Chapter 13:  Thermochemistry

 

 

Section 13-1: Enthalpy of Reaction, H

Section 13-2: Estimating H Using Bond Energies

Section 13-3: Calculating H Using Hesss Law

Section 13-4: Calculating H Using Standard Enthalpies of Formation, Hf

Section 13-5: Heat Transfer and the First Law of Thermodynamic

Section 13-6: Experiment - Determining the Specific Heat of a Metal Using Calorimetry

Section 13-7: Experiment Determining H Using Calorimetry

Section 13-8: H for Changes of Physical State and Heating/Cooling Curves

Section 13-9: Experiment Determining Hfusion of Ice

Section 13-10: Lattice Energy and the Born-Haber Method

Chapter 13 Practice Exercises and Review Quizzes

 

 

 

 

Section 13-1:  Enthalpy of Reaction, H

 

Thermochemistry is essentially the study of the heat released or absorbed by a reaction.  A reaction that releases heat is said to be exothermic, while a reaction that absorbs heat is said to be endothermic.  The heat released or absorbed by a reaction under constant pressure conditions is known as the enthalpy of reaction, H.  For an exothermic reaction, H will be negative, while H will be positive for an endothermic reaction.  To express information about the heat of a reaction, the value of H is typically written after the reaction in the unit kilojoules per mole as follows:

 

2 C2H6 (g) + 7 O2 (g) 6 H2O(g) + 4 CO2 (g)     H = -3119 kJ/mol

 

Since H is negative in this case, we know that the reaction is exothermic and releases heat.  We can use H as a conversion factor in the form

 

 

(where x is the coefficient of the particular reactant or product being considered) to convert between the actual amount reacted or produced and the actual amount of heat released in a given reaction situation:    

 

Sample Exercise 13A:

 

If 5.40 grams of ethane is burned according to the combustion equation shown above, how much heat will be released?

 

Solution:

 

Since we are considering the reactant ethane, which has a coefficient of 2 in the equation above, x = 2 in the conversion factor.  We first convert the grams of ethane to moles and then use the conversion factor to convert to the amount of heat released:

 

 

If instead we know the actual amount of heat in kJ involved in a given reaction situation and we wish to convert to the amount reacted or produced, we first use the conversion factor with kJ in the denominator and mol in the numerator, and then we can convert moles to any desired unit:

 

Sample Exercise 13B:

 

The formation of gaseous phosphorus trichloride is shown below:

 

P4 (g) + 6 Cl2 (g) 4 PCl3 (g)     H = -1207 kJ/mol

 

If 4.30 kJ of heat are released as the formation reaction above proceeds, how many milliliters of gaseous phosphorus trichloride, measured at 35C and 712 torr, will be produced?

 

Solution:

 

The question specifies that heat is released, so we will need to make the given amount of heat negative and begin the calculation with -4.30 kJ.  Since we are considering the product phosphorus trichloride, which has a coefficient of 4 in the equation above, x = 4 in the conversion factor.  We first use the conversion factor to convert to moles of phosphorus trichloride, then we use the Ideal Gas Law to find the volume of phosphorus trichloride produced:

 

 

 

 

Section 13-2:  Estimating H Using Bond Energies

 

The following table shows a variety of different average bond energies:

 

Bond

Average Bond Energy (kJ/mol)

C-C

347

C=C

614

CC

839

C-H

414

C-O

360

C=O

745

F-F

157

F-N

272

N-N

163

N=N

418

NN

941

O-H

464

O-O

142

O=O

498

 

Breaking covalent bonds requires energy to be absorbed and is endothermic, whereas forming covalent bonds releases energy and is exothermic.  For a chemical reaction in which all the reactants and products are gases, we can estimate H by adding the positive energies absorbed to break all bonds in the reactant molecules to the negative energies released by the formation of all bonds in the product molecules, as demonstrated in the following problem:

 

Sample Exercise 13C:

 

Use average bond energies to estimate H for the following reaction:

 

2 NF3 (g) N2 (g) + 3 F2 (g)

 

Solution:

 

First, draw a Lewis structure for each reactant and product molecule:

 

 

Then, being sure to take coefficients into account, add the positive energies absorbed to break all bonds in the reactant molecules to the negative energies released by the formation of all bonds in the product molecules to find H:

 

H (estimated) = (2x3)(N-F) – 1(NN) – 3(F-F)

= 6(272 kJ/mol) – 1(941 kJ/mol) – 3(157 kJ/mol) = 220 kJ/mol

 

 

 

Section 13-3:  Calculating H Using Hesss Law

 

When a reaction is reversed, the sign of H changes:

 

 

When all the coefficients in a reaction are multiplied by the same number, H is multiplied by that same number as well:

 

 

Hesss Law essentially states that the unknown H for a reaction of interest can be determined by adding together the H values from a series of two or more other reactions if the other reactions add up algebraically to the reaction of interest.  For example, if we wish to find H for the reaction A C, we can add together the H values from the following two reactions:

 

 

Reactions I and II add up algebraically to reaction III because an equal amount of B appears on both sides of the equation and, thus, can be canceled out.

 

In some cases, the given reactions with known H values may not add up to the reaction of interest with an unknown H.  However, we may be able to first reverse one or more of the given reactions (and, thus, switch the sign of H) and/or multiply all the coefficients of one or more of the given reactions (and, thus, multiply H by that same number) before adding the reactions to find the unknown H:

 

Sample Exercise 13D:

 

Calculate H for the reaction C2H4 (g) + Cl2 (g) C2H4Cl2 (l) using the following two reactions:

 

 

Solution:

 

Reactions I and II do not add up algebraically to the reaction of interest.  However, if we reverse Reaction I and also multiply all the coefficients in Reaction I by ½ and then reverse Reaction II, the new reactions will add up algebraically to the reaction of interest because H2O (l), 2 HCl (g), and ½ O2 (g) will cancel out.  Therefore, we can find the unknown H by adding (-½HI) + (-HII):

 

 

 

 

Section 13-4:  Calculating H Using Standard Enthalpies of Formation, Hf

 

The following table categorizes all elements into their standard states at 1 atm and 25C, which is essentially room temperature:

 

Gases

Liquids

Solids

Cl2, F2, H2, N2, O2

+

noble gases

Br2, Hg

I2

+

all other elements

 

C(graphite) is considered the standard state of solid carbon because it is more stable than the other allotrope, C(diamond). 

 

When a formation reaction is written, 1 mole of a given compound is produced on the right side and each element present in the compound is shown separately in its standard state as a reactant on the left side:

 

Formation Reaction:  standard state elements 1 mol compound

 

To balance the equation while keeping 1 mole of the compound on the right side, it may be necessary to use fractions on the left side.  For example, the formation reaction for solid ammonium iodide would be:

 

 

Sample Exercise 13E:

 

Write the balanced formation reaction, including physical states, for solid sucrose, C12H22O11.

 

Solution:

 

 

H for a formation reaction is known as the standard enthalpy of formation, Hf, and may be exothermic or endothermic.  However, Hf = 0 kJ/mol for all elements in their standard states.

 

H for a reaction carried out at 1 atm is known as the standard enthalpy of reaction, H.  We can determine H for a reaction using known Hf values and Hesss Law.  For example, consider the reaction aA + bB cC + dD, where a and b are the coefficients of the reactants A and B, and c and d are the coefficients of the products C and D.  Hesss Law suggests that we can determine H for this reaction by adding the following steps:

 

 

We can generalize this approach as follows:

 

 

Sample Exercise 13F:

 

Calculate H for the reaction 2 H2S (g) + 3 O2 (g) 2 H2O (l) + 2 SO2 (g) using the following information:

 

Compound

Hf (kJ/mol)

H2S (g)

-20.6

H2O (l)

-285.8

SO2 (g)

-296.8

 

Solution:

 

Hf = 0 kJ/mol for the standard state element O2 (g), so H = [2(-285.8) + 2(-296.8) - 2(-20.6) - 3(0)] kJ/mol = -1124.0 kJ/mol

 

 

 

Section 13-5:  Heat Transfer and the First Law of Thermodynamics

 

Calorimetry is essentially the measurement of heat in the laboratory.  To calculate the quantity of heat (q) in joules gained (positive value) when a substance increases in temperature or lost (negative value) when a substance decreases in temperature, without a change in physical state, we multiply the mass (m) of the substance in grams by the specific heat (s) of the substance in J/gC and also by the temperature change of the substance (tfinal – t initial) in C:

 

q = mst = ms(tfti)

 

Sample Exercise 13G:

 

The specific heat of copper metal is 0.385 J/gC.  How much heat in joules and in kilojoules is lost when a 165 gram sample of copper metal is cooled from 98C to 26C?

 

Solution:

 

 

The First Law of Thermodynamics has been referred to as the Law of Conservation of Energy and essentially states that energy can neither be created nor destroyed.  In calorimetry experiments, if one substance at a higher temperature is placed together with a second substance at a lower temperature in an insulated container that does not allow heat entry or exit, both substances will eventually reach the same final temperature.  Furthermore, the First Law of Thermodynamics suggests that the amount of heat lost by the higher temperature substance will be exactly equal to the amount of heat gained by the lower temperature substance:

 

qlost = -qgained

 

Note that, although the magnitude of heat will be equal, qlost is exothermic and will have a negative sign, whereas qgained is endothermic and will have a positive sign.

 

 

 

 

 

Section 13-6:  Experiment - Determining the Specific Heat of a Metal Using Calorimetry

 

The specific heat of water is known to be 4.18 J/gC.  If a metal with unknown specific heat but known mass and initial temperature is mixed in an insulated calorimeter, such as a securely-covered Styrofoam coffee cup, with a known mass of water that also has a known initial temperature, the specific heat of the metal can be determined as follows:

 

Sample Exercise 13H:

 

In an insulated calorimeter, a 175 gram piece of aluminum metal originally at 304C was added to 925 grams of water originally at 22C.  The final temperature of the aluminum-water mixture was 33C.  Determine the specific heat of aluminum.

 

Solution:

 

According to the First Law of Thermodynamics, the amount heat lost by the aluminum metal will be exactly equal to the amount of heat gained by the water (but opposite in sign):

 

qAl lost = -qwater gained

 

We first calculate the amount of heat the water gained and then switch the sign to find the amount of heat the aluminum lost, after which we can calculate the specific heat of aluminum:

 

 

 

If we know the masses, specific heats, and initial temperatures of both the metal and the water, we can calculate the final temperature in the calorimeter after mixing, as demonstrated below in Practice Exercise 13-9.

 

 

 

 

Section 13-7:  Experiment – Determining H Using Calorimetry 

 

When a solid ionic compound (salt) is dissolved in water, the cations and anions become separated from each other and are hydrated as water molecules surround the ions to create ion-dipole interactions. If this process is exothermic, the reaction particles will release heat into the surrounding mixed solution and increase the temperature of the solution.  If this process is endothermic, the reaction particles will absorb heat from the surrounding mixed solution and decrease the temperature of the solution.  We can determine the heat lost or gained by the reaction, qrxn, by first calculating the heat lost or gained by the surrounding solution, qsoln, and then switching the sign:

 

qrxn = -qsoln

 

After finding qrxn and converting to kilojoules, we can divide by the moles of salt dissolved to find H for the process in kJ/mol of salt dissolved:

 

Sample Exercise 13I:

 

In an insulated calorimeter, 2.6 grams of solid ammonium chloride at 23.4C was dissolved in 35.9 grams of water also at 23.4C, after which the final temperature of the mixed solution was 18.6C.  If the specific heat of the mixed solution was 3.94 J/gC, determine H for the dissolving process NH4Cl (s) NH4Cl (aq) in kJ/mol NH4Cl.   

 

Solution:

 

Since the temperature of the mixed solution decreased, the mixed solution lost heat and the reaction particles gained heat in an endothermic reaction with H > 0:

 

qrxn gained = -qsoln lost

 

We first calculate the amount of heat the mixed solution lost and then switch the sign to find the amount of heat the reaction particles gained, after which we can convert to kilojoules and divide by the moles of ammonium chloride to find H for the dissolving process NH4Cl (s) NH4Cl (aq):

 

 

For neutralization reactions between an aqueous acid and an aqueous metal hydroxide, the process of determining H by calorimetry is similar to that used for the dissolving of ionic salts, but the calculation typically differs in the following two ways:

 

1. Instead of masses being given, we may be given the volume of each solution being mixed, which we then can add to obtain the total volume of the mixed solution that absorbs heat from or releases heat into the reaction particles.  If we are given the density of the mixed solutions, we can multiply by the volume of the mixed solutions to obtain the mass of the mixed solutions that is necessary in the calculation of qsoln.

 

2. In the final step, we divide qrxn by the moles of a specified product formed in the reaction from the limiting reagent to obtain H for the reaction.      

 

Sample Exercise 13J:

 

In an insulated calorimeter, 45 mL of 1.0 M hydrochloric acid was mixed with 42 mL of 1.0 M sodium hydroxide, with both solutions originally at 19.4C.  The final temperature of the mixed solutions was 25.9C.  The density of the mixed solutions was 1.03 g/mL and the specific heat of the mixed solutions was 4.02 J/gC.  Write a balanced molecular equation, including physical states, and determine H for the neutralization reaction in kJ/mol of water formed.

 

Solution:

 

HCl (aq) + NaOH (aq) H2O (l) + NaCl (aq)

 

Since the temperature of the mixed solutions increased, the mixed solutions gained heat and the reaction particles lost heat in an exothermic reaction with H < 0:

 

 

 

Section 13-8:  H for Changes of Physical State and Heating/Cooling Curves

 

Changes of physical state occur at a constant temperature.  The endothermic change from solid to liquid is known as fusion and occurs at the melting point of the substance.  To calculate the amount of heat absorbed when a given amount of a substance is melted, we multiply the moles melted by H for the fusion process:

 

q = (nmelted)(Hfusion)

 

If the same amount of the substance was frozen from a liquid to a solid rather than melted, the magnitude of heat involved would be exactly the same, but heat would be released in the exothermic process and q would be negative:

 

q = (nfrozen)(-Hfusion)

 

The endothermic change from liquid to gas is known as vaporization and occurs at the boiling point of the substance.  To calculate the amount of heat absorbed when a given amount of a substance is vaporized, we multiply the moles vaporized by H for the vaporization process:

 

q = (nvaporized)(Hvaporization)

 

If the same amount of the substance was condensed from a gas to a liquid rather than vaporized, the magnitude of heat involved would be exactly the same, but heat would be released in the exothermic process and q would be negative:

 

q = (ncondensed)(-Hvaporization)

 

A heating curve is a graph that depicts the changes in temperature and physical state that occur over time as heat is absorbed by a substance.  The following sketch represents the general format of a heating curve for a substance originally below the melting point in the solid state being heated until it is a gas above the boiling point:

 

 

A cooling curve is a graph that depicts the changes in temperature and physical state that occur over time as heat is released by a substance.  The following sketch represents the general format of a cooling curve for a substance originally above the boiling point in the gas state being cooled until it is a solid below the melting point (same temperature as freezing point):

 

 

Note that heating and cooling curves may encompass a less narrow temperature range and, therefore, may not include all five of the distinct segments shown in the two sketches above.    

 

Sample Exercise 13K:

 

Consider the following data for H2O:

 

Melting Point = 0.0C

Boiling Point = 100.0C

 

Hfusion = 6.01 kJ/mol

Hvaporization = 40.8 kJ/mol

 

Specific Heat of Ice = 2.03 J/gC

Specific Heat of Water = 4.18 J/gC

Specific Heat of Water Vapor = 1.99 J/gC

 

Sketch a heating curve that depicts ice at -25.0C being heated until the temperature reaches 125.0C and then calculate the total amount of heat in kilojoules absorbed when 175 grams of H2O undergoes this process.

 

Solution:

 

 

We first calculate the amount of heat absorbed along each of the five distinct segments of the heating curve and then add the five heat values together to find the total heat absorbed:

 

 

qtotal = 8.881 kJ + 58.37 kJ + 73.15 kJ + 396.2 kJ + 8.706 kJ = 545 kJ

 

 

Section 13-9:  Experiment – Determining Hfusion of Ice

 

Given the specific heats of ice and water in the previous problem, we can determine Hfusion of ice in the laboratory by mixing in an insulated calorimeter a known mass of ice at a known original temperature below 0C with a known mass of water at a known original temperature above 0C.  If we use enough water and/or a high enough original water temperature, all the ice will melt and the final temperature in the calorimeter will be above 0C.  The sketch below depicts the heating curve for the ice as its temperature rises to its melting point, the ice melts, and then the temperature of the melted ice rises to the final temperature.  On the same sketch, the cooling curve for the water as it cools from its original temperature to the final temperature is also included:

 

 

The heat lost by the water as it cools to the final temperature will be equal in magnitude but opposite in sign to the total heat gained by the ice in the three segments of the heating curve above, which will allow us to determine Hfusion of ice:

 

Sample Exercise 13L:

 

In an experiment to determine Hfusion of ice, a student added 29.04 grams of ice originally at -15.1C to 97.91 grams of water originally at 41.9C in an insulated calorimeter.  If the final temperature in the calorimeter was 13.2C, calculate the experimentally determined Hfusion of ice.

  

Solution:

 

The heating curve for the ice + cooling curve for the water is:

 

 


 

Section 13-10:  Lattice Energy and the Born-Haber Method

 

Lattice energy (L.E.) is the energy associated with the process of forming one mole of a solid ionic compound from the separated gaseous metal cations and nonmetal anions:

 

metal cations (g) + nonmetal anions (g) 1 mol ionic (s)     H = L.E.

 

Lattice energy generally becomes more exothermic (negative) when the magnitudes of the ionic charges are larger and the sizes of the ions are smaller due to increased coulombic attraction.  The Born-Haber method utilizes Hess’s Law to calculate lattice energy by adding a combination of reactions from the following table:

 

Reaction Type

H

X (s) X (g)

Hsublimation

X (l) X (g)

Hvaporization

X2 (g) 2 X (g)

X2 bond energy

X (g) X+ (g) + e-

first ionization energy = I1

X (g) X2+ (g) + 2 e-

first + second ionization energy = I1 + I2

X (g) + e- X- (g)

first electron affinity = EA1

X (g) + 2 e- X2- (g)

first + second electron affinity = EA1 + EA2  

metal cations (g) + nonmetal anions (g) 1 mol ionic (s)

lattice energy = L.E.

standard state metal + nonmetal 1 mol ionic (s)

Hf° for ionic (s)

 

Reactions from the table above are added as follows:

 

1a) convert standard state metal to gas

1b) convert standard state nonmetal to gas (if necessary)

2) break bond in gaseous diatomic nonmetal to form neutral gaseous nonmetal atoms

3) remove electrons from neutral gaseous metal atoms to form gaseous metal cations

4) add electrons to neutral gaseous nonmetal atoms to form gaseous nonmetal anions

5) combine gaseous metal cations and nonmetal anions to form 1 mol solid ionic compound

____________________________________________________________________________

6) combine standard state metal and nonmetal to form 1 mol solid ionic compound

 

If we are given H for all reactions above except reaction 5, we can use Hess’s Law to solve for H of reaction 5, which is equal to the lattice energy of the ionic compound:

 

Sample Exercise 13M:


Calculate the lattice energy of sodium chloride using the information below.  Show all relevant reactions, including states of matter.

 

Hsublimation of sodium = 107 kJ/mol

Cl2 bond energy = 244 kJ/mol

first ionization energy of sodium = 496 kJ/mol

first electron affinity of chlorine = -349 kJ/mol

Hf° of solid sodium chloride = -411 kJ/mol

 

Solution:

 

1a) Na (s) Na (g)     Hsublimation = 107 kJ/mol

2) 1/2 Cl2 (g) Cl (g)     1/2(Cl2 bond energy) =1/2(244 kJ/mol)

3) Na (g) Na+ (g) + e-     I1 = 496 kJ/mol   

4) Cl (g) + e- Cl- (g)     EA1 = -349 kJ/mol

5) Na+ (g) + Cl- (g) NaCl (s)     L.E. = ?

______________________________________________________________

6) Na (s) + 1/2 Cl2 (g) NaCl (s)     Hf° = -411 kJ/mol

 

107 kJ/mol + 1/2(244 kJ/mol) + 496 kJ/mol + (-349 kJ/mol) + L.E. = -411 kJ/mol

L.E. of NaCl = -787 kJ/mol

 

If we are given the lattice energy but not Hf°, we can use the Born-Haber method to calculate Hf°, as demonstrated below in Practice Exercise 13-18.



Chapter 13 Practice Exercises and Review Quizzes:

 

13-1) The decomposition of ammonia gas into nitrogen gas and hydrogen gas is endothermic:

 

2 NH3 (g) N2 (g) + 3 H2 (g)     H = 92 kJ/mol

 

If 762 mL of hydrogen gas, measured at 325C and 792 mmHg, is formed in the reaction above, how much heat will be absorbed?

Click for Solution

 

13-1)

 

 

 

13-2) Consider the following reaction:

 

2 ZnS (s) + 3 O2 (g) 2 ZnO (s) + 2 SO2 (g)     H = -878 kJ/mol

 

What mass of solid ZnS, in kilograms, must react according to the equation above in order for 5.41 x 103 kJ of heat to be released?

Click for Solution

 

13-2) Heat is released = must start with -5.41 x 103 kJ:

 

 

 

 

13-3) Assuming all reactants and products are gases, write a balanced equation for the combustion of methanol using the smallest possible whole-number coefficients and estimate H for the reaction using average bond energies.

Click for Solution

 

13-3) 2 CH3OH (g) + 3 O2 (g) 2 CO2 (g) + 4 H2O (g)

 

Lewis structures:

 

 

H (estimated) = (2x3)(C-H) + (2x1)(C-O) + (2x1)(O-H) + (3x1)(O=O)

- (2x2)(C=O) - (4x2)(O-H)

= 6(414 kJ/mol) + 2(360 kJ/mol) + 2(464 kJ/mol) + 3(498 kJ/mol)

- 4(745 kJ/mol) – 8(464 kJ/mol) = -1066 kJ/mol

 

 

 

13-4) Calculate H for the reaction 4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (l) using the following three reactions:

 

Click for Solution

 

13-4) Reactions I + II + III do not add up algebraically to the reaction of interest.  However, if we multiply all the coefficients in Reaction I by 3, if we reverse Reaction II and multiply all coefficients in Reaction II by 2, and if we multiply all the coefficients in Reaction III by 2, the new reactions will add up algebraically to the reaction of interest because 2 N2 (g) and 6 H2 (g) will cancel out.  Therefore, we can find the unknown H by adding (3HI) + (-2HII) + (2HIII):

 

 

 

13-5) Write the balanced formation reaction, including physical states, for liquid tribromomethane, CHBr3.

Click for Solution

 

13-5)

 

 

 

13-6) Calculate H for the reaction 2 C4H10 (g) + 13 O2 (g) 10 H2O(g) + 8 CO2 (g) using the following information:

 

Compound

Hf (kJ/mol)

C4H10 (g)

-126.5

H2O (g)

-241.8

CO2 (g)

-393.5

 

Click for Solution

 

13-6) H = [10(-241.8) + 8(-393.5) - 2(-126.5) - 13(0)] kJ/mol = -5313.0 kJ/mol

 

 

 

13-7) The specific heat of lead metal is 0.130 J/gC.  How much heat in joules and in kilojoules is gained when a 225 gram sample of lead metal is heated from 125C to 241C?

Click for Solution

 

13-7)

 

 

 

13-8) In an insulated calorimeter, a 225 gram piece of iron metal originally at 463C was added to 845 grams of water originally at 27C.  The final temperature of the iron-water mixture was 39C.  Determine the specific heat of iron.

Click for Solution

 

13-8)

 

 

13-9) The specific heat of titanium metal is 0.54 J/g•°C.  In an insulated calorimeter, a 265 gram piece of titanium metal originally at 285°C was added to 645 grams of water originally at 26°C.  Determine the final temperature of the titanium-water mixture. 

Click for Solution

13-9)

 

 

13-10) In an insulated calorimeter, 12.7 grams of solid calcium chloride at 21.4C was dissolved in 129.1 grams of water also at 21.4C, after which the final temperature of the mixed solution was 38.3C.  If the specific heat of the mixed solution was 3.89 J/gC, determine H for the dissolving process CaCl2 (s) CaCl2 (aq) in kJ/mol CaCl2.    

Click for Solution

13-10) Since the temperature of the mixed solution increased, the mixed solution gained heat and the reaction particles lost heat in an exothermic reaction with H < 0:

 

 

 

 

13-11) In an insulated calorimeter, 85.6 mL of 2.55 M nitric acid was mixed with 94.3 mL of 2.51 M potassium hydroxide, with both solutions originally at 22.6C.  The final temperature of the mixed solutions was 39.1C.  The density of the mixed solutions was 1.04 g/mL and the specific heat of the mixed solutions was 3.98 J/gC.  Write a balanced molecular equation, including physical states, and determine H for the neutralization reaction in kJ/mol of water formed.

Click for Solution

 

13-11) HNO3 (aq) + KOH (aq) H2O (l) + KNO3 (aq)

 

Since the temperature of the mixed solutions increased, the mixed solutions gained heat and the reaction particles lost heat in an exothermic reaction with H < 0:

 

 

 

 

13-12) Consider the following data for ethanol, C2H5OH:

 

Melting Point = -114C

Boiling Point = 78C

 

Hfusion = 5.0 kJ/mol

Hvaporization = 42.6 kJ/mol

 

Specific Heat of Liquid Ethanol = 2.5 J/gC

 

Sketch a cooling curve that depicts ethanol vapor at 78C being cooled until solid ethanol is formed at -114C and then calculate the total amount of heat in kilojoules released when 55 grams of ethanol undergoes this process.

Click for Solution

 

13-12) In this case, there are only three distinct segments on the cooling curve because the ethanol vapor is already at the boiling point to start and the solid ethanol formed does not cool below the melting/freezing point:

 

 

We first calculate the amount of heat released along each of the three distinct segments of the cooling curve and then add the three heat values together to find the total heat released:

 


 

 

13-13) Given the reaction 2 F2 (g) + O2 (g) 2 OF2 (g)   H = 50. kJ/mol, use the table of average bond energies to calculate the F-O bond energy.

Click for Solution

13-13)

 

Lewis structures:

 

 

H = 2(F-F) + 1(O=O) – (2x2)(F-O)

50. kJ/mol = 2(157 kJ/mol) + 1(498 kJ/mol) – 4(F-O)

bond energy of F-O = 191 kJ/mol

 

 

 

13-14) Given the reaction N2H4 (l) + 2 H2O2 (l) N2 (g) + 4 H2O (g)   H° = -643 kJ/mol, use the information below to calculate the standard enthalpy of formation, Hf°, for H2O2 (l):

 

Compound

Hf° (kJ/mol)

H2O (g)

-242

N2H4 (l)

51

Click for Solution

13-14)

H° = -643 kJ/mol = 1(0 kJ/mol) + 4(-242 kJ/mol) – 1(51 kJ/mol) – 2(Hf° for H2O2 (l))

Hf° for H2O2 (l) = -188 kJ/mol

 

 

 

 

13-15) H for the dissolving process KOH (s) KOH (aq) is -57.6 kJ/mol KOH.  In an insulated calorimeter, 18.7 grams of solid potassium hydroxide at 21.3°C was dissolved in 115.5 grams of water also at 21.3°C.  If the specific heat of the mixed solution was 3.88 J/g•°C, determine the final temperature in the calorimeter.

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13-15)

 

 

13-16) Consider the reaction Ca(OH)2 (aq) + 2 HCH3COO (aq) 2 H2O (l) + Ca(CH3COO)2 (aq)   H = -108 kJ/mol.  In an insulated calorimeter, 93.6 mL of 1.09 M calcium hydroxide at 19.6°C was mixed with 84.8 mL of 1.78 M acetic acid also at 19.6°C.  If the density of the mixed solution was 1.03 g/mL and the specific heat of the mixed solution was 3.94 J/g•°C, determine the final temperature in the calorimeter.

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13-16)

 

 

13-17) If 5.5 grams of water vapor at 115°C is injected into an insulated calorimeter containing 175 grams of water at 22°C, and all the water vapor condenses, sketch a heating/cooling curve for the process and determine the final temperature of the liquid water in the calorimeter.

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13-17)

 

 

 

 

13-18) Calculate Hf° of solid calcium oxide using the information below.  Show all relevant reactions, including states of matter.

 

Hsublimation of calcium = 178 kJ/mol

O2 bond energy = 499 kJ/mol

first ionization energy of calcium = 589 kJ/mol

second ionization energy of calcium = 1145 kJ/mol

first electron affinity of oxygen = -141 kJ/mol

second electron affinity of oxygen = 744 kJ/mol

lattice energy of calcium oxide = -3401 kJ/mol

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13-18) 

1a) Ca (s) Ca (g)     Hsublimation = 178 kJ/mol

2) 1/2 O2 (g) O (g)     1/2(O2 bond energy) =1/2(499 kJ/mol)

3) Ca (g) Ca2+ (g) + 2 e-     I1 + I2 = (589 + 1145) kJ/mol   

4) O (g) + 2 e- O2- (g)     EA1 + EA2 = (-141 + 744) kJ/mol

5) Ca2+ (g) + O2- (g) CaO (s)     L.E. = -3401 kJ/mol

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6) Ca (s) + 1/2 O2 (g) CaO (s)     Hf° = ?

 

178 kJ/mol + 1/2(499 kJ/mol) + (589 + 1145) kJ/mol + (-141 + 744) kJ/mol + (-3401 kJ/mol) = -637 kJ/mol = Hf° of CaO (s)

 

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