature of the sample requires heat equal to the following:

Heat Capacity and Specific Heat

TCq D

Measuring Heats of Reaction

kJ = x oC

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A Problem to ConsiderSuppose a piece of iron has a heat capacity of 6.70 J/oC. The quantity of heat required to raise the temperature of the piece of iron from 25.0oC to 35.0oC is:

Basically, we are trying to determine the heat associated with the temperature increase of our iron system. We know the heat capacity of iron in J/oC and the temperature change; therefore, we can calculate the answer by following units or plugging into the previous equation:

TCq D ¿ (𝟔.𝟕𝟎 𝑱𝑪❑

𝒐 ) (𝟑𝟓 .𝟎−𝟐𝟓 .𝟎 𝑪❑𝒐 )=𝟔𝟕 .𝟎 𝐉

DT = Tf - Ti

Note: The positive sign on heat indicates that the iron (system in this problem) is absorbing the heat.

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Measuring Heats of ReactionHeat capacities are also compared for one gram amounts of substances. The specific heat capacity, s, (or “specific heat”) is the heat required to raise the temperature of one gram of a substance by one degree Celsius ().

Tmsq D

To find the heat required you must multiply the specific heat, s, of the substance times its mass in grams, m, and the temperature change, DT.

𝑱=( 𝑱𝒈 𝐶❑

𝑜 )(𝒈 ) (𝑻 𝒇−𝑻 𝒊 𝑪❑𝒐 )

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A Problem to ConsiderCalculate the heat when the temperature of 15.0 grams of water is raised from 20.0oC to 50.0oC. (The specific heat of water is 4.184 J/g.oC.)

HW 53

This problem involves calculating the heat associated with a mass of water being increased in temperature. You can follow the units on the specific heat of water and cancel them to solve for J or use the equation on the previous slide:

Tmsq D

𝒒=(𝟒 .𝟏𝟖𝟒 𝑱𝒈 𝑪❑

𝒐 )(𝟏𝟓 .𝟎𝒈 )(𝟓𝟎 .𝟎−𝟐𝟎 .𝟎 𝑪❑𝒐 )=¿𝟏 .𝟖𝟖𝒙𝟏𝟎𝟑 𝑱

Tf - Ti

Note: Positive heat indicates an endothermic reaction; also note in this problem water is the system. code: heat

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Heats of Reaction: CalorimetryA calorimeter is a device used to measure heats of reactions. The apparatus is designed with thermally insulated walls.

If a temperature change of DT is observed, then the heat flow that would be needed to reverse the temperature change is (Ccal)(-DT), where Ccal is the heat capacity and all of its contents (after the reaction has occurred). Therefore:

𝒒𝒓𝒙𝒏=−𝑪𝒄𝒂𝒍∆𝑻

Or another way to look at it is

𝒒𝒓𝒙𝒏=−𝒒𝒄𝒂𝒍

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Heats of Reaction: Calorimetry

An important concept to remember is that we typically are looking for the heat of a reaction (system) but often measure data involving the surroundings/calorimeter. This means we will need to change the sign of the heat value calculated to determine the heat of the reaction (system).

𝒒𝒓𝒙𝒏=−𝑪𝒄𝒂𝒍∆𝑻

(

(

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A Problem to ConsiderWhen 23.6 grams of calcium chloride, CaCl2, was dissolved in water in a calorimeter, the temperature rose from 25.0oC to 38.7oC. If the heat capacity of the solution and the calorimeter is 1258 J/oC, what is the enthalpy change per mole of calcium chloride?

First, this problem involves plugging the proper values into the equation to determine the ratio of heat generated in this reaction:

𝑞𝑟𝑥𝑛=−125 8 𝐽

𝐶❑𝑜 × (3 8.7−25. 0𝑜𝐶 )=−1.72 𝑥104 𝐽

This is the amount of heat released when 23.6 g of calcium chloride are dissolved, not per one mole of calcium chloride as requested in problem.

❑𝟐𝟑 .𝟔𝒈𝑪𝒂𝑪𝒍𝟐

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A Problem to ConsiderWhen 23.6 grams of calcium chloride, CaCl2, was dissolved in water in a calorimeter, the temperature rose from 25.0oC to 38.7oC. If the heat capacity of the solution and the calorimeter is 1258 J/oC, what is the enthalpy change per mole of calcium chloride?

HW 54

This problem is looking for a ratio that will give us a conversion factor that can be used to calculate the amount of heat associated with different quantities of calcium chloride in future experiments. Basically, we will convert g to moles giving us a conversion factor between heat and mols.

(−1.72 𝑥104 𝐽23.6𝑔𝐶𝑎𝐶𝑙2

)( 111.1𝑔𝐶𝑎𝐶𝑙2

1𝑚𝑜𝑙𝐶𝑎𝐶 𝑙2 )=−8.10𝑥 104 𝐽𝑚𝑜𝑙𝐶𝑎𝐶𝑙2

This factor can be use to calculate the heat released for any amount of calcium chloride.

code: calor

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Hess’ law states when a reaction can be expressed as the algebraic sum of two or more reactions, then the DH for the reaction is the algebraic sum of the DH of the reactions added.

Basically, R & P in individual steps can be added like algebraic quantities in determining overall equation and enthalpy change for that reaction.

Hess’ LawReaction enthalpies can be determined using calorimetry only for reactions that go to completion, are fast, and have no side reactions. For reactions that do not meet these criteria, the reaction enthalpy can be determined indirectly using Hess’ Law.

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Given: 1) A + D E + C DH = X kJ

2) 2A + B 2C DH = Y kJ

2A + 2D 2E + 2C DH = 2X kJ

Determine the enthalpy change for the equation below by using given equations.  2D B + 2E DH = ?

Typically, we look for a species in the desired equation that occurs in only one of the given equations and manipulate the equation to place the species on the correct side and correct number of moles. If we examine the desired equation, we will notice that D occurs in only equation 1. We ask ourselves two questions to determine how to manipulate equation 1: First, is the species on the correct side of the equation? In this case, D is on the correct side in equation 1 as compared to the desired equation; therefore, we will use the equation just as written. Second, does the given equation have the correct number of moles of the species as in the desired equation? In this case, D needs 2 moles and it has one mole in equation 1; therefore, we will multiply the whole reaction including DH by 2:

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Now we look for a species in equation 2 that we can examine and do the same process; we see that B only occurs in equation 2. We ask ourselves two questions to determine how to manipulate equation 2: First, is the species on the correct side of the equation? In this case, B is not on correct side in equation 2 as compared to the desired equation; therefore, we must reverse the equation and change the sign of DH. Second, does the given equation have the correct number of moles of the species as in the desired equation? In this case, B needs 1 mole and it has only one mole in equation 2; therefore, we do not need to multiply the equation by any factor:

Given: 1) A + D E + C DH = X kJ

2) 2A + B 2C DH = Y kJ

Determine the enthalpy change for the equation below by using given equations.  2D B + 2E DH = ?

2C 2A + B DH = -Y kJ

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Given: 1) A + D E + C DH = X kJ

2) 2A + B 2C DH = Y kJ

Determine the enthalpy change for the equation below by using given equations.  2D B + 2E DH = ?

We then add the two equations to determine if we obtained the desired equation. To obtain the DH for the desired equation, we add the manipulated DH’s. If we do this, we obtain

2A + 2D 2E + 2C DH = 2X kJ 2C 2A + B DH = -

Y kJ2D B + 2E DH = 2X – Y kJ

_______________________________________

Notice you can easily check the manipulation of the given equations. If we did things correctly, the overall equation will be exactly the same as the desired equation which is the case in this example.

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Let’s look at real chemistry example, suppose you are given the following GIVEN data:

Hess’ Law

kJ -297H );g(SO)g(O)s(S o22 D

kJ 198H );g(O)g(SO2)g(SO2 o223 D

use these data to obtain the enthalpy change for the following reaction?

?H );g(SO2)g(O3)s(S2 o32 D

x2

First, we will look for a species that is present in reaction in question and only in one of the given reaction. We find S (s) is such a species. We next ask ourselves two questions: 1.) Is it on the same side in the question reaction as in the given equation? The answer is yes (both on reactants side); therefore, we do not have to reverse the reaction. 2.) Does it have the same number of moles in both reactions? The answer is no; therefore, we will need to multiply the given equation by a factor of 2 to make it the same as in the question reaction.

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Let’s look at real chemistry example, suppose you are given the following GIVEN data:

Hess’ Law

kJ -297H );g(SO)g(O)s(S o22 D

kJ 198H );g(O)g(SO2)g(SO2 o223 D

use these data to obtain the enthalpy change for the following reaction?

?H );g(SO2)g(O3)s(S2 o32 D

x2

reverse

Second, we will look for a different species that is present in the question reaction and only in the given reaction remaining. We find SO3 (g) is such a species.

We next ask ourselves two questions: 1.) Is it on the same side in the question reaction as in the given equation? The answer is no; therefore, we will have to reverse the reaction.

2.) Does it have the same number of moles in both reactions? The answer is yes (both have 2 mols); therefore, we do not need use a multiplier.

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If we multiply the first equation by 2 and reverse the second equation, they will sum together to become the reaction in question.

(2)kJ) -297(H );g(SO2)g(O2)s(S2 o22 D

(-1)kJ) 198(H );g(SO2)g(O)g(SO2 o322 D

kJ -792H );g(SO2)g(O3)s(S2 o32 D

kJ -297H );g(SO)g(O)s(S o22 D

kJ 198H );g(O)g(SO2)g(SO2 o223 D

x2

reverse

HW 55note: HW may require the use of fractions as multipliers

Notice the overall equation is exactly the same as the desired equation; therefore, the sum of the DH equals the DH of the reaction.

code: hess

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The standard enthalpy of formation of something, denoted DHf

o, is the enthalpy change when one mole of that something at standard pressure (1 atm) and 298.15K (25oC) forms from its constituent elements in their standard state. The something can be a formula unit (molecular or ionic), an atom, or an ion. An element is in its standard state if it is in its most stable form at standard pressure and 298.15K.

i.e.

Note that the standard enthalpy of formation for a pure element in its standard state and H+ are zero. This means elements in their standard state have DHf

o = 0: metals - solids, diatomic gases, H+ ion.

Standard Enthalpies of Formation

Ag (s) + ½ Cl2 (g) AgCl (s) DHfo AgCl

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Another way to determine heat of reaction is the the law of summation of heats of formation which states that the enthalpy of a reaction is equal to the total formation energy of the products minus that of the reactants.

S is the mathematical symbol meaning “the sum of”, and n is the coefficients of the substances in the chemical equation.

)reactants()products( DDD of

of

o HnHnH

Standard Enthalpies of Formation

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aA + bB cC + dD

)reactants()products( DDD of

of

o HnHnH

Here’s how you would set up a calculation using the law of summations on the generic reaction below:

sum of nDHof of all products

sum of nDHof of all reactants

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A Problem to ConsiderWhat is the standard reaction enthalpy , DHo

rxn, for the following reaction?

)g(OH6)g(NO4)g(O5)g(NH4 223 molkJ /9.45 3.90 8.241:ofHD

)reactants()products( DDD of

of

o HnHnH

You will either be given or look up the DHof of the species in

question. Note: that species in their standard state will be “0” and not given or found in tables.

0

Next, we will use the law of summation and plug the values into the calculation:

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Using the summation law:

Be careful of arithmetic signs as they are a likely source of mistakes.

)reactants(Hm)products(HnH of

of

o DDD

D

)]/0(5)/9.45(4[ )]/8.241(6)/3.90(4[

2233

22

molOkJmolOmolNHkJmolNHOmolHkJOmolHmolNOkJmolNOH o

kJ 906Ho D

)g(OH6)g(NO4)g(O5)g(NH4 223 molkJ /9.45 0 3.90 8.241:ofHD

kJkJkJkJkJkJkJ

9066.1836.1089]0)6.183[()]8.1450(2.361[

HW 56code: formation