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Energy

Thermodynamics-study of energy and its interconversions

Labs#22 Calorimetry

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• EnergyEnergy• Capacity to do work (or produce heat)

• WorkWork• Energy used to cause object with mass to move

against a force• Force x distance (force acting over distance)

• HeatHeat• Involves transfer of energy between two objects• Energy used to cause temperature of object to

increase• Chemicals may store potential energy in bonds

that can be released as heat energy

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Potential energy-energy due to position or

composition• Depends on position relative to pull of gravity

• Stored energy released when process occurs that changes its position• When substances undergo chemical reactions, positions of atoms

change, and energy is released or absorbed• Chemical energyChemical energy stored within molecules as consequence of attractions

between electrons and atomic nuclei (electrostatic potential energy, Eelectrostatic potential energy, Eelel)

• Proportional to electrical charge on two interacting objects, Q1Q2, and inversely proportional to distance, d, separating them

• Based on magnitude of electron charge (1.60 x 10-19 C)• If both have same sign, Eel is positive

• If opposite signs, Eel is negative Eel = KQ1Q2/d• K = constant of proportionality, 8.99 x 109 J-m/C2

• Lower energy of system/more strongly opposite charges interact, more stable the of system

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Kinetic energy-energy due to the motion of an

object• KE= ½ mv2

• According to kinetic-molecular theory, all matter has thermal energy, because atoms and molecules are in constant motion

• Vibrate/rotate/move from one point to another

• More energy of molecular motion, higher temperature

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Temperature vs. Heat

• TemperatureTemperature• Random motion of particles in substance• Indicates direction heat energy will flow

• HeatHeat • Measure of energy content• What is transferred (due to temperature

difference between system and surroundings)• Always flows from hotter to colder body

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State Functions (state property)

property of system that depends only on its present state

• Independent of pathway (energy, temperature, pressure, enthalpy, entropy)

• Does not matter how you get there/how process carried out• Liter of water behind dam has same potential energy for work

regardless of whether it flowed downhill to dam, or was taken uphill to dam in bucket (PE state function dependent only on current position of water, not how it got there)

• If you climb mountain, you can take many paths to get to top, but once there, height is independent of path

• If NaCl is produced by one equation or many equations, change in heat still the same

• Work and heat are not state functions• Travel to top of mountain by many paths, so distance traveled

is dependent on path taken

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Exothermic Reactions

• Give off energy (PE in chemical bonds)• Products generally more stable (stronger bonds)

than reactants• Total chemical energy of products less than total

chemical energy of reactants• Heat released during chemical reaction comes

from decrease in chemical potential energy as reactants are converted to products

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Endothermic Reactions

• Energy absorbed from surroundings• Energy (heat) flows into system, increasing its PE• Products generally less stable (weaker bonds) than

reactants• Products have greater chemical energy than

reactants• Difference in energy between reactants/products is

equal to heat absorbed by system

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Thermodynamics

First law of thermodynamics First law of thermodynamics (law of conservation of energy) energy can be converted from one form to another, but cannot be created or destroyed

(Euniverse is constant)

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Thermodynamic quantities• ΔE = Efinal -Einitial

• Number/unit-gives magnitude of change• Sign-indicates direction of flow

• +q (ΔE)• Efinal > Einitial

• Endothermic reactions• Energy flows into system (gained energy)• System’s energy is increasing

• -q (ΔE)• Efinal < Einitial

• Exothermic reactions• Energy flows out of system (lost energy)• System’s energy is decreasing

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System Energy

• Internal energy, E, of system• Sum of KE/PE of all particles in system• Change energy of system by flow of work, heat or both

• Heat gained/word done on system both positive quantities

• Both increase internal energy of system, causing ΔE to be positive quantity

• ΔE = q + w• ΔE = change in system’s internal energy• q = heat• w = work

Sign conventions for q, w and ΔE

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For q + means system gainsgains heat

- Means system losesloses heat

For w + means work done onon system

- Means work done byby system

For ΔE + means net gainet gain of energy by system

- Means net loss net loss of energy by system

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Work

• Energy lost/gained by system by mechanical means, rather than by heat conduction

• Systems do not "contain" work or heat, but energy • SI unit for energy (work, heat)-joule (J)• Non-SI unit for heat-calorie (cal)-amount of

heat needed to raise temperature of 1 gram of water by 1°C from 14.5°C to 15.5°C

• Heat or energy in calories can be converted to joules

• 1 cal = 4.184 J and 1 kcal = 4.184 kJ

work = force distance

Since pressure = force / area,

work = pressure volume

wsystem = PV

• Units of pressure• P = F/A = kg m s2/m2 = kg m-1 s-2 = 1 Pascal

• Units of volume• P x V = kg m-1s-2 x m3 = kg m2 s-2 = 1 Joule

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Work done by gases:

• Work = P∆V for ideal gases (pressure constant)• From point of view of system, W = -P∆V • By gas (through expansion)

• System expands, positive work done on surroundings, negative work done on system

• ΔV is positive/w is negative• To gas (by compression)

• System contracts, surrounding have done work on system, positive work done on system

• ΔV is negative/w is positive

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• Calculate the work (with proper sign) associated with the contraction of a gas from 75 L to 30 L (work is done “on the system”) at a constant external pressure of 6.0 atm in:• L atm

• ∆V = Vfinal –Vinitial = 30 L – 75 L = -45 L• W = -P∆V = -6.0 atm (-45 L) = +270 L atm

• Joules (1 L atm = 101.3 J)• +270 L atm 101.3 J = +2.7 x 104 J 1 L atm

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• Calculate the change in energy of the system if 38.9 J of work is done by the system with an associated heat loss of 16.2 J.

• Get the sign correct:• Q = - because heat is lost• W = - because work is done by the system

• Solve:• ΔE = q + w = -16.2 J + (-38.9 J) = -55.1 J• The system has lost 55.1 J of energy.

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• A piston is compressed from a volume of 8.3 L to 2.8 L against a constant pressure of 1.9 atm. In the process, there is a heat gain by the system of 350 J. Calculate the change in energy of the system.

• w = -P∆V • w = -1.9 atm (-5.5 L) = 10.45 L atm

• 10.45 L atm 101.3 J = +1059 J (q) L atm• ΔE = q + w = 1059 + 350 = 1409 J = 1400 J

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Homework:

Read 6.1, pp. 241-248Q pp. 280-281, #9, 10, 18, 22, 24, 28

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Enthalpy (from Greek “to warm”)

Formerly called “heat content” so it is still symbolized as H

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Enthalpy

• H = E + PV• E = internal energy of system• P = pressure of system• V = volume of system

• Since internal energy, pressure and volume all state functions, enthalpy also state function (change in H independent of pathway)

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Enthalpy = ∆H = ∆E + ∆PVEnthalpy = ∆H = ∆E + ∆PV

E = qP + wSince w = -P∆V

E = qP PV (qP = heat at constant P)

qP = H only at constant pressureSo H = E + PV

Therefore, at constant pressure, qP = E + PV

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Enthalpy of reaction (rxn) or enthalpy change of system• ΔH

• Heat transferred to (absorbed) or from (evolved) system at constant pressure)

• Difference between enthalpies of products and reactants

• ΔH = enthalpy of system after reaction – enthalpy of system before reaction

• Difference between sum of enthalpies of products and sum of enthalpies of reactants

• ΔH = ∑Hproducts – ∑Hreactants

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Changes in enthalpy can be positive or negative• Endothermic reaction

• H(products) > H(reactants): H is positive, ΔH>0• Heat added (absorbed by system) is converted into

potential energy of system• Exothermic reaction

• H(reactants) > H(products): H is negative, ΔH<0• Loss of chemical potential energy accounts for the

amount of heat released (given off by system)

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Enthalpy Diagrams

• Each line represents set of reactants or products for balanced chemical reaction

• When going from one line to another, atoms must balance

• Enthalpy associated with condensation of water

• CO2(g) on both sides cancel to yield

• Relative distance of each line reflect relative enthalpy difference (ΔH) between reactants/products

• If enthalpy change in going from reactants to products is negative, line for products must be below reactants

• Length of distance must be proportional

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• Upon adding solid sodium hydroxide pellets to water, the following reaction takes place: NaOH(s) NaOH(aq). For this reaction at constant pressure, ∆H = -43 kJ/mol. Answer the following questions regarding the addition of 14 g of NaOH to water:• Does the beaker get warmer or colder?

• If ∆H < 0, heat is given off by system-beaker gets warmer.• Is the reaction exo- or endothermic?

• If heat is given off by system, reaction is exothermic.• What is the enthalpy change for the dissolution?

• ∆H = -43 kJ 1 mol NaOH 14 g NaOH = -15kJ mol 40.0 g NaOH

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Calorimetry

• Measurement of heat flow (released or absorbed) in reaction by calorimeter

• Experimental technique used to determine heat exchange (q) associated with reaction

• All calorimetry is based on measuring temperature change (t) of medium such as water

• Techniques/equipment used depend on nature of process being studied

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• At constant pressure, q = ∆H • At constant volume, q = ∆E• In both cases, heat gain or loss is determined

• Amount of heat exchanged in reaction depends upon

• Net temperature change during reaction• Amount of substances (more you have, more heat

can be exchanged)• Heat capacity (C) of substance

• Three ways to express heat capacity

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Heat Capacity (C)

• Amount of heat (q) required to raise temperature of given mass of substance by 1oC

• Units: J/oC or J/K• Ratio of heat absorbed to increase in T

• C = q/Δt (amt. of heat flowing in/out of substance)• Δt is temperature change given by tfinal – tinitial

C = heat absorbed

increase in temperature =

JC

or JK

Specific Heat Capacity (Specific heat)• Heat capacity of one

gram of substance• Unit = J/oC g or J/K g

• Energy required to raise temp of 1 gram of substance by 1°C

• Can calculate amount of heat absorbed by given amount of water/by object of given mass

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Molar Heat Capacity

• Heat capacity of one mole of substance

• Unit = J/oC-mol or J/K mol• Energy required to raise temp of 1

mole of substance by 1°C

Constant Pressure Calorimeter(solutions)• Determines changes in enthalpy (heats of reactions) for

reactions occurring in solutions• “coffee-cup” calorimeter

• Constant atmospheric pressure• No physical boundary between system/surroundings

• Reactants/products are system• Water in which they dissolve/calorimeter are surroundings

• Assume calorimeter prevents gain/loss of heat from solution (heat does not escape)

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• When reactants (each at same T) are mixed, T of mixed solution is observed to increase or decrease

• Calculating Heat of Rxn, ΔH• Energy released by reaction =

energy absorbed by solution• Energy released = s x m x ΔT • qsoln = (sp. Heat solution) x (g

of solution) x ΔT = -qrxn• Heat of rxn is extensive property-

dependent on amount of substance

• For dilute aqueous solutions, specific heat of solution approximately that of water (4.18 J/g-K)

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• If 10.0 g of solid NaOH is added to 1.00 L of water (specific heat capacity = 4.18 J/oC) at 25.0oC in a constant pressure calorimeter, what will be the final temperature of the solution? (assume the density of the final solution is 1.05 g/mL). We will use info from previous problem where enthalpy change per mole of NaOH is -43 kJ/mol.

• We need to know three things:• Mass of solution: 1.00 L x 1050 g/L = 1050 g• Heat capacity of solution: 4.18 J/oC• Enthalpy of reactant:

10.0 g NaOH 1 mol NaOH -43 kJ 1000 J = -10,750 J 40.0 g NaOH mol 1kJ

• We want to know the change in temperature: Energy released = s x m x ΔT

-10,750 J = (4.18 J/oC)(1050 g)(ΔT) = -2.45 25.0 + 2.45 = 27.45oC

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Constant Volume Calorimetry (Bomb Calorimetry)• Coffee cup calorimeter is not suitable for some reactions• Greater precision may be required, or a device that allows

higher temperatures• Bomb calorimeters are used to measure heat evolved in

combustion reactions of foods and fuels• Heat capacity of calorimeter must be known, generally in kJ/°C• Measurements of water/calorimeter related to heat evolved by

reaction q and to H• Energy released by reaction = energy absorbed by solution• =specific heat capacity x mass of solution x increase in temperature• Energy released = s x m x ΔT

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• Stainless steel "bomb“ loaded with small amount of combustible substance and O2 at 30 atm of pressure, and is immersed in known amount of water • Heat evolved during combustion absorbed by water/calorimeter assembly

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• The heat of combustion of glucose is 2800 kJ/mol. A sample of glucose weighing 5.00 g was burned with excess oxygen in a bomb calorimeter. The temperature of the bomb rose 2.4oC. What is the heat capacity of the calorimeter? A 4.40 g sample of propane was then burned with excess oxygen in the same bomb calorimeter. The temperature of the bomb increased 6.85oC. Calculate ΔEcombustion of propane.

• Calculate heat capacity of bomb using data for glucose:• Convert to moles: 5.00 g 1 mol = 2.78 x 10-2 moles glucose 180.0 g• Heat capacity = 2800 kJ/mol 2.78 x 10-2 moles 1 = 32.4 kJ/oC (kJ/ oC) 2.4oC

• Use this heat capacity to determine heat of combustion of propane: (kJ/mol)• 32.4 kJ 6.85oC 1 = -2200 kJ/mol oC 0.100 mol

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Homework:

Read 6.2, pp. 248-256Q pp. 280-282, #12, 32, 34, 36, 38, 42, 46

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Hess’s Law

In going from particular set of reactants to particular set of products, change in enthalpy (ΔH) is same whether reaction takes place in one step or in series of steps

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Using Hess’s Law

• Work backward from final reaction• Reverse reactions as needed, being sure to

also reverse ΔH• Identical substances on both sides of summed

equation cancel each other• Can multiply entire equation by factor (3, 2, ½,

or 1/3)-this includes ΔH

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Example:Step One:• N2(g) + 2O2(g) 2NO2(g) ΔH1 = 68 kJ

Step Two: • N2(g) + O2(g) 2NO(g) ΔH2 = 180 kJ

• 2NO(g) + O2 (g) 2NO2 g) ΔH3 = -112 kJ

• N2(g) + 2O2(g) 2NO2(g) ΔH2 + ΔH3 = 68 kJ

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Example:• S(s) + O2(g) SO2(g) H°rxn = –296.1 kJ

H2(g) + ½ O2(g) H2O(l) H°rxn = –285.8 kJ H2S(g) + 3/2O2(g) SO2(g) + H2O(l) H°rxn = –561.7 kJ

• These equations can be arranged so that their sum is the formation reaction of hydrogen sulfide.

• The first one is unchanged: S(s) + O2(g) SO2(g)The second one is unchanged: H2(g) + ½ O2(g) H2O(l)The third one is reversed: SO2(g) + H2O(l) H2S(g) + 3/2 O2(g) Their sum is overall equation: S(s) + H2(g) H2S(g)

• We must reverse the sign of the enthalpy change of the third reaction before summing the enthalpy changes.

• S(s) + O2(g) SO2(g) H°rxn = –296.1 kJ• H2(g) + O2(g) H2O(l) H°rxn = –285.8 kJ• SO2(g) + H2O(l) H2S(g) + 3/2O2(g) H°rxn = 561.7 kJ

S(s) + H2(g) H2S(g) H°f = –20.2 kJ• Summation of the three reaction steps yields the net reaction, and summation of the

H° values yields the enthalpy change of the net reaction.

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Homework:

Read 6.3, pp. 256-260Q pp. 280-283, #14, 52, 54, 56, 58

Enthalpy of formation (heat of formation)• Tables for

• Enthalpies of vaporization (ΔH for converting LG)• Enthalpies of fusion (ΔH for melting solids)• Enthalpies of combustion (ΔH for combusting substance in oxygen)

• Enthalpy (heat) of formation (ΔHf) associated with process of forming products from reactants

• Conditions influencing enthalpy changes• Temperature• Pressure• State of reactants/products (s/l/g/aq)

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Standard enthalpy of formation

•Change in enthalpy that accompanies formation of one mole of compound from its elements with all substances in their standard states

• Form most stable at 298K/1 atm

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Standard Enthalpy of Formation (ΔHf°)• “o” symbol indicates standard conditions.• Standard state for a substance is precisely

defined reference state• Elements/diatomic gases = 0• For a compound

• Gaseous substance at pressure of exactly 1 atmosphere• Condensed state (pure liquid/solid)• Substance present in solution of exactly 1M concentration

• For an element• Form in which the element exists under conditions of 1

atmosphere and 25oC [O2(g), Na(s), Hg(l), etc.].

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The enthalpy change in kilojoules applies to the reaction of the amount of substance shown in the balanced equation.SO2(g) + ½ O2(g) SO3(g) H° = –99 kJ • When 2 mol SO2 react with 1 mol O2, 198 kJ of heat

will be released• If the above equation was divided by 2 and written

then the enthalpy change (–198 kJ) must be divide by 2 also

• Remember that the heat evolved or absorbed by a chemical reaction is an extensive property: that is, it depends on the amount of reactants

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If we write the reverse of a chemical reaction, the magnitude of H is the same, but its sign changes.

• The following reaction absorbs heat• CaCO3(s) CaO(s) + CO2(g) H° = 178 kJ

• But the reverse reaction must release the same amount of heat

• CaO(s) + CO2(g) CaCO3(s) H° = –178 kJ

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We must always specify the physical states of all reactants and products, because they help determine the magnitude of the enthalpy change.

• In the combustion of methane, water is a product

• The enthalpy change is different when H2O is formed as a gas, than when H2O is a liquid

• CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H° = –802 kJ

• CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H° = –890 kJ

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C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)

3C(graphite) + 4H2(g) -> C3H8(g) DH�f = -103.85 kJ (reverse equation/sign)

3C(graphite) + 3O2(g) -> 3CO2(g) DH�f = 1180.5 kJ�

ΔH = ∑Hproducts – ∑Hreactants [3(-393.5) + 4(-285.8)] – [(-103.85) + 0] = -2219.8 kJ

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• Calculate ΔH° for the following reaction:2C3H6(g) + 9O2(g) 6CO2 (g) + 6H2O

• ΔHreactionO = ΣΔHf

O (Products) - ΣΔHfO (Reactants)

• Appendix 4 in the textbook lists them• [6(-286 kJ/mol) + 6(-393.5 kJ/mol)] – [2(20.9 kJ/mol) +

9(0 kJ/mol)] –remember, elements are 0• ΔH° = -4119 kJ

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Homework:

Read 6.4 –6.6, pp. 260-278Q pg. 284, #60, 62, 64, 66, and try 72 (extra credit)Submit quizzes by email to me:http://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch06_ace1.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch06_ace2.xmlhttp://www.cengage.com/chemistry/book_content/0547125321_zumdahl/ace/launch_ace.html?folder_path=/chemistry/book_content/0547125321_zumdahl/ace&layer=act&src=ch06_ace3.xml