Welcome to Chem 1B Thermo-Chemistry Chapter 9 E-mail: benzene4president@gmail Web-site:

Welcome to Chem 1BThermo-Chemistry

Chapter 9E-mail: benzene4president@gmail

Web-site: http://clas.sa.ucsb.edu/staff/terri/

To Aleks or not to Aleks?

Ave Midterm score with 100% Aleks

Ave Midterm score without Aleks

50% 60%

60% 68%

70% 76%

80% 84%

90% 92%

100% 100%

Ave Midterm score with 80% Aleks

Ave Midterm score with no Aleks

50% 56%

60% 64%

70% 72%

80% 80%

90% 88%

100% 96%

Thermodynamics - ch 9

1. For a particular process q = – 10 kJ and w = 25 kJ. Which of the following statements is true?

a. Heat flows from the surroundings to the system.

b. The system does work on the surroundings.

c. ∆E = – 15 kJ

d. All of these are true.

e. None of these is true.


Sign Notation

q > 0 ⇒ endothermic ⇒ heat is flowing from the surroundings into the system

q < 0 ⇒ exothermic ⇒ heat is flowing out of the system to the surroundings

w > 0 ⇒ work is being done on the system by the surroundings when a gas is getting compressed by an external pressure

w < 0 ⇒ the system is doing work on the surroundings when a gas is expanding against an external pressure

∆E > 0 ⇒ internal energy of a gas increases when the temperature of the gas increases

∆E < 0 ⇒ internal energy of a gas decreases when the temperature of the gas decreases

∆E = 0 ⇒ internal energy of a gas remains constant if isothermal


2. A gas absorbs 4.8 J of heat and then performs 13.0 J of work. What is the change in internal energy of the gas?

Thermodynamics - ch 93. Which of the following statements is correct?

a. The internal energy of a system increases when more work is done by the system than heat is flowing into the system.b. The internal energy of a system decreases when work is done on

the system and heat is flowing into the system.c. The system does work on the surroundings when an ideal gas expands against a constant external pressure.d. All the statements are true.e. All the statements are false.


4. Which of the following are always endothermic?

a. Melting

b. Combustion

c. Condensation

d. All of the above


Breaking bonds requires energy ⇒ q > 0

Making bonds releases energy ⇒ q < 0

Thermodynamics - ch 95. The following reaction occurs at constant temperature and pressure.

4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)

a. The system does work on the surroundings and w < 0b. The surroundings do work on the system and w < 0c. The system does work on the surroundings and w > 0d. The surroundings do work on the system and w > 0e. No work is done, w = 0


3 Equations for Work

1) w = – Pext∆V2) w = – ∆nRT3) w = – nR∆Tequations 2 & 3

are derived from PV = nRTchoose the equation based on the

variables that are given in the problem


6. There are two containers, one has 1 mole of CO2 (g) and the other has 1 mole of Ne (g). Both gases are heated from 25 oC to 30 oC at constant pressure. For each of the following indicate true or false.

a. No work is done in heating either gas

b. More work is done in heating CO2 (g) than for Ne (g)

c. The change in internal energy of CO2 (g) is greater than it is for Ne (g)

Thermodynamics - ch 97. For nitrogen gas at 25 oC, Cv = 20.8 J K-1 mol-1 and Cp = 29.1 J K-1

mol-1. When a sample of nitrogen gas is heated at constant pressure, what fraction of the energy is converted to heat? a. 1.00 b. 0.417 c. 0.285 d. 0.715 e. 0.588


At constant pressure the fraction of energy converted to heat

⇒ Cv /Cp

At constant volume the fraction of energy converted to heat

⇒ Cv /Cv or 1


8. Which of the following are state functions?

a. work, heat

b. work, heat, enthalpy, energy

c. enthalpy, energy

d. work, heat, enthalpy

e. heat, enthalpy, energy


State Functions are pathway independent ⇒ since ∆ is defined as (final state –

initial state) then ∆ anything can not depend on the states in between the initial and final states a.k.a. the “path” ⇒ if

dealing with a state function you can use any path to solve for it ⇒ the easiest paths for ∆E and ∆H are…

∆E = qv

∆H = qp


State FunctionsThink of all the

possible ways youcould go from peak A

to peak B – no matter whichpath you choose the value of

Δh will be the same Peak A

Peak B

Δh

Thermodynamics - ch 99. Which one of the following statements is false?

a. The change in internal energy, ∆E, for a process is equal to the amount of heat absorbed at constant volume, qv.b. The change in enthalpy, ∆H, for a process is equal to the amount of heat absorbed at constant pressure, qp.c. A bomb calorimeter measures ∆H directly.

d. If qp for a process is negative, the process is exothermic.e. The freezing of water is an example of an exothermic reaction.

Thermodynamics - ch 910. When a gas is cooled at constant volume from an initial

temperature of 50 oC to 25 oC it loses 60 J of heat. When the same gas is cooled from 50 oC to 25 oC at constant pressure, it loses 100 J of heat. What is the value of the enthalpy change, ∆H, when the gas is heated from 25 oC to 50 oC at constant volume? a. + 60 J b. – 60 J c. – 100 J d. + 100 J e. – 40 J

Thermodynamics - ch 911. Suppose you add 45 J of heat to a system, let it do 10. J of

expansion work, and then return the system to its initial state by cooling and compression. Which statement is true for this process?a. ∆H < ∆Eb. The work done in compressing the system must exactly equal the work done by the system in the expansion step.c. ∆H = 70. Jd. The change in the internal energy for this process is zero.


12. Consider a gas in a 1.0-L bulb at STP that is connected via a valve to another bulb that is initially evacuated. After the valve is opened…

a. q > 0 q < 0 q = 0

b. w > 0 w < 0 w = 0

c. ∆H > 0 ∆H < 0 ∆H = 0

d. ∆E > 0 ∆E < 0 ∆E = 0


13. For the following reactions at constant pressure, predict if ∆H > ∆E, ∆H < ∆E, or ∆H = ∆E. Why is there a difference? Is work being done by the system or the surroundings?

a. H2 (g) + Cl2 (g) → 2 HCl (g)

b. N2 (g) + 3 F2 (g) → 2 NF3 (g)


14. Consider a sample containing 2 moles of xenon that undergoes the following changes:

Pa = 10.0 atmVa = 10.0 L

Pb = 10.0 atmVb = 5.0 L

Pc = 20.0 atmVc = 25.0 L

Step 1 Step 2

For each step, assume that the external pressure is constant and equals the final pressure of the gas for that step. Calculate q, w, ΔE and ΔH (in kJ) for each step and calculate the total values. Which values would change if you went directly from state a to state c?


Definition Easiest Path Substitution

ΔE = q + wConstant Volume

ΔE = qv = nCvΔT ΔE = Δ(PV) monoatomic gases

ΔE = Δ(PV) polyatomic gases

ΔH = ΔE + Δ(PV)Constant Pressure

ΔH = qp = nCpΔT

ΔH = Δ(PV) monoatomic gases

ΔE = Δ(PV) polyatomic gases


15. Calculate ∆E (in kJ) when 1 mol of liquid is vaporized at 1 atm and its boiling point (73°C). ∆Hvap = 28.6 kJ/mol at 73°C.

Thermodynamics - ch 916. Consider the combustion of ethanol:

C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l) ∆H = –1.37×103 kJWhen a 15.6-g sample of ethanol (molar mass = 46.1 g/mol) is burned,

how much energy is released as heat?A) 87.8 kJB) 2.14 × 104 kJC) 4.64 × 102 kJD) 4.05 × 103 kJE) 4.75 kJ


17. Consider the following reaction:

CH4 + 4 Cl2 (g) → CCl4 (g) + 4 HCl (g), ∆H = – 434 kJ

How many grams of methane must be react to raise the temperature of 1.2 L of water 50°C (CH2O = 4.184 J/g°C)


18. A 25.0 g piece of unknown metal was transferred from an oven at 115 °C into a coffee cup calorimeter containing 150. mL of water at 24 °C and allowed to come to equilibrium where the temperature was measured to be 28 °C. Calculate the specific heat capacity of the metal. (Ccal = 25 J/K and Cwater = 4.18 J/g°C)


19. A 5.00 g of aluminum pellets (CAl = 0.89 J°C-1g-1) and 10.0 g of iron pellets (CFe = 0.45 J°C-1g-1) are heated to 100.0°C. The mixture of hot aluminum and iron is then dropped into 97.3 mL of water at 22.0°C. Calculate the equilibrium temperature assuming no heat is lost to the surroundings. CH2O = 4.18 Jg-1°C-1


20. In a coffee cup calorimeter 50.0 mL of 0.10 M NaOH and 20.0 mL of 0.30 M HCl are mixed. The temperatures before and after the reaction were measured to be 23.0°C and 31.0°C. Calculate the molar enthalpy change for the neutralization reaction. Assume no heat is absorbed by the calorimeter. The density of each solution is 1.0 g/mL and the specific heat capacity for each solution is 4.18 J/g°C.


21. Calculate the final temperature when 38 mL of 0.15 M Pb(NO3)2 is mixed with 42 mL of 0.23 M KI both initially at 23 °C. Assume no heat is lost to the surroundings. The specific heat capacity for the solution is 4.18 J/g°C and the density is 1.0 g/mL.

Pb2+ + 2 I- → PbI2 ∆H = – 297 kJ/mol


22. Given the following data calculate the ∆H° for the reaction: NO + O → NO2

Rxn 1: 2 O3 → 3 O2 ∆H° = – 427 kJ

Rxn 2: O2 → 2 O ∆H° = +495 kJ

Rxn 3: NO + O3 → NO2 + O2 ∆H° = – 195 kJ


Hess’ 3 Laws

1. If a reaction is flipped ⇒

change the sign of ∆H

2. If a reaction is multiplied by a number ⇒

multiply ∆H by the same number

3. If two or more reactions are added ⇒

take the sum of the ∆Hs

Thermodynamics - ch 9 - Feldwinn23. Calculate ΔH for the reaction

N2H4 (l) + O2 (g) → N2 (g) + 2 H2O (l)

given the following data:

2 NH3 (g) + 3 N2O (g) → 4 N2 (g) + 3 H2O (l) ΔH = – 1010. kJ

N2O (g) + 3 H2 (g) → N2H4 (l) + H2O (l) ΔH = – 317 kJ

2 NH3 (g) + ½ O2 (g) → N2H4 (l) + H2O (l) ΔH = – 143 kJ

H2 (g) + ½ O2 (g) → H2O (l) ΔH = – 286 kJ


24. Choose the correct equation for the standard enthalpy of formation of CO (g), where ∆H°f for CO = –110.5 kJ/mol.

a. 2 Cgraphite (s) + O2 (g) → 2 CO (g) ∆H° = –110.5 kJ

b. Cgraphite (s) + O (g) → CO (g) ∆H° = –110.5 kJ

c. Cgraphite (s) + ½ O2 (g) → CO (g) ∆H° = –110.5 kJ

d. Cgraphite (s) + CO2 (g) → 2 CO (g) ∆H° = –110.5 kJ

e. CO (g) → Cgraphite (s) + O (g) ∆H° = –110.5 kJ

Thermodynamics - ch 925. The combustion of methanol takes place according to the reaction

2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l)Calculate the molar ∆H for the combustion under standard conditions. Use the following standard enthalpies of formation:

∆H°f for CH3OH (l) = – 238.5 kJ/mol ∆H°f for CO2 (g) = – 393.5 kJ/mol ∆H°f for H2O (l) = – 285.6 kJ/mol


Hess’ Equation:

∆H°rxn = Σ ∆H°f (products) – Σ ∆H°f (reactants)

Thermodynamics - ch 926. The heat combustion of acetylene, C2H2 (g), at 25°C, is –1299

kJ/mol. At this temperature, ∆H°f values for CO2 (g) and H2O (l) are –393 and –286 kJ/mol, respectively. Calculate ∆H°f for acetylene.a. 2376 kJ/molb. 625 kJ/molc. 227 kJ/mold. –625 kJ/mole. none of these


You have completed Ch.9

Answer key – ch. 91. For a particular process q = – 10 kJ and w = 25 kJ. Which of

the following statements is true?

a. Heat flows from the surroundings to the system.

b. The system does work on the surroundings.

c. ∆E = – 15 kJ

d. All of these are true.

e. None of these is true. If q < 0 heat flows out of the system. If w > 0 work is done on the system by the

surroundings. ∆E = q + w = -10 kJ + 25 kJ = 15 kJ

Answer key – ch. 92. A gas absorbs 4.8 J of heat and then performs 13.0 J of work.

What is the change in internal energy of the gas?

∆E = q + w

∆E = (4.8 J) + (-13.0 J) = -8.2 J

Answer key – ch. 93. Which of the following statements is correct?

a. The internal energy of a system increases when more work is done by the system than heat is flowing into the

system.b. The internal energy of a system decreases when work is done on the system and heat is flowing into the system.c. The system does work on the surroundings when an ideal

gas expands against a constant external pressure.

w = -Pext∆V (“expand” => ∆V > 0 => w < 0)d. All the statements are true.e. All the statements are false.

Answer key – ch. 94. Which of the following are always endothermic?

a. Melting => heat is absorbed when bonds are broken

b. Combustion => always exothermic

c. Condensation => heat is released when bonds are formed

d. All of the above

Answer key – ch. 95. The following reaction occurs at constant temperature

and pressure.4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g)

a. The system does work on the surroundings and w < 0The reaction goes from 9 moles of gas to 10 moles of gas => ∆V > 0 => w < 0

b. The surroundings do work on the system and w < 0c. The system does work on the surroundings and w > 0d. The surroundings do work on the system and w > 0e. No work is done, w = 0

Answer key – ch. 96. There are two containers; one has 1 mole of CO2 (g) and the other

has 1 mole of Ne (g). Both gases are heated from 25 oC to 30 oC at constant pressure. For each of the following indicate true or false.

a. No work is done when heating either gas False => w = 0 only at constant volume or if expanding in a vacuum

b. More work is done in heating CO2 (g) than for Ne (g) False => work is dependent on the change in temperature which is the same for both gases

c. The change in internal energy of CO2 (g) is greater than it is for Ne (g) True => the heat capacity for CO2 is higher than Ne due to the bonds in the molecule

Answer key – ch. 97. For nitrogen gas at 25 oC, Cv = 20.8 J K-1 mol-1 and Cp = 29.1 J

K-1 mol-1. When a sample of nitrogen gas is heated at constant pressure, what fraction of the energy is converted to heat? a. 1.00 b. 0.417 c. 0.285 d. 0.715 e. 0.588

Cv/Cp = 20.8JK-1mol-1/29.1JK-1mol-1 = 0.715

Answer key – ch. 98. Which of the following are state functions?

a. work, heat

b. work, heat, enthalpy, energy

c. enthalpy, energy

d. work, heat, enthalpy

e. heat, enthalpy, energy

Answer key – ch. 99. Which one of the following statements is false?

a. The change in internal energy, ∆E, for a process is equal to the amount of heat absorbed at constant volume, qv.

b. The change in enthalpy, ∆H, for a process is equal to the amount of heat absorbed at constant pressure, qp.c. A bomb calorimeter measures ∆H directly. In a bomb calorimeter the ∆H is measured indirectly by observing how much heat is absorbed by the calorimeter

d. If qp for a process is negative, the process is exothermic.e. The freezing of water is an example of an exothermic reaction.

Answer key – ch. 910. When a gas is cooled at constant volume from an initial

temperature of 50 oC to 25 oC it loses 60 J of heat. When the same gas is cooled from 50 oC to 25 oC at constant pressure, it loses 100 J of heat. What is the value of the enthalpy change, ∆H, when the gas is heated from 25 oC to 50 oC at constant volume? a. + 60 J b. – 60 J c. – 100 J d. + 100 J e. – 40 J

∆H is a state function which means we can pick any path we

want to calculate ∆H => the easiest pathis constant pressure =>

∆H = qp = +100 J

Answer key – ch. 911. Suppose you add 45 J of heat to a system, let it do 10. J of

expansion work, and then return the system to its initial state by cooling and compression. Which statement is true for this process?a. ∆H < ∆Eb. The work done in compressing the system must exactly equal the work done by the system in the expansion step.c. ∆H = 70. Jd. The change in the internal energy for this process is zero.

Since it has returned to it’s initial state there has been no overall change to the system

Answer key – ch. 912. Consider a gas in a 1.0-L bulb at STP that is connected via a

valve to another bulb that is initially evacuated. After the valve is opened…a. q > 0 q < 0 q = 0b. w > 0 w < 0 w = 0c. ∆H > 0 ∆H < 0 ∆H = 0d. ∆E > 0 ∆E < 0 ∆E = 0

No energy is required toexpand into a vacuum

Answer key – ch. 913. For the following reactions at constant pressure, predict if

∆H > ∆E, ∆H< ∆E, or ∆H = ∆E. Why is there a difference? Is work being done by the system or the surroundings?

a. H2 (g) + Cl2 (g) 2 HCl (g)

∆H = ∆E + ∆(PV) or ∆H = ∆E + ∆(nRT)since the moles of gas remain constant ∆H =

∆E

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

since the moles of gas decrease => ∆H < ∆E

Answer key – ch. 914. Consider a sample containing 2 moles of xenon that

undergoes the following changes:

Pa = 10.0 atmVa = 10.0 L

Step 1 Pb = 10.0 atmVb = 5.0 L

Pc = 20.0 atmVc = 25.0 L

Step 2

For each step, assume that the external pressure is constant and equals the final pressure of the gas for that step. Calculate q, w, ΔE and ΔH (in kJ) for each step and calculate the total values. Which values would change if you went directly from state a to state c?

Step 1 => constant pressure => ΔH = q = nCp ΔT => if you use PV=nRT to find the temperatures you get Ta = 609K, Tb = 305K and Tc = 3047 K (for step 2) => continued on next slide…

Answer key – ch. 914. …continued

ΔH = q = (2mol)(5/2)(8.3145x10-3kJmol-1K-1)(305K-609K) = -12.6 kJw = -PΔV => w = -(10atm)(5.0L-10L)(0.1013kJatm-1L-1) = 5.1 kJ

ΔE = q + w = (-12.6 kJ) + (5.1 kJ) = -7.5 kJ

Step 2 => neither constant pressure or constant volume =>

w = -(20atm)(25L-5L)(0.1013kJatm-1L-1) = -40.5 kJ

ΔH = nCp∆T =(2mol)(5/2)(8.3145x10-3kJmol-1K-1)(3047K-305K)= 114 kJ

ΔE = nCv ∆T =(2mol)(3/2)(8.3145x10-3kJmol-1K-1)(3047K-305K)= 68.4 kJ

q = ΔE – w = (68.4 kJ) – (-40.5 kJ) = 109 kJ

Answer key – ch. 915. Calculate ∆E (in kJ) when 1 mol of liquid is vaporized at 1

atm and its boiling point (73°C). ∆Hvap = 28.6 kJ/mol at 73°C.

∆E = ∆H - ∆(PV) => ∆E = ∆H - ∆(nRT)

since T and R are constants

∆E = ∆H - ∆nRT

∆E = (1 mol)(28.6 kJ/mol) –

(1mol-0mol )(8.3145x10-3kJmol-1K-1)(73 + 273K)

= 25.7 kJ

Answer key – ch. 916. Consider the reaction:C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l) ∆H = –1.37×103 kJWhen a 15.6-g sample of ethyl alcohol (molar mass = 46.1 g/mol)

is burned, how much energy is released as heat?A) 87.8 kJB) 2.14 × 104 kJC) 4.64 × 102 kJD) 4.05 × 103 kJE) 4.75 kJ

15.6 g C2H5OH x1 mol C2H5OH

46.1 gx

1370 kJ

1 mol C2H5OH= 464 kJ

Answer key – ch. 917. Consider the following reaction:

CH4 + 4 Cl2 (g) → CCl4 (g) + 4 HCl (g), ∆H = –434 kJ

How many grams of methane must be burned to raise the temperature of 1.2 L of water 50°C (CH2O = 4.184 J/g°C)

qH2O = mC ∆T = (1.2 kg)(4.184 J/g°C)(50°C) = 251 kJ

251 kJ x434 kJ

1 mol CH4 x1 mol CH4

16 g CH4 = 9.3 g CH4

Answer key – ch. 918. A 25.0 g piece of unknown metal was transferred from an

oven at 115 °C into a coffee cup calorimeter containing 150. mL of water at 24 °C and allowed to come to equilibrium where the temperature was measured to be 28 °C. Calculate the specific heat capacity of the metal. (Ccal = 25 J/K and Cwater = 4.18 J/g°C)

heat lost by metal = heat absorbed by water and the calorimeter

-qmetal = qwater + qcal

-mC∆T = mC∆T + C∆T -(25g)(C)(28°C-115°C) = (150g)(4.18 J/g°C)(28°C- 24°C) + (25

J/K)(28°C- 24°C)

Cmetal = 1.2 J/g°C

Answer key – ch. 919. A 5.00 g of aluminum pellets (CAl = 0.89 J°C-1g-1) and 10.0 g

of iron pellets (CFe = 0.45 J°C-1g-1) are heated to 100.0°C. The mixture of hot aluminum and iron is then dropped into 97.3 mL of water at 22.0°C. Calculate the equilibrium temperature assuming no heat is lost to the surroundings. CH2O = 4.18 Jg-1°C-1

-(qAl +qFe) = qwater

-(mC∆T + mC∆T) = mC∆T

-((5g)(0.89 J°C-1g-1)(Tf-100°C) + (10g)(0.45 J°C-1g-1)(Tf-100°C)) = (97.3g)(4.18 Jg-1°C-1)(Tf-22°C)

Tf = 23.7°C

Answer key – ch. 920. In a coffee cup calorimeter 50.0 mL of 0.10 M NaOH and

20.0 mL of 0.30 M HCl are mixed. The temperatures before and after the reaction were measured to be 23.0°C and 31.0°C. Calculate the molar enthalpy change for the neutralization reaction. Assume no heat is absorbed by the calorimeter. The density of each solution is 1.0 g/mL and the specific heat capacity for each solution is 4.18 J/g°C.

-qrxn = +qsoln

-qrxn = mC∆T = (50g+20g)(4.18 J/g°C)(31°C-23°C) = 2341J

qrxn = -2.34 kJcontinued on next slide…


HCl = > (20 mL)(0.3M) = 6 mmol

NaOH => (50 mL)(0.1M) = 5 mmol => Limiting reagent

since HCl + NaOH NaOH + H2O

qrxn = -2.34 kJ/0.005mol = -468 kJ/mol

Answer key – ch. 921. Calculate the final temperature when 38 mL of 0.15 M

Pb(NO3)2 is mixed with 42 mL of 0.23 M KI both initially at 23 °C. Assume no heat is lost to the surroundings. The specific heat capacity for the solution is 4.18 J/g°C and the density is 1.0 g/mL.

Pb2+ + 2 I- PbI2 ∆H = -297 kJ/mol

-qrxn = qsoln

-qrxn = mC∆T to find qrxn we need to identify the limiting reagent

Continued on next slide…


Pb2+ => (38 mL)(0.15 M) = 5.7 mmol

I- => (42 mL)(0.23 M) = 9.7 mmol => Limiting Reagent

qrxn = (0.0097 mol I-)(-297kJ/2 mol I-) = 1440J

now using -qrxn = mC∆T

-(-1440J) = (38g + 42g)(4.18 J/g°C)(Tf - 23°C)

Tf = 27.3 °C

Answer key – ch. 922. Given the following data calculate the ∆H° for the reaction:

NO + O NO2

2 O3 3 O2 ∆H° = -427 kJ

O2 2 O ∆H° = +495 kJ

NO + O3 NO2 + O2 ∆H° = -195 kJ

Rxn 1 => flip and divide by 2

Rxn 2 => flip and divide by 2

Rxn 3 => keep as is

∆H°total = (-427 kJ)(-0.5) + (495 kJ)(-0.5) + (-195 kJ) = -49 kJ

Answer key – ch. 923. Calculate ΔH for the reaction

N2H4 (l) + O2 (g) N2 (g) + 2 H2O (g)

given the following data:

2 NH3 (g) + 3 N2O (g) 4 N2 (g) + 3 H2O (l) ΔH = -1010. kJ

N2O (g) + 3 H2 (g) N2H4 (l) + H2O (l) ΔH = -317 kJ

2 NH3 (g) + ½ O2 (g) N2H4 (l) + H2O (l) ΔH = -143 kJ

H2 (g) + ½ O2 (g) H2O (l) ΔH = -286 kJ

continued on next slide…


Rxn 1 => multiply by 1/4

Rxn 2 => flip and multiply by 3/4

Rxn 3 => flip and multiply by 1/4

Rxn 4 => multiply by 9/4

∆Htotal = (1/4)(-1010. kJ) + (-3/4)(-317 kJ) + (-1/4)(-143 kJ) + (9/4)(-286 kJ) = -622.5 kJ

Answer key – ch. 924. Choose the correct equation for the standard enthalpy of

formation of CO (g), where ∆H°f for CO = –110.5 kJ/mol.

a. 2 Cgraphite (s) + O2 (g) → 2 CO (g) ∆H° = –110.5 kJ

b. Cgraphite (s) + O (g) → CO (g) ∆H° = –110.5 kJ

c. Cgraphite (s) + ½ O2 (g) → CO (g) ∆H° = –110.5 kJ

d. Cgraphite (s) + CO2 (g) → 2 CO (g) ∆H° = –110.5 kJ

e. CO (g) → Cgraphite (s) + O (g) ∆H° = –110.5 kJ

Answer key – ch. 925. The combustion of methanol takes place according to the

reaction2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l)

Calculate ∆H° for the combustion of 1 mol of methanol under standard conditions. Use the following standard enthalpies of formation:

∆H°f for CH3OH (l) = -238.5 kJ/mol ∆H°f for CO2 (g) = -393.5 kJ/mol ∆H°f for H2O (l) = -285.6 kJ/mol

∆H°rxn = Σ∆H°(products) - Σ∆H°(reactants)∆H°rxn = (2 mol CO2)(-393.5 kJ/mol) + (4 mol H2O)(-285.6

kJ/mol) – (2 mol CH3OH)(-238.5 kJ/mol) = -1452 kJif only 1 mol of CH3OH => (0.5)(-1452 kJ) = -726 kJ

Welcome to Chem 1B Thermo-Chemistry Chapter 9 E-mail: benzene4president@gmail Web-site:

Documents

Transcript of Welcome to Chem 1B Thermo-Chemistry Chapter 9 E-mail: benzene4president@gmail Web-site: