thermochemistry

ThermochemistryThermochemistryBy By

Sumana RamakrishnanSumana RamakrishnanAndAnd

Reema SilReema Sil

ThermochemistryThermochemistry is is the study of heat change the study of heat change in chemical reactionsin chemical reactions

ExothermicExothermicis any process that gives off

heat ndash transfers thermal energy from the system to the surroundings

Endothermic is any process in which heat

has to be supplied to the system from the surroundings

DCHeatBA

HeatDCBA

EnthalpyEnthalpy is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

tsreacproductsreaction HHH tan

Hproducts lt Hreactants Hproducts gt Hreactants

∆H lt 0

∆H gt 0

Exothermic(system gives off

heat)

Endothermic (system absorbs

heat)Heat (q) absorbed or released Tmcq

m mass of a given quantity in grams (g)c specific heat the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius (Jg-1degC-1)∆t = temperaturefinal ndash temperatureinitial (degC)

CC66HH55OHOH(s)(s) + 7 O + 7 O22(g)(g) rarr 6 CO rarr 6 CO22(g)(g) + 3 H + 3 H22OO(l)(l)

When a 2000-gram sample of pure phenol CWhen a 2000-gram sample of pure phenol C66HH55OHOH(s)(s) is completely burned is completely burned according to the equation above 6498 kilojoules of heat is released Use according to the equation above 6498 kilojoules of heat is released Use the information in the table below to answer the questions that followthe information in the table below to answer the questions that follow

Substance

Standard Heat of Formation ∆H˚ at

25˚C (kJmol) Absolute Entropy S˚

at 25˚C (JmolK)

C(graphite) 000 569

CO2(g) -3935 2136

H2(g) 000 1306

H2O(l) -28585 6991

O2(g) 000 2050

C6H5OH(s) 1440

(a) Calculate the molar heat of combustion of phenol in kilojoules per mole at 25˚C

(b) Calculate the standard heat of formation ∆H˚ of phenol in kilojoules per mole at 25˚C

(c) Calculate the value of the standard free-energy change ∆G˚ for the combustion of phenol at 25˚C

(d) If the volume of the combustion container is 100 liters calculate the final pressure in the container when the temperature is changed to 110˚C (Assume no oxygen remains unreacted and that all products are gaseous)

CC66HH55OHOH(s)(s) + 7 O + 7 O22(g)(g) rarr 6 CO rarr 6 CO22(g)(g) + 3 H + 3 H22OO(l)(l) When a 2000-gram sample of pure phenol CWhen a 2000-gram sample of pure phenol C66HH55OHOH(s)(s) is completely is completely burned according to the equation above 6498 kilojoules of heat is burned according to the equation above 6498 kilojoules of heat is

released released


combH

molarmassmass

kJ

The molar heat of combustion of The molar heat of combustion of phenol isphenol is

-3058 kilojoules-3058 kilojoules

mol

g

g

kJ

11394

0002

9864

molkJ3058

CC66HH55OHOH(s)(s) + 7 O + 7 O22(g)(g) rarr 6 CO rarr 6 CO22(g)(g) + 3 H + 3 H22OO(l)(l) When a 2000-gram sample of pure phenol CWhen a 2000-gram sample of pure phenol C66HH55OHOH(s)(s) is completely burned is completely burned

according to the equation above 6498 kilojoules of heat is released Use the according to the equation above 6498 kilojoules of heat is released Use the information in the table below to answer the questions that followinformation in the table below to answer the questions that follow

(From (a) ∆Hrxn = -3058 kJ)

Substance

Standard Heat of Formation ∆H˚ at 25˚C

(kJmol)

C(graphite) 000

CO2(g) -3935

H2(g) 000

H2O(l) -28585

O2(g) 000

C6H5OH(s)


Equation

Variables

Numbers

tsreacproductsreaction HHH tan xOHCOH reaction 22 36 x 52853539363058

molkJx 161



(c) (c) Calculate the value of the standard free-energy change ∆G˚ for the combustion of phenol at 25˚C

)tan()( tsreacproducts SSS First you must solve for ∆S

25622 736 OOHHCOHCO

0205701449169362136 J6787Then plug values into

STHG kJ0876702983058

kJ3032

STHG




01907 moles of

gaseous products

nRTPV

V

nRTP

L

Kmol molKatmL

10

38308206019070

atm5990

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

molCO

56

2

1

6

21270 molCO

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

OmolH

56

2

1

3 OmolH206370

nRTPV

2 NO2 NO(g)(g) + O + O22(g)(g) rarr 2 NOrarr 2 NO22(g)(g) ∆H˚= -1141 kJ ∆S˚= -1465 J K∆H˚= -1141 kJ ∆S˚= -1465 J K-1-1

The reaction represented above is one that contributes significantly to the The reaction represented above is one that contributes significantly to the formation of photochemical smogformation of photochemical smog

(a) Calculate the quantity of heat released when 731 g of NO(a) Calculate the quantity of heat released when 731 g of NO(g)(g) is converted to NO is converted to NO22(g)(g)(b) For the reaction at 25˚C the value of the standard free-energy change ∆G˚ is -704 kJ(b) For the reaction at 25˚C the value of the standard free-energy change ∆G˚ is -704 kJ

(i) Calculate the value of the equilibrium constant K(i) Calculate the value of the equilibrium constant Keqeq for the reaction at 25˚C for the reaction at 25˚C(ii) Indicate whether the value of ∆G˚ would become more negative less negative or (ii) Indicate whether the value of ∆G˚ would become more negative less negative or remain unchanged as the temperature is increased Justify your answerremain unchanged as the temperature is increased Justify your answer

(c) Use the data in the table to calculate the value of the standard molar entropy S˚ for (c) Use the data in the table to calculate the value of the standard molar entropy S˚ for OO22(g)(g) at 25˚C at 25˚C

(d) Use the data in the table to calculate the bond energy in kJ mol(d) Use the data in the table to calculate the bond energy in kJ mol-1-1 of the of the nitrogen-oxygen bond in NOnitrogen-oxygen bond in NO22 Assume that the bonds in the NO Assume that the bonds in the NO22 molecule are molecule are

equivalent (ie they have the same energy)equivalent (ie they have the same energy)

Standard Molar Standard Molar Entropy S˚ Entropy S˚

(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401

Bond EnergyBond Energy

(kJ mol(kJ mol-1-1))

Nitrogen-oxygen bond in NONitrogen-oxygen bond in NO 607607

Oxygen-oxygen bond in OOxygen-oxygen bond in O22 495495

Nitrogen-oxygen bond in NONitrogen-oxygen bond in NO22



(a) Calculate the quantity of heat released when 731 g of NO(a) Calculate the quantity of heat released when 731 g of NO(g)(g) is converted to is converted to NONO22(g)(g)

gNO173

gNO

molNO

0130

1

molNO

kJ

2

1114 kJ97138

The reason our answer is negative is because the reaction is exothermic

Even though it results in a negative change in heat for the reaction

the amount of heat released is 13897kJ (a positive quantity)

So the answer to the question in part (a) is the positive quantity

13897 kJ



(b) For the reaction at 25˚C the value of the standard free-energy change ∆G˚ is -704 kJ(b) For the reaction at 25˚C the value of the standard free-energy change ∆G˚ is -704 kJ

(i) Calculate the value of the equilibrium constant K(i) Calculate the value of the equilibrium constant Keqeq for the reaction at 25˚C for the reaction at 25˚C

eqKmolJ KKJ ln298314840070

eqKln428

eqKe 428

12101582 eqK

eqKRTG ln




(ii) Indicate whether the value of ∆G˚ would become more negative less negative or (ii) Indicate whether the value of ∆G˚ would become more negative less negative or remain unchanged as the remain unchanged as the temperature is increased Justify your answertemperature is increased Justify your answer

As T increases the quantity T∆S becomes As T increases the quantity T∆S becomes more and more negativemore and more negative

According to the equation subtracting a According to the equation subtracting a large negative number (T∆S) from another large negative number (T∆S) from another

negative number (∆H) is the same as negative number (∆H) is the same as addingadding a value to the negative ∆H making a value to the negative ∆H making

∆G˚ less negative∆G˚ less negative

∆∆G˚ will become less negative because of G˚ will become less negative because of the roles of ∆H T and ∆S in the equationthe roles of ∆H T and ∆S in the equation

STHG



(c) Use the data in the table below to calculate the value of the standard molar (c) Use the data in the table below to calculate the value of the standard molar entropy S˚ for Oentropy S˚ for O22(g)(g) at 25˚C at 25˚C


(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401

∆S˚ = sum S˚products - sum S˚reactants

The standard molar entropy for O2(g) is 2051 J K-1 mol-1

22 22 OSNOSNOSS xS 82102124025146

molKJx 1205642124805146



(d) Use the data in the table below to calculate the bond energy in kJ mol(d) Use the data in the table below to calculate the bond energy in kJ mol -1-1 of of the nitrogen-oxygen bond in NOthe nitrogen-oxygen bond in NO22 Assume that the bonds in the NO Assume that the bonds in the NO22 molecule are equivalent (ie they have the same energy)molecule are equivalent (ie they have the same energy)






(bonds of reactants are broken bonds of products are formed)

dbondsformenbondsbroke HHH

ctsbondsprodutsbondsreac HHH tan

molkJx 5911rarr

In this equation x represents the total bond energy in an NO2 molecule Since there are

two N-O bonds in NO2 x divided by two equals the bond energy of each N-O bond in

the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2

Answer the following questions that relate to the Answer the following questions that relate to the chemistry of nitrogenchemistry of nitrogen

(a) Two nitrogen atoms combine to (a) Two nitrogen atoms combine to form a nitrogen molecule as form a nitrogen molecule as represented by the below equation represented by the below equation Using the table of average bond Using the table of average bond energies determine the enthalpy energies determine the enthalpy change ∆H for the reactionchange ∆H for the reaction

)(2)(2 gg NN

BondBond Average Bond Average Bond Energy (kJ molEnergy (kJ mol-1-1))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950

(b) The reaction between nitrogen and hydrogen to from ammonia is (b) The reaction between nitrogen and hydrogen to from ammonia is represented below Predict the sign of the standard entropy change represented below Predict the sign of the standard entropy change ∆S˚ for the reaction Justify your answer∆S˚ for the reaction Justify your answer

)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ

(c) The value of ∆G˚ for the reaction represented in part (b) is (c) The value of ∆G˚ for the reaction represented in part (b) is negative at low temperatures but positive at high temperatures negative at low temperatures but positive at high temperatures ExplainExplain

(d) When N(d) When N22(g)(g) and Hand H22(g)(g) are placed in a sealed contained at a low are placed in a sealed contained at a low temperature no measurable amount of NHtemperature no measurable amount of NH33(g)(g) is produced Explain is produced Explain

(a) Two nitrogen atoms combine to form a nitrogen molecule (a) Two nitrogen atoms combine to form a nitrogen molecule as represented by the below equation Using the table of as represented by the below equation Using the table of

average bond energies determine the enthalpy change ∆H average bond energies determine the enthalpy change ∆H for the reactionfor the reaction

)(2)(2 gg NN

BondBondAverage Bond Average Bond Energy (kJ molEnergy (kJ mol--

11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500

The enthalpy change when two nitrogen atoms combine is -950 kJ The reaction only forms bonds which requires heat

energy

This makes the equation endothermic because it requires a heat supply to form N2(g)

(b) The reaction between nitrogen and hydrogen to from (b) The reaction between nitrogen and hydrogen to from ammonia is represented below Predict the sign of the ammonia is represented below Predict the sign of the

standard entropy change ∆S˚ for the reaction Justify your standard entropy change ∆S˚ for the reaction Justify your answeranswer

∆∆S˚S˚ (change in entropy) is the measure of disorder in a reaction This measure of entropy can be judged by the change in moles of gaseous particles

In this reaction since there are fewer gaseous moles of products than there are gaseous moles of reactants there is a decrease in disorder - meaning ∆S is negative for this reaction

)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ

productstsreac MolesGasMolesGas tan


SinEntropyMolesGas

SinEntropyMolesGas

(c) The value of ∆G˚ for the reaction represented in part (b) is (c) The value of ∆G˚ for the reaction represented in part (b) is negative at low temperatures but positive at high negative at low temperatures but positive at high

temperatures Explaintemperatures Explain

The value of ∆G free energy is determined by the value of ∆H ndash T(∆S) We determined that both ∆H and ∆S are negative (from part b) and T will always be positive (because temperature in Kelvin is never negative

If T is a small value T(∆S) will not be large enough to overcome ∆H Thus ∆G will be negative

If T is a large value then T(∆S) will be large enough to overcome ∆H and make ∆G positive

So subtracting the negative value of T(∆S) from ∆H can result in either a negative or positive number

STHG

eL

Small

arg

(d) When N(d) When N22(g) (g) and Hand H22(g)(g) are placed in a sealed are placed in a sealed contained at a low temperature no measurable contained at a low temperature no measurable

amount of NHamount of NH33(g)(g) is produced Explain is produced Explain

As we determined in Part C when at a low temperature ∆G is negative When ∆G is negative it means that the reaction is spontaneous proceeding in the forward direction on its own

However if the reaction does not seem to visibly proceed it does not mean that the reaction is not spontaneous

There are two reasons for why a measurable amount of NH3(g) is not observed- All reactions require a certain activation energy to initiate the

reaction If a reaction does not have that required energy the reaction will remain inert even if it is spontaneous

- The reaction might require an extensive amount of time to proceed Because the reaction is so slow the amount of NH3(g) produced may not be immediately visible but the reaction continues to occur

)(3)(2)(2 23 ggg NHHN

A student is asked to determine the molar enthalpy of neutralization ∆Hneut for the reaction represented above The student combines equal volumes of 10M HCl and 10 M NaOH in an open polystyrene cup calorimeter The heat released by the reaction is determined by using the equation q=mc ∆TAssume the following

Both solutions are at the same temperature before they are combinedThe densities of all the solutions are the same that of waterAny heat lost to the calorimeter of to the air is negligibleThe specific heat capacity of the combined solutions is the same as that of

water

(a) Give appropriate units for each of the terms in the equation q = mc ∆T(b) List the measurements that must be made in order to obtain the value of q(c) Explain how to calculate each of the following

I The number of moles of water formed during the experimentII The value of the molar enthalpy of neutralization ∆Hneut for the reaction between HCl(aq) and NaOH(aq)

(d) The student repeats the experiment with the same equal volumes as before but this time uses 20 M HCl and 20M NaOHI Indicate whether the value of q increases decreases or stays the same when compared to the first experiment Justify your predictionII Indicate whether the value of the molar enthalpy of neutralization ∆Hneut increases decreases or stays the same when compared to the first experiment Justify your prediction

(e) Suppose that a significant amount of heat were lost to the air during the experiment What effect would that have on the calculated value of the molar enthalpy of neutralization ∆Hneut Justify your answer

)(2)()( laqaq OHOHH


Both solutions are at the same temperature before they are combined

The densities of all the solutions are the same that of waterAny heat lost to the calorimeter of to the air is negligibleThe specific heat capacity of the combined solutions is the same as

that of water

Tmcq (a) Give appropriate units for each of the terms in the equation q = mc ∆T

CaloriesKilojoules GramsKilo

Kg

Jor

Cg

J

KC




that of water

(b) List the measurements that must be made in order to obtain the value of q

1 Volume or mass of the HCl and NaOH solutions2 Initial temperature of HCl and NaOH before mixing

3 Final temperature of HCl and NaOH after mixing




that of water

(c) Explain how to calculate each of the followingI The number of moles of water formed during the experiment

)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01

You could also reason that both HCl and NaOH react in a 11 ratio to form the same number of moles of water




that of water

(c) Explain how to calculate each of the followingII The value of the molar enthalpy of neutralization ∆Hneut for the reaction between HCl(aq) and NaOH(aq)

First you must solve for the heat produced using q=cm ∆T Then dividing this value by the moles of H2O (from part I)

OmolH

qH neut

2

This gives you the molar enthalpy of neutralization for the reaction


Both solutions are at the same temperature before they are combinedThe densities of all the solutions are the same that of waterAny heat lost to the calorimeter of to the air is negligibleThe specific heat capacity of the combined solutions is the same as that of water

(d) The student repeats the experiment with the same equal volumes as before but this time uses 20 M HCl and 20M NaOHI Indicate whether the value of q increases decreases or stays the same when compared to the first experiment Justify your prediction

Tmcq You can reason that since more moles of HCl and NaOH are reacting

the final temperature of the mixture will be higher

So ∆T will be greater which will increase q

(d) The student repeats the experiment with the same equal volumes as before but this time uses 20 M HCl and 20M NaOHII Indicate whether the value of the molar enthalpy of neutralization ∆Hneut increases decreases or stays the same when compared to the first experiment Justify your prediction




that of water

As done in part C II the molar enthalpy is found by dividing q by the moles of water In the reaction both q and water increase proportionally by x

OmolHx

qxH neut

2

OmolH

q

2

Since x drops out you can see that the value of the molar enthalpy of neutralization remains the same as in the first experiment




that of water(e) Suppose that a significant amount of heat were lost to the air during the

experiment What effect would that have on the calculated value of the molar enthalpy of neutralization ∆Hneut Justify your answer

Heat lost to the air will decrease ∆T ndash which in turn equals a smaller q

Thus when you divide q by the number of moles of H2O ∆Hneut will be a smaller value

Tmcq

OmolH

qH neut

2

The EndThe End

Voices by Sumana Ramakrishnan Voices by Sumana Ramakrishnan and Reema Siland Reema Sil

Powerpoint by Reema SilPowerpoint by Reema Sil

AP Problems painstakingly solved by bothAP Problems painstakingly solved by both

Special Thanks toSpecial Thanks toMr GangluffMr Gangluff

Mr AmendolaMr AmendolaEric LiuEric Liu

Jack WangJack Wang

ThermochemistryThermochemistry is is the study of heat change the study of heat change in chemical reactionsin chemical reactions

ExothermicExothermicis any process that gives off

heat ndash transfers thermal energy from the system to the surroundings

Endothermic is any process in which heat

has to be supplied to the system from the surroundings

DCHeatBA

HeatDCBA




∆H lt 0

∆H gt 0


heat)






Substance



at 25˚C (JmolK)

C(graphite) 000 569

CO2(g) -3935 2136

H2(g) 000 1306

H2O(l) -28585 6991

O2(g) 000 2050

C6H5OH(s) 1440






released released


combH

molarmassmass

kJ



mol

g

g

kJ

11394

0002

9864

molkJ3058




Substance


(kJmol)

C(graphite) 000

CO2(g) -3935

H2(g) 000

H2O(l) -28585

O2(g) 000

C6H5OH(s)


Equation

Variables

Numbers


molkJx 161





25622 736 OOHHCOHCO


STHG kJ0876702983058

kJ3032

STHG




01907 moles of

gaseous products

nRTPV

V

nRTP

L

Kmol molKatmL

10

38308206019070

atm5990

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

molCO

56

2

1

6

21270 molCO

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

OmolH

56

2

1

3 OmolH206370

nRTPV









(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401









gNO173

gNO

molNO

0130

1

molNO

kJ

2

1114 kJ97138





13897 kJ






eqKln428

eqKe 428

12101582 eqK

eqKRTG ln










STHG





(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401



22 22 OSNOSNOSS xS 82102124025146

molKJx 1205642124805146












molkJx 5911rarr



the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang




∆H lt 0

∆H gt 0


heat)






Substance



at 25˚C (JmolK)

C(graphite) 000 569

CO2(g) -3935 2136

H2(g) 000 1306

H2O(l) -28585 6991

O2(g) 000 2050

C6H5OH(s) 1440






released released


combH

molarmassmass

kJ



mol

g

g

kJ

11394

0002

9864

molkJ3058




Substance


(kJmol)

C(graphite) 000

CO2(g) -3935

H2(g) 000

H2O(l) -28585

O2(g) 000

C6H5OH(s)


Equation

Variables

Numbers


molkJx 161





25622 736 OOHHCOHCO


STHG kJ0876702983058

kJ3032

STHG




01907 moles of

gaseous products

nRTPV

V

nRTP

L

Kmol molKatmL

10

38308206019070

atm5990

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

molCO

56

2

1

6

21270 molCO

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

OmolH

56

2

1

3 OmolH206370

nRTPV









(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401









gNO173

gNO

molNO

0130

1

molNO

kJ

2

1114 kJ97138





13897 kJ






eqKln428

eqKe 428

12101582 eqK

eqKRTG ln










STHG





(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401



22 22 OSNOSNOSS xS 82102124025146

molKJx 1205642124805146












molkJx 5911rarr



the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang



Substance



at 25˚C (JmolK)

C(graphite) 000 569

CO2(g) -3935 2136

H2(g) 000 1306

H2O(l) -28585 6991

O2(g) 000 2050

C6H5OH(s) 1440






released released


combH

molarmassmass

kJ



mol

g

g

kJ

11394

0002

9864

molkJ3058




Substance


(kJmol)

C(graphite) 000

CO2(g) -3935

H2(g) 000

H2O(l) -28585

O2(g) 000

C6H5OH(s)


Equation

Variables

Numbers


molkJx 161





25622 736 OOHHCOHCO


STHG kJ0876702983058

kJ3032

STHG




01907 moles of

gaseous products

nRTPV

V

nRTP

L

Kmol molKatmL

10

38308206019070

atm5990

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

molCO

56

2

1

6

21270 molCO

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

OmolH

56

2

1

3 OmolH206370

nRTPV









(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401









gNO173

gNO

molNO

0130

1

molNO

kJ

2

1114 kJ97138





13897 kJ






eqKln428

eqKe 428

12101582 eqK

eqKRTG ln










STHG





(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401



22 22 OSNOSNOSS xS 82102124025146

molKJx 1205642124805146












molkJx 5911rarr



the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang


released released


combH

molarmassmass

kJ



mol

g

g

kJ

11394

0002

9864

molkJ3058




Substance


(kJmol)

C(graphite) 000

CO2(g) -3935

H2(g) 000

H2O(l) -28585

O2(g) 000

C6H5OH(s)


Equation

Variables

Numbers


molkJx 161





25622 736 OOHHCOHCO


STHG kJ0876702983058

kJ3032

STHG




01907 moles of

gaseous products

nRTPV

V

nRTP

L

Kmol molKatmL

10

38308206019070

atm5990

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

molCO

56

2

1

6

21270 molCO

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

OmolH

56

2

1

3 OmolH206370

nRTPV









(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401









gNO173

gNO

molNO

0130

1

molNO

kJ

2

1114 kJ97138





13897 kJ






eqKln428

eqKe 428

12101582 eqK

eqKRTG ln










STHG





(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401



22 22 OSNOSNOSS xS 82102124025146

molKJx 1205642124805146












molkJx 5911rarr



the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang




Substance


(kJmol)

C(graphite) 000

CO2(g) -3935

H2(g) 000

H2O(l) -28585

O2(g) 000

C6H5OH(s)


Equation

Variables

Numbers


molkJx 161





25622 736 OOHHCOHCO


STHG kJ0876702983058

kJ3032

STHG




01907 moles of

gaseous products

nRTPV

V

nRTP

L

Kmol molKatmL

10

38308206019070

atm5990

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

molCO

56

2

1

6

21270 molCO

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

OmolH

56

2

1

3 OmolH206370

nRTPV









(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401









gNO173

gNO

molNO

0130

1

molNO

kJ

2

1114 kJ97138





13897 kJ






eqKln428

eqKe 428

12101582 eqK

eqKRTG ln










STHG





(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401



22 22 OSNOSNOSS xS 82102124025146

molKJx 1205642124805146












molkJx 5911rarr



the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang





25622 736 OOHHCOHCO


STHG kJ0876702983058

kJ3032

STHG




01907 moles of

gaseous products

nRTPV

V

nRTP

L

Kmol molKatmL

10

38308206019070

atm5990

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

molCO

56

2

1

6

21270 molCO

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

OmolH

56

2

1

3 OmolH206370

nRTPV









(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401









gNO173

gNO

molNO

0130

1

molNO

kJ

2

1114 kJ97138





13897 kJ






eqKln428

eqKe 428

12101582 eqK

eqKRTG ln










STHG





(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401



22 22 OSNOSNOSS xS 82102124025146

molKJx 1205642124805146












molkJx 5911rarr



the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang




01907 moles of

gaseous products

nRTPV

V

nRTP

L

Kmol molKatmL

10

38308206019070

atm5990

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

molCO

56

2

1

6

21270 molCO

OHHgC 560002

OHHgC

OHHmolC

56

56

1294

1

OHHmolC

OmolH

56

2

1

3 OmolH206370

nRTPV









(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401









gNO173

gNO

molNO

0130

1

molNO

kJ

2

1114 kJ97138





13897 kJ






eqKln428

eqKe 428

12101582 eqK

eqKRTG ln










STHG





(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401



22 22 OSNOSNOSS xS 82102124025146

molKJx 1205642124805146












molkJx 5911rarr



the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang









(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401









gNO173

gNO

molNO

0130

1

molNO

kJ

2

1114 kJ97138





13897 kJ






eqKln428

eqKe 428

12101582 eqK

eqKRTG ln










STHG





(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401



22 22 OSNOSNOSS xS 82102124025146

molKJx 1205642124805146












molkJx 5911rarr



the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang




gNO173

gNO

molNO

0130

1

molNO

kJ

2

1114 kJ97138





13897 kJ






eqKln428

eqKe 428

12101582 eqK

eqKRTG ln










STHG





(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401



22 22 OSNOSNOSS xS 82102124025146

molKJx 1205642124805146












molkJx 5911rarr



the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang






eqKln428

eqKe 428

12101582 eqK

eqKRTG ln










STHG





(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401



22 22 OSNOSNOSS xS 82102124025146

molKJx 1205642124805146












molkJx 5911rarr



the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang










STHG





(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401



22 22 OSNOSNOSS xS 82102124025146

molKJx 1205642124805146












molkJx 5911rarr



the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang





(J K(J K-1-1 mol mol-1-1))

NONO(g)(g) 21082108

NONO22(g)(g) 24012401



22 22 OSNOSNOSS xS 82102124025146

molKJx 1205642124805146












molkJx 5911rarr



the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang












molkJx 5911rarr



the molecule

xkJH 249560721114

molkJmol

kJx775455

2

5911

2



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang



)(2)(2 gg NN


N-NN-N 160160

N=NN=N 420420

NN==NN 950950


)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ





)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang



)(2)(2 gg NN


11))

N-NN-N 160160

N=NN=N 420420

NN==NN 950950


kJH molkJ 9509500


energy






)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang





)(3)(2)(2 23 ggg NHHN ∆H˚= -922 kJ



SinEntropyMolesGas

SinEntropyMolesGas







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang







STHG

eL

Small

arg








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang








)(3)(2)(2 23 ggg NHHN



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang



water





)(2)()( laqaq OHOHH




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang




that of water



Kg

Jor

Cg

J

KC




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang




that of water







that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang




that of water


)(2)()( laqaq OHOHH

- Or -

OmolH

molHCl

OmolH

L

molHClvolumeHCl 2

2

1

1

1

01

OmolH

molNaOH

OmolH

L

molNaOHvolumeNaOH 2

2

1

1

1

01





that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang




that of water



OmolH

qH neut

2












that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang











that of water


OmolHx

qxH neut

2

OmolH

q

2









Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang








Tmcq

OmolH

qH neut

2

The EndThe End






Jack WangJack Wang

The EndThe End






Jack WangJack Wang