chemistry: the molecular science moore, stanitski and jurs

© 2008 Brooks/Cole 1

Chemistry: The Molecular ScienceChemistry: The Molecular ScienceMoore, Stanitski and Jurs

Chapter 6: Energy and Chemical ReactionsChapter 6: Energy and Chemical Reactions


Energy (Energy (EE)) = the capacity to do work.

All energy is either KineticKinetic or PotentialPotential energy.

Work (w)Work (w) occurs when an object moves against a resisting force:

w = −(resisting force) x (distance traveled)

w = −F d

The Nature of Energy


Potential energy (Potential energy (EEpp)) – Energy of position. Stored E.

It may arise from: gravity: Ep = m g h (mass x gravity x height).

charges held apart. bond energy.

Kinetic energy (Kinetic energy (EEkk)) - Energy of motion macroscale = mechanical energy random nanoscale = thermal energy periodic nanoscale = acoustic energy

Ek = ½mv2 (m = mass, v = velocity of object)

The Nature of Energy


2.0 kg mass moving at 1.0 m/s (~2 mph):

Ek = ½ mv2 = ½ (2.0 kg)(1.0 m/s)2

= 1.0 kg m2 s-2

= 1.0 J

1 J is a relatively small amount of energy.

1 kJ (1000 J) is more common in chemical problems.

joule (J)joule (J) - SI unit (1 J = 1 kg m2s-2)

Energy Units


calorie (cal)calorie (cal)

Originally:

“The energy needed to heat of 1g of water from 14.5 to 15.5 °C.”

Now: 1 cal = 4.184 J (exactly)

Dietary Calorie (Cal)Dietary Calorie (Cal) - the “big C” calorie

Used on food products. 1 Cal = 1000 cal

= 1 kcal

Energy Units


““Energy can neither be created nor destroyed”Energy can neither be created nor destroyed”

E can only change form.

Total E of the universe is constant.

Also called the 11stst Law of Thermodynamics Law of Thermodynamics

Conservation of Energy


Conservation of Energy

A diver:• Has Ep due to macroscale position.

• Converts Ep to macroscale Ek.

• Converts Ek,macroscale to Ek,nanoscale (motion of water, heat)


Energy and Working

If an object moves against a force, work is done.

• Lift a book you do work against gravity. The book’s Ep increases.

• Drop the book: Ep converts into Ek

The book does work pushing the air aside.

• The book hits the floor no work is done on the floor (it does not move). Ek converts to a sound wave and T of the book and floor

increase (Ek converts to heat).


Energy and WorkingIn a chemical process, work occurs whenever something expands or contracts.

Imagine the gas inside a balloon heat the gas. the gas expands and the balloon grows. the gas does work pushing back the rubber and

the air outside it.

Expansion pushes back the surrounding air.


TemperatureTemperature is a measure of the thermal energy of a sample.

Thermal energyThermal energy

• E of motion of atoms, molecules, and ions.

• Atoms of all materials are always in motion.

• Higher T = faster motion.

Energy, Temperature and Heating



Consider a thermometer. As T increases: Atoms move faster, and on average get farther apart. V of the material increases. Length of liquid column increases.


HeatHeat • Thermal E transfer caused by a T difference.• Heat flows from hotter to cooler objects until

they reach thermal equilibriumthermal equilibrium (have equal T ).



Systems, Surroundings & Internal Energy

SurroundingsSurroundings = rest of the universe (or as much as needed…) the flask. perhaps the flask and this classroom. perhaps the flask and all of the building, etc.

UniverseUniverse = System + Surroundings

SystemSystem = the part of the universe under study chemicals in a flask. the coffee in your coffee cup. my textbook.


Internal energyInternal energy = E within the system because of nanoscale position or motion

Einternal= sum of all nanoscale Ek and Ep

nanoscale Ek = thermal energy

nanoscale Ep

• ion/ion attraction or repulsion

• nucleus/electron attraction

• proton/proton repulsion …..



Internal energy depends on• Temperature

higher T = larger Ek for the nanoscale particles.

• Type of material nanoscale Ek depends upon the particle mass.

nanoscale Ep depends upon the type(s) of particle.

• Amount of material number of particles. double sample size, double Einternal, etc.



Einitial

SURROUNDINGS

SYSTEM

Efinal

Efinal

SURROUNDINGS

SYSTEM

Einitial

Calculating Thermodynamic Changes

ΔE > 0

ΔE positive: internal energy increases

Energy change of system = final E – initial E

ΔΔEE = = EEfinalfinal – – E Einitialinitial

A system can gain or lose E

E in ΔE < 0

ΔE negative: internal energy decreases

E out


• No subscript? Refers to the system: E = Esystem

• E is transferred by heat or by work.• Conservation of energy becomes: ΔE = q + w

Note the same sign convention for q and w

heat work

Calculating Thermodynamic Changes

SURROUNDINGS

SYSTEM

ΔE = q + w

Heat transfer outq < 0

Work transfer inw > 0

Heat transfer inq > 0

Work transfer outw < 0


Heat Capacity

Heat capacityHeat capacity = E required to raise the T of an object by 1°C. Varies from material to material.

Specific heat capacity (Specific heat capacity (cc))• E needed to heat 1 g1 g of substance by 1°C.

Molar heat capacity (Molar heat capacity (ccmm))

• E needed to heat 1 mole1 mole of substance by 1°C.


Heat Capacity

E required to change the T of an object is:

Heat required = mass x specific heat x ΔT

q = m c ΔT

or…

Heat required = moles x molar heat capacity x ΔT

q = n cm ΔT


Heat Capacities

Substance c (J g-1 °C-1) cm (J mol-1 °C-1)

ElementsAl(s) 0.902 24.3C (graphite) 0.720 8.65Fe(s) 0.451 25.1Cu(s) 0.385 24.4Au(s) 0.129 25.4 CompoundsNH3(l) 4.70 80.1H2O(l) 4.184 75.3C2H5OH(l) 2.46 112.(CH2OH)2(l) 2.42 149.H2O(s) 2.06 37.1CCl4(l) 0.861 132.CCl2F2(l) 0.598 72.3Common solidswood 1.76concrete 0.88glass 0.84granite 0.79


Heat Capacity

Example Example

How much energy will be used to heat 500.0 g of iron from 22°C to 55°C? cFe = 0.451 J g-1 °C-1.

Heat required = mass x specific heat x ΔT

q = m c ΔT

q = 500.0 g (0.451 J g-1 °C-1)(55−22)°C

q = 7442 J = +7.4 kJ

+ sign, E added to the system (the iron)


Heat Capacity

ExampleExample

24.1 kJ of energy is lost by a 250. g aluminum block. If the block is initially at 125.0°C what will be its final temperature? (cAl = 0.902 J g-1 °C-1)

q = m c ΔT

ΔT = q / (m c)

Thus Tfinal = ΔT + Tinital = −107 + 125°C = 18°C

heat is lost, q is negativenegative

ΔT = −24.1 x 103 J /(250. g x 0.902 J g-1 °C-1)

ΔT = Tfinal – Tinital = −107 °C


Heat CapacityA 200. g block of Cu at 500.°C is plunged into 1000. g of water (T = 23.4 °C) in an insulated container. What will be the final equilibrium T of water and Cu? (cCu = 0.385 J g-1 °C-1)

Cu cools (−q); water heats (+q); q = m c ΔT

Heat lost by Cu = −−(200. g)(0.385 J g-1 °C-1)(Tfinal− 500)

Heat gained by H2O = ++(1000. g)(4.184 J g-1 °C-1)(Tfinal− 23.4)

So: −77.0(Tfinal – 500) = 4184(Tfinal – 23.4)

(4184 + 77.0)Tfinal = 38500 + 97906

Tfinal = 32.0°C

(Note: Tfinal must be between Thot and Tcold)


Conservation of Energy and Changes of State

When heat is: AddedAdded to a system q is positive the change is endothermicendothermic

Removed Removed from a system q is negative the change is exothermicexothermic.

Water Boils: H2O(l) H2O(g) endothermic

Steam Condenses: H2O(g) H2O(l) exothermic

Work occurs as the sample expands or contracts. Overall: ΔE = q + w


A liquid cools from 45°C to 30°C, transferring 911 J to the surroundings. No work is done on or by the liquid. What is ΔEliquid?

ΔEliquid = qliquid + wliquid here wliquid = 0

Heat transfers from the liquid to the surroundings:

qliquid = -911 J (qsurroundings = +911 J)

ΔEliquid = -911J



A system does 50.2 J of work on its surroundings and there is a simultaneous 90.1 J heat transfer from the surroundings to the system. What is ΔEsystem?

Work done on the surroundings by the system

Heat transfers from the surroundings to the system

wsystem = -50.2 J qsystem = +90.1 J

ΔEsystem = qsystem + wsystem

ΔEsystem = -50.2 J +90.1 J = +39.9 J



Because ΔE = q + w:

At Constant P: ΔE = qP + watm= ΔH + watm

• Subscript P shows fixed P.

• watm = work done to push back the atmosphere

• H = enthalpy. ΔH = qp

At Constant V: ΔE = qV • subscript V shows fixed V• work requires motion against an opposing force.• constant V = no motion, so w = 0.

Enthalpy: Heat Transfer at Constant P


T

empe

ratu

re (

°C)

-50

-25

0

25

50

0 100 200 300 400 500 600 Quantity of energy transferred (J)

Water warms from 0 to 50°C

During freezing (or melting)

Example:Example:

Convert ice at -50°C to water at +50°C

• Substance loses (or gains) E, but…• T remains constant.

Ice is melting. T remains at 0°C

Ice warms from -50 to 0°C

Freezing and Melting (Fusion)


Change Name value for H2O (J/g)

solid → liq enthalpy of fusion 333

liq → gas enthalpy of vaporization 2260

ΔHfusion = qP = heat to melt a solid.

liq → solid enthalpy of freezing −333

gas → liq enthalpy of condensation −2260

Note: ΔHfusion = − ΔHfreezing etc.

qfusion = −qfreezing

Changes of State


State functionsState functions

Always have the same value whenever the system is in the same state.

Two equal mass samples of water produced by:

1. Heating one from 20°C to 50°C.

2. Cooling the other from 100°C to 50°C.

have identical final H (and V, P, E…).

State functions

HH E P V T etc.

State Functions and Path Independence

State function changes are path independentpath independent.

ΔH = Hfinal – Hinitial is constant.


CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) ΔH° = −803.05 kJ

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ΔH° = −890.36 kJ

ΔH = qP can be added to a balanced equation.

ΔH° is the standard enthalpy change P = 1 bar. T must be stated (if it isn’t, assume 25°C). ΔH° is a molar value. Burn 1 mol of CH4 with 2 mol O2 to

form 2 mol of liquid waterliquid water and release 890 kJ of heat Change a physical state, change ΔH° : H2O(l) vs. H2O(g)

Thermochemical Expressions


Enthalpy Changes for Chemical Reactions

Calculate the heat generated when 500. g of propane burns in excess O2.

C3H8(l) + 5 O2(g) 3 CO2(g) + 4 H2O(l) ΔH° = – 2220. kJ

Molar mass of C3H8 = 44.097 g/mol.

nC3H8 = (500. g) / (44.097 g/mol)

= 11.34 mol C3H8

Since ΔH° = qp = –2220. kJ/(1 mol C3H8)

q = (11.34 mol C3H8) (–2220. kJ / mol C3H8 )

= –2.52 x 104 kJ


Where Does the Energy Come From?

Bond Enthalpy (bond energy)Bond Enthalpy (bond energy)• Equals the strength of 1 mole of bonds

• Always positive It takes E to break a bond Separated parts are less stable than the molecule. Less stable = higher E

• E is always released when a bond forms Product is more stable than the separated parts. More stable = lower E


Bond Enthalpies

During a chemical reaction:During a chemical reaction:

Old bonds break: requires E (endothermic)

Both typically occur:

H2(g) + Cl2(g) 2 H(g) + 2 Cl(g) 2 HCl(g)

endothermicΔH= +678 kJ/mol

exothermic ΔH= -862 kJ/mol

New bonds form: releases E (exothermic)


Bond Enthalpies

Endothermic reactionsEndothermic reactions (ΔH > 0) E is absorbed. New bonds are less stable than the old,

or Fewer bonds are formed than broken

Overall, heat may be absorbed or released:

Exothermic reactionsExothermic reactions (ΔH < 0) E is released. New bonds are more stable than the old,

or More bonds are formed than broken.

reactants

products

ener

gyle

ss s

tabi

lity

reactants

productsener

gyle

ss s

tabi

lity


Measuring Enthalpy Changes

Heat transfers are measured with a calorimetercalorimeter.

Common types:

• Bomb calorimeter. rigid steel container. filled with O2(g) and a small sample to be burnt.

constant V, so qV = ΔE

• Flame calorimeter. samples burnt in an open flame. constant P, so qp = ΔH

• Coffee-cup calorimeter in lab (constant P).


or

−qreaction = qbomb + qwater

with

qbomb = mcalccalΔT = CcalΔT

Bomb CalorimeterBomb Calorimeter

Measure ΔT of the water. Constant V: qV = ΔE

Conservation of E:

qreaction + qbomb + qwater = 0


A constant for a calorimeter


Octane (0.600 g) was burned in a bomb calorimeter containing 751 g of water. T increased from 22.15°C to 29.12°C. Calculate the heat evolved per mole of octane burned. Ccal = 895 J°C-1.

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(l)

−qreaction = qbomb + qwater

qbomb = CcalΔT = 895 J°C-1 (29.12 – 22.15)°C = +6238 J

qwater = m c ΔT = 751 g (4.184 J g-1 °C-1)(29.12 – 22.15)°C

= +2.190 x 104 J

So −qreaction = +6238 + 2.190 x 104 J = 2.81 x 104 J = 28.1 kJ

qreaction = −28.1 kJ



Octane (0.600 g) was burned in a bomb calorimeter… Calculate the heat evolved per mole of octane burned

Molar mass of C8H18 = 114.23 g/mol.

nC8H18 = (0.600 g) / (114.23 g/mol)

= 0.00525 mol C8H18

Heat evolved /mol octane = −28.1 kJ 0.00525 mol

= −5.35 x 103 kJ/mol = −5.35 MJ/mol



Coffee-cup calorimeterCoffee-cup calorimeterNested styrofoam cups prevent heat transfer with the surroundings.

Constant P. ΔT measured.q = qp = ΔH

Assume the cups do not absorb heat. (Ccal = 0 and qcal = 0)



0.800g of Mg was added to 250. mL of 0.40 M HCl in a coffee-cup calorimeter at 1 bar. Tsolution increased from 23.4 to 37.9°C. Assume csolution = 4.184 J g-1°C-1 and complete the equation:

Mg(s) + 2 HCl(aq) H2(g) + MgCl2(aq) ΔH = ?

nMg = 0.800 g = 0.03291 mol

nHCl = 0.250 L = 0.100 mol

1 mol24.31 g

0.4 mol1 L

1Mg ≡ 2HCl Mg is limitingMg is limiting



qsolution = msolution c ΔT (msolution = macid + mMg )

= 250.8 g (4.184 J g-1 °C-1)(37.9 − 23.4)°C

= 15,220 J

So = or ΔH = – 462 kJΔH1 mol Mg

–15.22 kJ0.03291mol

Heat fromfrom the reaction went intointo the solution. So:qsolution = – qreaction

qreaction = –15.22 kJ = ΔH

exothermicexothermic

… 0.800g of Mg was added to 250. mL of 0.40 M HCl. T increased from 23.4 to 37.9°C. Assume the solution has c = 4.184 J g-1°C-1. ΔH = ?



Hess’s Law

“If the equation for a reaction is the sum of the equations for two or more other reactions, then ΔH° for the 1st reaction must be the sum of the ΔH° values of the other reactions.”

Another version:Another version:“ΔH° for a reaction is the same whether it takes place in a single step or several steps.”

HH is a state function is a state function


Hess’s LawMultiply a reaction, multiply ΔH.

Reverse a reaction, change the sign of ΔH.

Then

2 CO2(g) → 2 CO(g) + O2(g) ΔH = –1(–566.0 kJ)

= + 566.0 kJ

4 CO2(g) → 4 CO(g) + 2 O2(g) ΔH = –2(–566.0 kJ)

= +1132.0 kJ

2 CO(g) + O2(g) → 2 CO2 (g) ΔH = −566.0 kJ


Use Hess’s Law to find ΔH for unmeasured reactions.

ExampleExample

It is difficult to measure ΔH for:

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

Some CO2 always forms. Calculate ΔH given:

C(graphite) + O2(g) CO2(g) ΔH = −393.5 kJ

2 CO(g) + O2(g) 2 CO2(g) ΔH = −566.0 kJ

Hess’s Law


Rearrange:

or:

+2[C + O2 CO2] +2(−393.5) = −787.0

−1[2 CO + O2 2 CO2] −1(−566.0) = +566.0

Calculate ΔH for the reaction: 2C(graphite) + O2(g) → 2CO(g)

2 C + 2 O2 2 CO2 ΔH° = −787.0 kJ

2 CO2 2 CO + O2 ΔH° = +566.0 kJ

Hess’s Law

Add, then cancel:

2 C + 2 O2 + 2 CO2 2 CO2 + 2 CO + O2 −221.0

2 C + O2 2 CO ΔH° = −221.0 kJ


Determine ΔH° for the production of coal gas:2 C(s) + 2 H2O(g) CH4(g) + CO2(g)

C(s) + H2O(g) CO(g) + H2(g) ΔH° = 131.3 kJ

CO(g) + H2O(g) CO2(g) + H2(g) ΔH° = −41.2 kJ

CH4(g) + H2O(g) CO(g) + 3 H2(g) ΔH° = 206.1 kJ

AA

BB

CC

Using:

Hess’s Law


Want: 2 C(s) + 2 H2O(g) CH4(g) + CO2(g)

2 C on left, use 2 x AA

2 C(s) + 2 H2O(g) 2 CO(g) + 2 H2(g) +262.6

1 CO2 on right, so use 1 x BB

CO(g) + H2O(g) CO2(g) + H2(g) −41.2

1 CH4 on right, use −1 x CC

CO(g) + 3 H2(g) CH4(g) + H2O(g) −206.1

Add and cancel:2C + 3H2O + 2CO + 3H2 2CO + 3H2 + CH4 + H2O + CO2 15.3 kJ

change to 2 H2Ochange to 2 H2O 2 C + 2 H2O → CH4 + CO2 ΔH = 15.3 kJ2 C + 2 H2O → CH4 + CO2 ΔH = 15.3 kJ

Hess’s Law


1

but the formation reaction is:

H2(g) + ½ O2(g) H2O(l) ΔHf° = −285.83 kJ

Hess’s law problems often use a combustion or …

H2 combustion:

2 H2(g) + O2(g) 2 H2O(l) ΔH° = −571.66 kJ

f = formation

Standard Molar Enthalpy of Formation

Formation reactionFormation reaction

Make 1 mol of compound from its elements in their standard states.


Standard stateStandard state = most stable form of the pure element at P = 1 bar.

e.g. C standard state = graphite (not diamond)

ΔHf° for any element in its standard state is zero.

(take 1 mol of the element and make… 1 mol of element)

ΔHf° (Br2(l) ) = 0 at 298 K

ΔHf° (Br2(g) ) ≠ 0 at 298 K

Standard Molar Enthalpy of Formation


Formula Name Hf°, kJ/mol

Al2O3(s) aluminum oxide −1675.7

CaO(s) calcium oxide −635.09

CH4(g) methane −74.81

C2H2(g) acetylene +226.73

C2H4(g) ethylene +52.26

C2H6(g) ethane −84.68

C2H5OH(l) ethanol −277.69

H2O(g) water vapor −241.818

H2O(l) liquid water −285.830

NaF(s) sodium fluoride −573.647

NaCl(s) sodium chloride −411.153

SO2(g) sulfur dioxide −296.830

SO3(g) sulfur trioxide −395.72

Appendix J

NotesNotes• Most are negative

(formation releases E), but can be positive.

• If the physical state changes, ΔHf° changes.

Standard Enthalpy of Formation (25°C)


ΔH° ={(nproducts)(ΔHf° products)}

– {(nreactants)(ΔHf° reactants)}

ΔH° = [1ΔHf°(HCN) + 33ΔHf°(H2)] −

[1ΔHf°(NH3) + 1ΔHf°(CH4)]

Standard Molar Enthalpies of Formation

= [+134 + 3(0)] − [− 46.11 + (−74.85)] = 255 kJ/mol

ExampleExample

Calculate ΔH° for:

CH4(g) + NH3(g) HCN(g) + 3 H2(g)


Hydrazine

N2H4(g) + O2(g) → N2(g) + 2H2O(g)

Chemical Fuels for Home and Industry

Chemical FuelChemical Fuel – reacts exothermically with O2 in air.

A good fuel has weak bonds and/or strong product bonds.

Ebond(kJ/mol) 160(NN), 391(NH) 498(OO) 946(NN) 467(OH)

Emolecule 1724 498 946 1864Ereagents 2222 2814

ΔE -592 kJ/mol


• Fuel value = (E released) / (mass of fuel in g)

• Energy density = (E released) / Vfuel

U.S. sources of E:

Mostly fossil fuels.

Chemical Fuels for Home and Industry

Good fuels have large:


Foods: Fuels for Our BodyCarbohydrates, Cx(H2O)y, are converted to glucose, C6H12O6

Glucose is the body’s fuel

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O ΔH° = −2801.6 kJ

(average carbohydrate =17 kJ/g or 4 Cal/g)

Excess glucose fat. Fat is metabolized when needed:

2 C57H110O6 + 163 O2 → 114 CO2 + 110 H2O ΔH° = −75,520 kJ

(average fat = 38 kJ/g or 9 Cal/g)

Metabolism of dietary protein releases 17 kJ/g or 4 Cal/g.


Approximate Composition

Food per 100. g Caloric Value

Fat Carbohydrate Protein Cal/g kJ/g

All-purpose flour 0.0 73.3 13.3 3.33 13.95

Apple 0.5 13.0 0.4 0.59 2.47

Brownie with nuts 16.0 64.0 4.0 4.04 16.9

Egg 0.7 10.0 13.0 1.40 5.86

Hamburger 30.0 0.0 22.0 3.60 15.06

Peanuts 50.0 21.4 28.6 5.71 23.91

Rice 1.0 77.6 8.2 3.47 14.52

Basal metabolic rate (BMR) = minimum energy to maintain a body at rest.

≈ 1 Cal kg-1 hr-1

Metabolizing food requires energy and this adds ≈ 10% to the BMR.

Foods: Fuels for Our Body


Summary ProblemSO2 is a major pollutant emitted by coal-fired electric power plants. A large plant can produce 8.64 x 1013 J of electricity each day by burning 7000 tons of coal (1 ton = 9.08 x 105 g).

(a) Assume coal ≈ graphite. Calculate the energy transferred per day to the surroundings by coal combustion.

(b) What is the efficiency of the plant; what % of thermal energy becomes electrical energy?

(c) SO2 can be trapped by MgO in the smokestack to form MgSO4: SO2 + MgO + ½ O2 → MgSO4

If 140 tons of SO2 is emitted each year, how much MgO is needed? How much MgSO4 is produced?

(d) How much heat does the reaction above add/remove?


Summary Problem…8.64 x 1013 J/day from 7000 tons of coal (1 ton = 9.08 x 105 g).

(a) Calculate the E transferred/day to the surroundings.

C + O2 → CO2 ΔH° = ΔHf° = −393.509 kJ/mol

ncoal = 7000 tons = 5.292 x 108 mol1 mol

12.011 g

9.08 x 105g

1 ton

Heat released = 5.292x108 mol = −2.08 x 1011 kJ

= −2.08 x 1014 J

–393.509 kJ

1 mol

(b) What % of thermal E becomes electrical E?

Efficiency = (Eout/Ein) x 100% = x100% = 41.5%8.64 x 1013 J

2.08 x 1014 J


Summary Problem(c) SO2 can be trapped: SO2 + MgO + ½O2 → MgSO4

140 tons of SO2 is emitted/year. MgO needed? MgSO4 produced?

140 tons = 140 x 9.08 x 105g = 1.271 x 108g

nSO2 = 1.271 x 108g /64.07 g mol-1 = 1.984 x 106 mol

1 MgO ≡ 1 SO2

So 1.984 x 106 mol MgO required

= 1.984 x 106 mol x (40.30 g/mol) = 7.996 x 107 g

= 88.1 tons of MgO

1 MgO ≡ 1MgSO4 So 1.984 x 106 mol MgSO4 produced

= 1.984 x 106 mol x (120.4 g/mol) = 2.39 x 108 g

= 263 tons of MgSO4


Summary ProblemSO2 + MgO + ½O2 → MgSO4 ΔH° = ?

(d) How much heat does the reaction add/remove?

nSO2 = 1.984 x 106 mol

1 ΔH° ≡ 1 SO2

ΔH°? Use ΔHf° values

ΔH° = ΔHf°(MgSO4) − ΔHf°(SO2 ) − ΔHf°(MgO) − ½ΔHf°(O2 )

= −1284.9 −(−296.83) − (−601.70) − ½(0) = −386.4 kJ

Heat produced = 1.984 x 106 x (-386.4) kJ

= −7.67 x 108 kJ

7.67 x 1011 J of heat are addedadded to the combustion

chemistry: the molecular science moore, stanitski and jurs

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