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The energy aspect of chemistry. Thermochemistry. Thermochemical Data and Services. Energy is a quantity that can be stored in a chemical form, and chemical reaction involve energy. There is an extensive thermochemical data for various scientific and engineering applications.

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thermochemical data and services
Thermochemical Data and Services

Energy is a quantity that can be stored in a chemical form, and chemical reaction involve energy. There is an extensive thermochemical data for various scientific and engineering applications.

NIST site has thermochemical data for over 6000 organic and small inorganic compounds. It also has reaction thermochemistry data for over 9000 reactions and ion energetics data for over 16,000 compounds.

Thermochemical services you may need or provide Calculation services Consulting Experimental services Technical literature search service

Thermochemistry

temperature energy heat
Temperature Energy & Heat

Temperature is a measure of hot and cold – an intensive property. Thermometers in various scales are used to compare temperature.Recall: urms = ( 3 k T / m)1/2 = (3 RT / M)1/2

Energy is the driving force for changes. A change often is associated with a certain amount of energy and amounts of energy can be measured according to the quantities changed. Like other quantities, energy is an extensive property, unlike temperature.

Heat is energy in transfer or energy transferred and heat is often used as another term for energy. In this aspect, it is the quantity of energy transferred or involved.

Describe hotness and heat

Thermochemistry

a system for thermochemistry
A system for Thermochemistry

For the study of energy, there must be a reference for its amounts transferred into or out. Such a reference is called a system.

The boundary isolates the system from its surrounding, so that the amount of energy or heat transferred into or out can be identified.

Surrounding

System

Environment

Picture a system with boundary, surrounding, and heat transfer in your mind.

Thermochemistry

measuring heat energy heat energy work
Measuring Heat (Energy)Heat = Energy = Work

Heat is energy – the ability to do work. Units for work, heat or energy is J 1 J = 1 N m (in honor of James Joule 1818 – 1889) 1 cal = 4.184 J (defined. The amount of heat to raise 1 g water for 1 K)

Gaining heat increase T, as the kinetic energy of molecules increase.

Remember these from discussion of gases?Kinetic energy = ½ m u 23 R T = 1/3NAm u2

Heat capacity: the amount of heat required to raise T by 1 K

Specific heat: heat capacity (no unit) compare to water (1.000 cal K-1 g-1)

Meaning of T?

Thermochemistry

conservation of energy
Conservation of Energy

Energy can be converted from one form to another at a fixed rate, and it cannot be destroyed or created. In other words, energy is conserved. Quantity of heat q followsqsystem + qsurrounding = 0

What can be created and destroyed? 1 cal = 4.184 J (heat-work conversion) 1 J = 1 kPa L

Has this principle been proved?

What is a perpetual motion?

How can this principle be disproved?

What’s the principle of conservation of matter?

no creation & no destructiongain of q is +veloss of q is -ve

Thermochemistry

calorimeter experiment
Calorimeter Experiment

q = C * T

Heat gained

Temperature difference

Heat capacity

Heat capacity of water = 1 cal deg-1 g-1molar heat capacity of water = 18 * 4.184 J deg-1 mol-1 = 75.3 J deg-1 mol-1

Specific Heat (no unit) water 1 copper 0.385 aluminum 0.903

Thermochemistry

What to use for a miracle thaw?

bomb and coffee cup calorimeter
Bomb and Coffee-cup Calorimeter

q = C * T

Use a known amount of q and measure T to determine C.

By measuring T , calculate C = q / T (corection)

Once C is known, and a known amount of reactants, heat of reaction, Hreaction, can be evaluated

Thermochemistry

how do you solve this
How do you solve this?

Two 100.0-mL solutions of NaOH and HCl are both 1.00 mol L-1. When they are mixed in a coffee cup, the temperature raised from 20.0 to 26.7o C. What is the heat released.

Solution:

What’s the reaction?

What assumptions to make?

Steps?

Thermochemistry

determine heat of reaction
Determine Heat of Reaction

Two 100.0-mL solutions of NaOH and HCl are both 1.00 mol L-1. When they are mixed in a coffee cup, the temperature raised from 20.0 to 26.7o C. What is the heat released.

Solution:Assumptions: no heat loss, consider solution as water, C = 200.0 g * 4.18 J g–1 K –1 = 836 J K–1.

Heat of reaction = 836 J K–1.* (26.7 – 20.0) K = 5601 J = 5.6 kJ.

Mole of H+ and OH–1 reacted = 1.0 mol L–1 * 0.100 L = 0.10 mol

Molar heat of neutralization, Hreaction = 5.6 kJ / 0.1 mol = 65 kJ mol –1

Reaction: H+ + OH–1 = H2O Hreaction = 65 kJ mol –1

Thermochemistry

pressure volume work
Pressure-Volume Work

P

Pressure = P (N m-2)

V = AArea * hdistance (m3 = 1000 L)

Fforce = P * A

WWork = F * hdistance

= P * A * hdistance

= P V

1 J = 1 N m-2 m3 ( )

= 1 1000 Pa L

= 1 kPa L

A

h

1000 L1 m3

Show P V is work and 1 J = 1 kPa L

Thermochemistry

the gas constant r
The Gas Constant R

The ABC laws of gases can be combined into one and the result is an ideal gas equation.

P V = n R T

1 atm * 22.4140 L R = ---------------------------- = 0.082057 L atm mol–1 K–1 1 mol 273.15 K

101.325 kPa * 22.414 L R = ----------------------------------- = 8.3145 L kPamol–1 K–1 = 8.3145J mol–1 K–1 1 mol 273.15 K

1 kPa L = __?__ J 1 L = __?__ m3

1 L atm = __?__ J 1 m3 = __?__ L

Thermochemistry

the gas constant r1
The Gas Constant R

You blow a balloon for a volume of 1.00 L per breath. If the pressure you are blowing against is 1.10 atm, how much work, in J, have you done?

Solution:

P * V = 1.10 atm * 1.00 L = 1.10 L atm1.10 L atm

= 111.4 J

8.3145 J mol–1 K–10.08205 L atm mol–1 K–1

Ris used for unit conversion

8.3145 J mol–1 K–10.08205 L atm mol–1 K–1

= 101.33

Thermochemistry

first law of thermodynamics
First Law of Thermodynamics

To accommodate the principle of conservation of energy, we define an Internal EnergyU, as the total kinetic and potential energy.

The U cannot be measured or determined, thus we usually are only interested in U, the difference in internal energy before and after a change.

U = q + w

The w: work done onto the system

The q: heat absorbed by the system

The heat absorbed by the system and the work done onto the system are all used to increase the internal energy of the system.

Thermochemistry

evaluating internal energy
Evaluating Internal Energy

A balloon expanded against the atmosphere pressure performed25 J of work, and absorbed 30 J of heat, what is the change in internal energy?

Solution:

U = q + w

= (30– 25) J

= 5 J

Discussion:A change in internal energy results in change in temperature T.How does the average kinetic energy of the molecule change?

Thermochemistry

state and state function
State and State Function

The state of a system is the description of T, P, amounts and kinds of substances in the system, etc. The state defines all these quantities, and these quantities are specific for the state.

A property that has a specific value for a state is called a state function (or function of state).

The internal energy, U, is a state function, and the difference U is evaluated from the final and initial state. It is independent of the path.

U = U f – Ui

I will further explain the meaning of path independence using examples.Heat q and work w depend on path, for example.

Explain state functions!

Thermochemistry

heats of reaction u h
Heats of Reaction, U & H

For a reaction Reactants  ProductsThe change in internal energy is U = UReactants– Uproducts = qrxn + w

The change in internal energyU is the heat of reaction under constant volume, qv, U = qv, because w = 0, no P-V work.

Enthalpy H is the heat of reaction under constant pressure qp, H =qp Under this condition, work done on or by the system, is w = – P V.

Thus, U = qp - PV = H – PV

or H= U + PV

reactants

HU + PV

products

Explain U and H .

Thermochemistry

an u h example
An U, H Example

For the reaction, 2CO (g) + O2 (g)  2 CO2 (g), at 298 K qp = -566.0 kJ, calculate qv and w.

Solution:The work is due to change of volume, w = PV = (nf – ni) R T = ( 2 – 3) * 8.3145 J mol –1 K–1 * 298 K = – 2500 J = – 2 .5 kJ mol –1

H = – 566.0 kJ mol –1(of equation) U = – 566.0 – (– 2.5) kJ = – 563.5 kJ

Note: 3 moles of gas is reduced to 2 moles, work must be done to the system to compressit back to the same P, w ispositive.

2CO (g) + O2 (g)

U

H

w

2CO2 (g)

Thermochemistry

application of enthalpy
Application of Enthalpy

Given per rxn eqn

Given,C12H22O11 (s) + 12 O2 (g)  12 CO2 (g) + 11 H2O (l), H = –5650 kJ

How much heat is released when 10.0 g of sucrose, (molar mass 342.3), after you have ingested, digested, and completely oxidized.?

Solution: The condition given here applies to the human utilization of sucrose (S),

10.0 g S = – kJ

1 mol S342.3 g S

– 5650 kJ1 mol S

–ve, energy released

Evaluate U and w

Thermochemistry

enthalpy diagram for states
Enthalpy Diagram for States

vapor

Explain these changes of state:

H2O (s)  H2O (l) H = 6.01 kJ at 273 K

H2O (l)  H2O (g) H = 44.0 kJ at 298 K

How much energy is required to convert 72 kg of ice at 273 K to vapor at 298 K?

Solution:72e3 g ice = __x_ ?___

72e3 g water * 4.184 J (298 – 273) K = __y_ ?___4e3 mol * 44.0e3 J / mol = __z__ ?___

Total = _x_ + _y_ + _z_ = ________________

44.0 kJ

Water at 298 K

Water at 273 K

6.01 kJ

18 g Ice at 273 K

1 mol18 g

6.01 kJ1 mol

Thermochemistry

endothermic and exothermic reactions
Endothermic and Exothermic Reactions

Use an energy level diagram to help visualize endothermic and exothermic reactions.

Explain the sign changes in the following:

H2O (s)  H2O (l) H = 6.01 kJ at 298 K

H2O (l)  H2O (s) H = – 6.01 kJ at 298 K

H2O (l)  H2O (g) H = 44.0 kJ at 273 K

H2O (g)  H2O (l) H = – 44.0 kJ at 273 K

2 NO (g)  N2 (g) + O2 (g) H = 181 kJ

N2 (g) + O2 (g)  2 NO (g) H = – 181 kJ

Explain why the sign changes when the process is reversed

Thermochemistry

hess s law

Germain Hess(1802 - 1850)

Hess’s Law

Like internal energy, enthalpy is also a state function.

Hess’s Law is based on the principle of conservation of energy.

It implies that if a process occurs in stages (even if only hypothetical), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps.

Another statement: the heat evolved or absorbed in a chemical process is the same whether the process takes place in one or in several steps.

Explain Hess’s law

Thermochemistry

hess s law illustrated
Hess’s Law Illustrated

A + 2 B

A + B  AB H1

AB + B = AB2H2__add___________________A + 2 B = AB2 H1 + H2 = H1 2

H1

AB + B

H1 2

H 2

AB2

Thermochemistry

chemical energy and hess s law
Chemical Energy and Hess’s Law

C(graphite) + 0.5 O2 CO H° = – 110 kJ/mol.

2 C(graphite) + O2 2 CO H° = – 220 kJ/mol (multiplied by 2)

6 C(graphite) + 3 O2 6 CO H° = – 660 kJ/mol (multiplied by 6)

2 CO  C(graphite) + O2H° = 220 kJ/mol (+ve)

CO (g) + 0.5 O2 (g)  CO2 (g) H° = – 283 kJ/mol.

1 mol C and O2

A quantity difficult to measure: – 110 kJ

1 mol CO & 0.5 mol O2

– 393 kJ

– 283 kJ

1 mol CO2

Thermochemistry

standard state
Standard State

Standard temperature and pressure (STP) for gases are 273.15 K and 1 atm.

Standard state refers to states at pressure of 1.0 bar, and temperature must be specified for data at standard state. Standard enthalpy of reaction, standard enthalpy of formation, standard enthalpy of combustion, and in general standard energy of changes are referring to standard states.

Reaction at other states

Reactants at standard states

Products at standard states

Thermochemistry

standard enthalpy of formation
Standard Enthalpy of Formation

The standard enthalpy of formation, Hf°, is the energy of reaction for the formation of the substance at its standard state from its elements at their standard states.

Enthalpies of elements are 0 at their standard state, these are reference points.

Hf° kJ mol–1 of compounds

H2O (l) – 285.8 H2O (g) – 241.8 CO (g) – 110.5 CO2 (g) – 393.5 CH4 (g) – 74.8 C2H2 (g) + 226.7 C2H4 (g) + 52.4 C2H6 (g) – 84.7 CH3OH (l) – 238.7 C2H5OH (l) – 277.7

Discuss the significance and value in everyday life of the data in the Table on the right!

Thermochemistry

meaning of some special enthalpies of formation
Meaning of Some Special Enthalpies of Formation

Cpd Hf.kJ Reaction (Explain)

Cdiamond 1.9 C (graphite)  C (diamond) (phase transition) Br2(g) 30.9 Br2 (l)  Br2 (g) (evaporation at 298 K)P4(red) -17.6 P4 (s, white)  P4 (s, red) (White phosphorus is considered the standard state)H (g) 217.8 ½ H2 (g)  H (g) (½ H–H bond energy)

Thermochemistry

find standard enthalpy of formation
Find Standard Enthalpy of Formation

The enthalpy of combustion for H2, C(graphite) and CH4 are - 285.8, - 393.5, and - 890.4 kJ/mol respectively. Calculate the standard enthalpy of formation dHf for CH4. (correction made)

Solution: C (graphite) + 2 H2 CH4 Hf° = _____Explain these from the data 2 *(-285.8) 2 H2(g) + O2(g)  2 H2O(l) H = - 571.6 kJ

C(graphite) + O2(g)  CO2(g) H = - 393.5 kJ

CO2(g) + 2H2O(l)  CH4(g) + 2O2(g) H= 890.4 kJ (+ve)

(1) + (2) + (3)

C (graphite) + 2 H2 CH4 Hf° = - 74.7 kJ

Make an energy level diagram from these to see the relationship.

Thermochemistry

energy level diagram
Energy Level Diagram

C(graphite) + 2 H2(g) + 2 O2(g)

Hf° = - 74.7 kJ

CH4(g) + 2 O2(g)

H = - 965.3 kJ

H = - 965.3 kJ

CO2 + 2 H2O

Thermochemistry

application of energy level diagram and enthalpies of formation
Application of Energy Level Diagram and Enthalpies of Formation

C(graphite) + 2 H2(g) + 2 O2(g)

Hf° = - 74.7 kJ

CH4(g) + 2 O2(g)

If you know any two of the three thermodynamic data, you can evaluate the third.

H = - 965.3 kJ

H = - 965.3 kJ

CO2 + 2 H2O

Thermochemistry

applications of h f
Applications of Hf°

2 C (s) + 3 H2 (g) + 7/2O2(g)

Hf° can be very useful for the calculation of enthalpy of reaction, by applying Hess’s Law.

Calculate the H for the reaction:C2H6 (g) + 7/2 O2 (g)  2 CO2 (g) + 3 H2O (g) H = ____?

Solution: look up Hf°s for C2H6 etcHrxn = products Hf° – reacts Hf°

H= 2*(– 393.5) + 3*(– 285.8) – {1 * (– 84.7) + (7/2)*(0)} = _______ kJ

Hf C2H6 = - 84.7

2*Hf CO22*(-393.5)= - 787.0

C2H6 + 7/2 O2

2 CO2 + 3 H2 + 3/2 O2

H = ______

3*Hf H2O3*(-285.8)= - 857.4

Thermochemistry

2 CO2(g)+ 3 H2O (g)

Explain the generalized formula

enthalpy of reaction and fuel
Enthalpy of Reaction and Fuel

ChlorophyllSun light h v

6 CO2 + 6 H2O

C6H12O6 + 6 O2, H = 2.8e3 kJ

Animal digestion Oxidation

Fuels:coal, petroleum, natural gas, methanol, ethanol, hydrogen, gasified coal, waste, organic matter, etc.

Please read about fuel and environmental impacts.

Thermochemistry

chemistry relationship of material and energy
Chemistry: relationship of material and energy

Chemistry is the study of behavior of materials under the influence of energy, which drives all changes. Without energy, there is no behavior to study, and energy not only causes all changes, it is also the source of life. Thus, life must constantly utilizes energy, almost exclusively chemical energy.

Furthermore, energy is also required to keep and maintain a comfortable and pleasant living. Chemical energy, usually known as food and fuel, is one of the most important energy sources. Thus, the skill to use thermodynamic data is important.

Thermochemistry

review questions 0
Review questions – 0

For an endothermic reaction, how does the internal energy, U, change, positive, zero, or negative?

What is the units for heat capacity?

Look for the data and name H for the processes H2O (l)  H2O(g), H2O (g)  H2O(l), H2O (s)  H2O(g), H2O (s)  H2O(l) at STP? For which of the above has the smallest difference of U - H?

The reaction in a bomb calorimeter is carried out at what condition?constant P, constant V, constant T, at STP or at 1 atm and 25 K

When 2.00 g of CH4 are burned in a bomb calorimeter (Cv = 2.677e3 J / K), the temperature rises from 24.00 to 27.06o C. What is the q absorbed by the calorimeter?What is the molar heat of combustion of CH4 under this condition?

Thermochemistry

review questions 1
Review questions - 1

When 0.0500 mol HCl reacts with 0.100 mol NaOH contained in 25.0 mL of solution, the temperature of the water increases by 13.7o C. What is the heat of the reaction? (limiting reagent, heat capacity)What is the molar heat of neutralization reaction? (stoichiometry and thermal energy)

Which of the following are state functions?U, H, qv, P, T, w (work), V

The standard enthalpy of formation for C and CO are –393 and 287 kJ respectively. What is the standard enthalpy of formation for CO? (Hess’s law, standard enthalpy of formation, energy level diagram slide 24, given any two of these three, you should be able to find the third one.) How much energy is released when 100.0 g CO is formed from reaction of carbon and oxygen? (stoichiometry)

Thermochemistry

review questions 2
Review questions - 2

The standard enthalpy of formation of water (liquid H2O) is –286 kJ mol–1.

What is standard enthalpy of combustion (or oxidation) of H2?

What is the energy for the reaction, H2O (l)  H2(g) + ½O2(g)?

How much energy is release when 0.15 g H2 at 273K and 1 bar is ignited in excess O2 in a bomb calorimeter in which the water vapor so formed is condensed into liquid at at the same temperature? (calculation gives –21.5 kJ, and the negative sign indicated energy released)

What are DU,DW and DH for the reaction 0.1 H2O (l) = 0.1 H2O (g, 300 K and 1 bar) ?(Note, the internal energy increases, W = P V = n R T; n = 0.1 mol; if hydrogen is burned in air, this amount of energy and this energy is not released immediately from the burning of H2 gas.)

Thermochemistry

review questions 3
Review questions - 3

When 4.765 g of MgCl2 is formed by burning Mg(s) in Cl2(g), 32.1 kJ of energy is released. What is the enthalpy of formation of MgCl2?

The standard enthalpy of formation is zero for which ones of the following, O2(g), O3(g), O(g), C(graphite), C(diamond), C(g), P4(white), P4(red), Br2(s), Br2(l), Br2(g), I2(s), I2(l), I2(g), Hg(s), Hg(l), S(s), S(g)? (standard states, which ones have positive enthalpies of formation?)

Given these data at STP, S(s) + O2(g)  SO2 (g) H = –3 95 kJ S(g) + O2(g)  SO2 (g) H = –618 kJWhat is the standard enthalpy of formation of S(g)? (+223 kJ mol-1)

Thermochemistry

review questions 4
Review questions - 4

For applications of standard enthalpy of formation and enthalpy of reaction, try these:

The standard enthalpy of formation for CH3CHO(g), H2O(l) and CO2(g) are –166, –286, and –394 kJ mol-1 respectively. Calculate the heat of combustion of CH3CHO(g). (-1194 kJ mol-1, Hess’s law)

The standard enthalpy of formation of C2H2(g) is 227 kJ mol-1. What is the enthalpy of combustion of C2H2(g)? (use data from above, Hess’s law)

Calculate the energy evolved in the complete oxidation of 5.55 g of Al at 298 K and 1 atm pressure. (DHof of Al2O3 is –1767 kJ mol-1)

What is the energy required to reduce 5.55 g of Al2O3? (use data from the above question)

Thermochemistry

review questions 5
Review questions - 5

When concentrated sulfuric acid is diluted with water, the solution becomes hot. Is the reaction of H2SO4 with H2O exothermic or endothermic?

A synthetic gas contains 55.0% CO, 33.0% H2, and 12% CO2. If all the energy from the combustion of 1.0 L of this gas is absorbed by 1.0 L of H2O initially at 300 K, what is the final temperature? (find the required data and carried out the calculation)

A and B are two different compounds with the same molecular formula, C5H10. Both compounds burn in oxygen to give CO2(g) and H2O(g). The enthalpies of combustion for A and B are –3050 and 3025 kJ mol-1, respectively, at 298 K. What is the DHo for the reaction A  B?

Thermochemistry

review 6
Review – 6

(see slide 27 and 28)(see slide 7)(review DHo andDuo difference, see slides 16 - 18)(see slide 25-29)

The standard enthalpy of formation of diamond is 1.9 kJ mol-1, and the standard enthalpy of combustion at 298 K is 293.5 kJ mol-1 for graphite. What is the theoretical enthalpy of combustion of diamond at 298 K?

The heat capacity of platinum and liquid water are 0.133 J g-1, K-1 and 4.185 respectively. What mass of liquid water at 283 K, is needed to cool 50.0 g of platinum from 323 K to 300 K when the platinum is placed in the water?

The thermochemical equation of the reaction at 298 K is, Al(s) + 3/2O2 (g)  Al2O3(s) DHo = –1767 kJWhat is the standard internal energy change, DUo, at 298 K for this reaction?

With the following information, what is the standard enthalpy of ormation of C2H4(g) C2H4(g) + 3 O2 2CO2 + 2 H2O(l), DHo = – x kJData: DHof = – y kJ mol for CO2DHof = – z kJ mol for H2O

Solution: DHof = _______ kJ mol for C2H4

Thermochemistry

weakness shown in test 2 problems
Weakness shown in Test 2 problems

3. A sample of gas absorbs exactly 200 J from its surroundings. The gas expands from 2.75 L to 13.5 L against a constant 0.860 atm pressure. What is the internal energy change for the gas?

5. Radiation with a wavelength of 217 nm causes electrons to be ejected from the surface of lithium metal. If the maximum kinetic energy of ejected electrons is 4.420e-19 J, what is the longest wavelength radiation that can be used to disloge electrons from the surface of lithium?

8. Number of humps p, d orbitals in 4r22 versus r2 plot. (4r2R2 vs r2)

11. What is the electronic configuration of the transition ions, Fe3+?

Thermochemistry

final exam info
Final Exam info

Chem120 Dec. 11, 14 – 17 pmPAC all areas and MC4020, 4021, 4045, 4058, 4059, 4060, 4061

Thermochemistry

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