Thermochemistry
24
Thermochemistry Sections 17.1, 17.2
description
Thermochemistry. Sections 17.1, 17.2. Observing Heat Flow. Test the temperature difference of a rubber band (by touching to forehead) Control temp = non-stretched Test 1 = stretch and hold Test 2 = stretch and release Continue tests until certain of observations When did it feel warm?. - PowerPoint PPT Presentation
Transcript of Thermochemistry
Thermochemistry
Sections 171 172
Observing Heat Flow
Test the temperature difference of a rubber band (by touching to forehead)Control temp = non-stretched Test 1 = stretch and hold Test 2 = stretch and releaseContinue tests until certain of observations
When did it feel warm
Energy Changes
The kinds of atoms and their arrangement in the substance determine the amount of energy storedGasoline (hydrocarbon) has high chemical
potential energyControlled explosions to make it run (do WORK)
also makes the engine HOT
Energy changes occur either as heat work or a combination of the two
Heat
Always flows from a warmer object to a cooler objectUntil both temperatures are the same (equilibrium)
SystemSurroundingsLaw of conservation of energyEndothermic reaction
System gains (requires) heat from surroundingsExothermic reaction
System loses (gives) heat to surroundings
UNITS
Common = calorie and joule (SI unit)1 Calorie = 1 kilocalorie = 1000 calories
1 calorie (cal) = 4184 Joules (J)1 J = 02390 cal
1 calorie is defined as the quantity of heat required to raise the temperature of 1 gram of pure H2O by 1degC
Heat Capacity
Heat capacity depends on both the mass and chemical composition of the substance (extensive property)Greater mass = greater heat capacity
(larger pots of water take longer to boil)
Specific heat capacity (specific heat C ) is the amount of heat needed to raise the temperature of 1 gram of the substance by 1degC (intensive property)
Heat Capacity Calculations
Heat = mass x specific heat x change in temperature in Celsius
q = m C ΔT Units for specific heat= J(gdegC) OR cal(gdegC)
Practice When 435 J of heat is added to 34 g olive oil at 21degC the temperature increases to 85 degC What is Colive oil
C=qm ΔT = 435 J (34g x (85-21degC))bull Colive oil = 1999 = 20 J(gdegC)
More Heat Calculations
How much heat is required to raise the temperature of 2500 g of Hg by 52ordmC (CHg = 014 J(gordmC))q = m C ΔT
= (2500g)x(014 JgordmC)x(52ordmC)
= 1820 J
= 18 kJ
Calorimetry
Measures heat flow OUT of a systemHeat released by system = heat absorbed by
surroundings In aq rxns system = chemicals surroundings = water
Good bc know Cwater and Dwater to calc mwater
CalorimeterEnthalpy (H) Δ = q at constant pressureqsys = ΔH = -qsurr - ΔH = exothermic rxn+ ΔH = endothermic rxn
Lab Chem skip this slideCalorimetry CalculationsPractice
When 500 mL of water containing 050 mol HCl at 225degC is mixed with 500 mL of water containing 050 mol of NaOH at 225 degC in a calorimeter the temperature of the solution increases to 260degC How much heat (in kJ) was released by this reactionWater = surroundings rxn= systemMass water = (500mL + 500mL) x 100gmL
= 1000 g H2OΔT = 260 ndash 225 = 35degCHwater = m x C x ΔT
= (1000g)x( 4184 JgdegC)x(35degC)= 14644 J = 146 kJ (endothermic)
Hsys = - Hwater = -146 kJ (exothermic)
More Cal Calculations A small pebble is heated up and placed in a
foam cup calorimeter containing 250 mL of water at 250degC The water reaches a max temp of 264degC How many joules of heat were released by the pebbleSystem = pebble surroundings = waterqwater = -qpebble
q = mCΔT
= (250mL x 100 gmL) x (4184 JgdegC) x (264 ndash 250 degC)
= (250 g) x (4184 JgdegC) x (14degC) = 14644 J = 146 J
qpebble = - qwater = - (146 J) = -146 J (exothermic)
Thermochemical Equations
In a chemical equation the heat (enthalpy) change for the reaction can be written as either a reactant (endothermic) or product (exothermic)
Thermochemical equation States MUST be listed since different ΔH
Ex H2O (l) requires 44 kJ more than H2O (g) to separate into constituent gases
Heat of reactionThe amount of heat absorbed or released during a
reaction depends on the number of MOLES of reactants involved
Heat of combustion ΔH given in kJ but understood to be kJmol
Figure 177
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Observing Heat Flow
Test the temperature difference of a rubber band (by touching to forehead)Control temp = non-stretched Test 1 = stretch and hold Test 2 = stretch and releaseContinue tests until certain of observations
When did it feel warm
Energy Changes
The kinds of atoms and their arrangement in the substance determine the amount of energy storedGasoline (hydrocarbon) has high chemical
potential energyControlled explosions to make it run (do WORK)
also makes the engine HOT
Energy changes occur either as heat work or a combination of the two
Heat
Always flows from a warmer object to a cooler objectUntil both temperatures are the same (equilibrium)
SystemSurroundingsLaw of conservation of energyEndothermic reaction
System gains (requires) heat from surroundingsExothermic reaction
System loses (gives) heat to surroundings
UNITS
Common = calorie and joule (SI unit)1 Calorie = 1 kilocalorie = 1000 calories
1 calorie (cal) = 4184 Joules (J)1 J = 02390 cal
1 calorie is defined as the quantity of heat required to raise the temperature of 1 gram of pure H2O by 1degC
Heat Capacity
Heat capacity depends on both the mass and chemical composition of the substance (extensive property)Greater mass = greater heat capacity
(larger pots of water take longer to boil)
Specific heat capacity (specific heat C ) is the amount of heat needed to raise the temperature of 1 gram of the substance by 1degC (intensive property)
Heat Capacity Calculations
Heat = mass x specific heat x change in temperature in Celsius
q = m C ΔT Units for specific heat= J(gdegC) OR cal(gdegC)
Practice When 435 J of heat is added to 34 g olive oil at 21degC the temperature increases to 85 degC What is Colive oil
C=qm ΔT = 435 J (34g x (85-21degC))bull Colive oil = 1999 = 20 J(gdegC)
More Heat Calculations
How much heat is required to raise the temperature of 2500 g of Hg by 52ordmC (CHg = 014 J(gordmC))q = m C ΔT
= (2500g)x(014 JgordmC)x(52ordmC)
= 1820 J
= 18 kJ
Calorimetry
Measures heat flow OUT of a systemHeat released by system = heat absorbed by
surroundings In aq rxns system = chemicals surroundings = water
Good bc know Cwater and Dwater to calc mwater
CalorimeterEnthalpy (H) Δ = q at constant pressureqsys = ΔH = -qsurr - ΔH = exothermic rxn+ ΔH = endothermic rxn
Lab Chem skip this slideCalorimetry CalculationsPractice
When 500 mL of water containing 050 mol HCl at 225degC is mixed with 500 mL of water containing 050 mol of NaOH at 225 degC in a calorimeter the temperature of the solution increases to 260degC How much heat (in kJ) was released by this reactionWater = surroundings rxn= systemMass water = (500mL + 500mL) x 100gmL
= 1000 g H2OΔT = 260 ndash 225 = 35degCHwater = m x C x ΔT
= (1000g)x( 4184 JgdegC)x(35degC)= 14644 J = 146 kJ (endothermic)
Hsys = - Hwater = -146 kJ (exothermic)
More Cal Calculations A small pebble is heated up and placed in a
foam cup calorimeter containing 250 mL of water at 250degC The water reaches a max temp of 264degC How many joules of heat were released by the pebbleSystem = pebble surroundings = waterqwater = -qpebble
q = mCΔT
= (250mL x 100 gmL) x (4184 JgdegC) x (264 ndash 250 degC)
= (250 g) x (4184 JgdegC) x (14degC) = 14644 J = 146 J
qpebble = - qwater = - (146 J) = -146 J (exothermic)
Thermochemical Equations
In a chemical equation the heat (enthalpy) change for the reaction can be written as either a reactant (endothermic) or product (exothermic)
Thermochemical equation States MUST be listed since different ΔH
Ex H2O (l) requires 44 kJ more than H2O (g) to separate into constituent gases
Heat of reactionThe amount of heat absorbed or released during a
reaction depends on the number of MOLES of reactants involved
Heat of combustion ΔH given in kJ but understood to be kJmol
Figure 177
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Energy Changes
The kinds of atoms and their arrangement in the substance determine the amount of energy storedGasoline (hydrocarbon) has high chemical
potential energyControlled explosions to make it run (do WORK)
also makes the engine HOT
Energy changes occur either as heat work or a combination of the two
Heat
Always flows from a warmer object to a cooler objectUntil both temperatures are the same (equilibrium)
SystemSurroundingsLaw of conservation of energyEndothermic reaction
System gains (requires) heat from surroundingsExothermic reaction
System loses (gives) heat to surroundings
UNITS
Common = calorie and joule (SI unit)1 Calorie = 1 kilocalorie = 1000 calories
1 calorie (cal) = 4184 Joules (J)1 J = 02390 cal
1 calorie is defined as the quantity of heat required to raise the temperature of 1 gram of pure H2O by 1degC
Heat Capacity
Heat capacity depends on both the mass and chemical composition of the substance (extensive property)Greater mass = greater heat capacity
(larger pots of water take longer to boil)
Specific heat capacity (specific heat C ) is the amount of heat needed to raise the temperature of 1 gram of the substance by 1degC (intensive property)
Heat Capacity Calculations
Heat = mass x specific heat x change in temperature in Celsius
q = m C ΔT Units for specific heat= J(gdegC) OR cal(gdegC)
Practice When 435 J of heat is added to 34 g olive oil at 21degC the temperature increases to 85 degC What is Colive oil
C=qm ΔT = 435 J (34g x (85-21degC))bull Colive oil = 1999 = 20 J(gdegC)
More Heat Calculations
How much heat is required to raise the temperature of 2500 g of Hg by 52ordmC (CHg = 014 J(gordmC))q = m C ΔT
= (2500g)x(014 JgordmC)x(52ordmC)
= 1820 J
= 18 kJ
Calorimetry
Measures heat flow OUT of a systemHeat released by system = heat absorbed by
surroundings In aq rxns system = chemicals surroundings = water
Good bc know Cwater and Dwater to calc mwater
CalorimeterEnthalpy (H) Δ = q at constant pressureqsys = ΔH = -qsurr - ΔH = exothermic rxn+ ΔH = endothermic rxn
Lab Chem skip this slideCalorimetry CalculationsPractice
When 500 mL of water containing 050 mol HCl at 225degC is mixed with 500 mL of water containing 050 mol of NaOH at 225 degC in a calorimeter the temperature of the solution increases to 260degC How much heat (in kJ) was released by this reactionWater = surroundings rxn= systemMass water = (500mL + 500mL) x 100gmL
= 1000 g H2OΔT = 260 ndash 225 = 35degCHwater = m x C x ΔT
= (1000g)x( 4184 JgdegC)x(35degC)= 14644 J = 146 kJ (endothermic)
Hsys = - Hwater = -146 kJ (exothermic)
More Cal Calculations A small pebble is heated up and placed in a
foam cup calorimeter containing 250 mL of water at 250degC The water reaches a max temp of 264degC How many joules of heat were released by the pebbleSystem = pebble surroundings = waterqwater = -qpebble
q = mCΔT
= (250mL x 100 gmL) x (4184 JgdegC) x (264 ndash 250 degC)
= (250 g) x (4184 JgdegC) x (14degC) = 14644 J = 146 J
qpebble = - qwater = - (146 J) = -146 J (exothermic)
Thermochemical Equations
In a chemical equation the heat (enthalpy) change for the reaction can be written as either a reactant (endothermic) or product (exothermic)
Thermochemical equation States MUST be listed since different ΔH
Ex H2O (l) requires 44 kJ more than H2O (g) to separate into constituent gases
Heat of reactionThe amount of heat absorbed or released during a
reaction depends on the number of MOLES of reactants involved
Heat of combustion ΔH given in kJ but understood to be kJmol
Figure 177
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Heat
Always flows from a warmer object to a cooler objectUntil both temperatures are the same (equilibrium)
SystemSurroundingsLaw of conservation of energyEndothermic reaction
System gains (requires) heat from surroundingsExothermic reaction
System loses (gives) heat to surroundings
UNITS
Common = calorie and joule (SI unit)1 Calorie = 1 kilocalorie = 1000 calories
1 calorie (cal) = 4184 Joules (J)1 J = 02390 cal
1 calorie is defined as the quantity of heat required to raise the temperature of 1 gram of pure H2O by 1degC
Heat Capacity
Heat capacity depends on both the mass and chemical composition of the substance (extensive property)Greater mass = greater heat capacity
(larger pots of water take longer to boil)
Specific heat capacity (specific heat C ) is the amount of heat needed to raise the temperature of 1 gram of the substance by 1degC (intensive property)
Heat Capacity Calculations
Heat = mass x specific heat x change in temperature in Celsius
q = m C ΔT Units for specific heat= J(gdegC) OR cal(gdegC)
Practice When 435 J of heat is added to 34 g olive oil at 21degC the temperature increases to 85 degC What is Colive oil
C=qm ΔT = 435 J (34g x (85-21degC))bull Colive oil = 1999 = 20 J(gdegC)
More Heat Calculations
How much heat is required to raise the temperature of 2500 g of Hg by 52ordmC (CHg = 014 J(gordmC))q = m C ΔT
= (2500g)x(014 JgordmC)x(52ordmC)
= 1820 J
= 18 kJ
Calorimetry
Measures heat flow OUT of a systemHeat released by system = heat absorbed by
surroundings In aq rxns system = chemicals surroundings = water
Good bc know Cwater and Dwater to calc mwater
CalorimeterEnthalpy (H) Δ = q at constant pressureqsys = ΔH = -qsurr - ΔH = exothermic rxn+ ΔH = endothermic rxn
Lab Chem skip this slideCalorimetry CalculationsPractice
When 500 mL of water containing 050 mol HCl at 225degC is mixed with 500 mL of water containing 050 mol of NaOH at 225 degC in a calorimeter the temperature of the solution increases to 260degC How much heat (in kJ) was released by this reactionWater = surroundings rxn= systemMass water = (500mL + 500mL) x 100gmL
= 1000 g H2OΔT = 260 ndash 225 = 35degCHwater = m x C x ΔT
= (1000g)x( 4184 JgdegC)x(35degC)= 14644 J = 146 kJ (endothermic)
Hsys = - Hwater = -146 kJ (exothermic)
More Cal Calculations A small pebble is heated up and placed in a
foam cup calorimeter containing 250 mL of water at 250degC The water reaches a max temp of 264degC How many joules of heat were released by the pebbleSystem = pebble surroundings = waterqwater = -qpebble
q = mCΔT
= (250mL x 100 gmL) x (4184 JgdegC) x (264 ndash 250 degC)
= (250 g) x (4184 JgdegC) x (14degC) = 14644 J = 146 J
qpebble = - qwater = - (146 J) = -146 J (exothermic)
Thermochemical Equations
In a chemical equation the heat (enthalpy) change for the reaction can be written as either a reactant (endothermic) or product (exothermic)
Thermochemical equation States MUST be listed since different ΔH
Ex H2O (l) requires 44 kJ more than H2O (g) to separate into constituent gases
Heat of reactionThe amount of heat absorbed or released during a
reaction depends on the number of MOLES of reactants involved
Heat of combustion ΔH given in kJ but understood to be kJmol
Figure 177
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
UNITS
Common = calorie and joule (SI unit)1 Calorie = 1 kilocalorie = 1000 calories
1 calorie (cal) = 4184 Joules (J)1 J = 02390 cal
1 calorie is defined as the quantity of heat required to raise the temperature of 1 gram of pure H2O by 1degC
Heat Capacity
Heat capacity depends on both the mass and chemical composition of the substance (extensive property)Greater mass = greater heat capacity
(larger pots of water take longer to boil)
Specific heat capacity (specific heat C ) is the amount of heat needed to raise the temperature of 1 gram of the substance by 1degC (intensive property)
Heat Capacity Calculations
Heat = mass x specific heat x change in temperature in Celsius
q = m C ΔT Units for specific heat= J(gdegC) OR cal(gdegC)
Practice When 435 J of heat is added to 34 g olive oil at 21degC the temperature increases to 85 degC What is Colive oil
C=qm ΔT = 435 J (34g x (85-21degC))bull Colive oil = 1999 = 20 J(gdegC)
More Heat Calculations
How much heat is required to raise the temperature of 2500 g of Hg by 52ordmC (CHg = 014 J(gordmC))q = m C ΔT
= (2500g)x(014 JgordmC)x(52ordmC)
= 1820 J
= 18 kJ
Calorimetry
Measures heat flow OUT of a systemHeat released by system = heat absorbed by
surroundings In aq rxns system = chemicals surroundings = water
Good bc know Cwater and Dwater to calc mwater
CalorimeterEnthalpy (H) Δ = q at constant pressureqsys = ΔH = -qsurr - ΔH = exothermic rxn+ ΔH = endothermic rxn
Lab Chem skip this slideCalorimetry CalculationsPractice
When 500 mL of water containing 050 mol HCl at 225degC is mixed with 500 mL of water containing 050 mol of NaOH at 225 degC in a calorimeter the temperature of the solution increases to 260degC How much heat (in kJ) was released by this reactionWater = surroundings rxn= systemMass water = (500mL + 500mL) x 100gmL
= 1000 g H2OΔT = 260 ndash 225 = 35degCHwater = m x C x ΔT
= (1000g)x( 4184 JgdegC)x(35degC)= 14644 J = 146 kJ (endothermic)
Hsys = - Hwater = -146 kJ (exothermic)
More Cal Calculations A small pebble is heated up and placed in a
foam cup calorimeter containing 250 mL of water at 250degC The water reaches a max temp of 264degC How many joules of heat were released by the pebbleSystem = pebble surroundings = waterqwater = -qpebble
q = mCΔT
= (250mL x 100 gmL) x (4184 JgdegC) x (264 ndash 250 degC)
= (250 g) x (4184 JgdegC) x (14degC) = 14644 J = 146 J
qpebble = - qwater = - (146 J) = -146 J (exothermic)
Thermochemical Equations
In a chemical equation the heat (enthalpy) change for the reaction can be written as either a reactant (endothermic) or product (exothermic)
Thermochemical equation States MUST be listed since different ΔH
Ex H2O (l) requires 44 kJ more than H2O (g) to separate into constituent gases
Heat of reactionThe amount of heat absorbed or released during a
reaction depends on the number of MOLES of reactants involved
Heat of combustion ΔH given in kJ but understood to be kJmol
Figure 177
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Heat Capacity
Heat capacity depends on both the mass and chemical composition of the substance (extensive property)Greater mass = greater heat capacity
(larger pots of water take longer to boil)
Specific heat capacity (specific heat C ) is the amount of heat needed to raise the temperature of 1 gram of the substance by 1degC (intensive property)
Heat Capacity Calculations
Heat = mass x specific heat x change in temperature in Celsius
q = m C ΔT Units for specific heat= J(gdegC) OR cal(gdegC)
Practice When 435 J of heat is added to 34 g olive oil at 21degC the temperature increases to 85 degC What is Colive oil
C=qm ΔT = 435 J (34g x (85-21degC))bull Colive oil = 1999 = 20 J(gdegC)
More Heat Calculations
How much heat is required to raise the temperature of 2500 g of Hg by 52ordmC (CHg = 014 J(gordmC))q = m C ΔT
= (2500g)x(014 JgordmC)x(52ordmC)
= 1820 J
= 18 kJ
Calorimetry
Measures heat flow OUT of a systemHeat released by system = heat absorbed by
surroundings In aq rxns system = chemicals surroundings = water
Good bc know Cwater and Dwater to calc mwater
CalorimeterEnthalpy (H) Δ = q at constant pressureqsys = ΔH = -qsurr - ΔH = exothermic rxn+ ΔH = endothermic rxn
Lab Chem skip this slideCalorimetry CalculationsPractice
When 500 mL of water containing 050 mol HCl at 225degC is mixed with 500 mL of water containing 050 mol of NaOH at 225 degC in a calorimeter the temperature of the solution increases to 260degC How much heat (in kJ) was released by this reactionWater = surroundings rxn= systemMass water = (500mL + 500mL) x 100gmL
= 1000 g H2OΔT = 260 ndash 225 = 35degCHwater = m x C x ΔT
= (1000g)x( 4184 JgdegC)x(35degC)= 14644 J = 146 kJ (endothermic)
Hsys = - Hwater = -146 kJ (exothermic)
More Cal Calculations A small pebble is heated up and placed in a
foam cup calorimeter containing 250 mL of water at 250degC The water reaches a max temp of 264degC How many joules of heat were released by the pebbleSystem = pebble surroundings = waterqwater = -qpebble
q = mCΔT
= (250mL x 100 gmL) x (4184 JgdegC) x (264 ndash 250 degC)
= (250 g) x (4184 JgdegC) x (14degC) = 14644 J = 146 J
qpebble = - qwater = - (146 J) = -146 J (exothermic)
Thermochemical Equations
In a chemical equation the heat (enthalpy) change for the reaction can be written as either a reactant (endothermic) or product (exothermic)
Thermochemical equation States MUST be listed since different ΔH
Ex H2O (l) requires 44 kJ more than H2O (g) to separate into constituent gases
Heat of reactionThe amount of heat absorbed or released during a
reaction depends on the number of MOLES of reactants involved
Heat of combustion ΔH given in kJ but understood to be kJmol
Figure 177
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Heat Capacity Calculations
Heat = mass x specific heat x change in temperature in Celsius
q = m C ΔT Units for specific heat= J(gdegC) OR cal(gdegC)
Practice When 435 J of heat is added to 34 g olive oil at 21degC the temperature increases to 85 degC What is Colive oil
C=qm ΔT = 435 J (34g x (85-21degC))bull Colive oil = 1999 = 20 J(gdegC)
More Heat Calculations
How much heat is required to raise the temperature of 2500 g of Hg by 52ordmC (CHg = 014 J(gordmC))q = m C ΔT
= (2500g)x(014 JgordmC)x(52ordmC)
= 1820 J
= 18 kJ
Calorimetry
Measures heat flow OUT of a systemHeat released by system = heat absorbed by
surroundings In aq rxns system = chemicals surroundings = water
Good bc know Cwater and Dwater to calc mwater
CalorimeterEnthalpy (H) Δ = q at constant pressureqsys = ΔH = -qsurr - ΔH = exothermic rxn+ ΔH = endothermic rxn
Lab Chem skip this slideCalorimetry CalculationsPractice
When 500 mL of water containing 050 mol HCl at 225degC is mixed with 500 mL of water containing 050 mol of NaOH at 225 degC in a calorimeter the temperature of the solution increases to 260degC How much heat (in kJ) was released by this reactionWater = surroundings rxn= systemMass water = (500mL + 500mL) x 100gmL
= 1000 g H2OΔT = 260 ndash 225 = 35degCHwater = m x C x ΔT
= (1000g)x( 4184 JgdegC)x(35degC)= 14644 J = 146 kJ (endothermic)
Hsys = - Hwater = -146 kJ (exothermic)
More Cal Calculations A small pebble is heated up and placed in a
foam cup calorimeter containing 250 mL of water at 250degC The water reaches a max temp of 264degC How many joules of heat were released by the pebbleSystem = pebble surroundings = waterqwater = -qpebble
q = mCΔT
= (250mL x 100 gmL) x (4184 JgdegC) x (264 ndash 250 degC)
= (250 g) x (4184 JgdegC) x (14degC) = 14644 J = 146 J
qpebble = - qwater = - (146 J) = -146 J (exothermic)
Thermochemical Equations
In a chemical equation the heat (enthalpy) change for the reaction can be written as either a reactant (endothermic) or product (exothermic)
Thermochemical equation States MUST be listed since different ΔH
Ex H2O (l) requires 44 kJ more than H2O (g) to separate into constituent gases
Heat of reactionThe amount of heat absorbed or released during a
reaction depends on the number of MOLES of reactants involved
Heat of combustion ΔH given in kJ but understood to be kJmol
Figure 177
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
More Heat Calculations
How much heat is required to raise the temperature of 2500 g of Hg by 52ordmC (CHg = 014 J(gordmC))q = m C ΔT
= (2500g)x(014 JgordmC)x(52ordmC)
= 1820 J
= 18 kJ
Calorimetry
Measures heat flow OUT of a systemHeat released by system = heat absorbed by
surroundings In aq rxns system = chemicals surroundings = water
Good bc know Cwater and Dwater to calc mwater
CalorimeterEnthalpy (H) Δ = q at constant pressureqsys = ΔH = -qsurr - ΔH = exothermic rxn+ ΔH = endothermic rxn
Lab Chem skip this slideCalorimetry CalculationsPractice
When 500 mL of water containing 050 mol HCl at 225degC is mixed with 500 mL of water containing 050 mol of NaOH at 225 degC in a calorimeter the temperature of the solution increases to 260degC How much heat (in kJ) was released by this reactionWater = surroundings rxn= systemMass water = (500mL + 500mL) x 100gmL
= 1000 g H2OΔT = 260 ndash 225 = 35degCHwater = m x C x ΔT
= (1000g)x( 4184 JgdegC)x(35degC)= 14644 J = 146 kJ (endothermic)
Hsys = - Hwater = -146 kJ (exothermic)
More Cal Calculations A small pebble is heated up and placed in a
foam cup calorimeter containing 250 mL of water at 250degC The water reaches a max temp of 264degC How many joules of heat were released by the pebbleSystem = pebble surroundings = waterqwater = -qpebble
q = mCΔT
= (250mL x 100 gmL) x (4184 JgdegC) x (264 ndash 250 degC)
= (250 g) x (4184 JgdegC) x (14degC) = 14644 J = 146 J
qpebble = - qwater = - (146 J) = -146 J (exothermic)
Thermochemical Equations
In a chemical equation the heat (enthalpy) change for the reaction can be written as either a reactant (endothermic) or product (exothermic)
Thermochemical equation States MUST be listed since different ΔH
Ex H2O (l) requires 44 kJ more than H2O (g) to separate into constituent gases
Heat of reactionThe amount of heat absorbed or released during a
reaction depends on the number of MOLES of reactants involved
Heat of combustion ΔH given in kJ but understood to be kJmol
Figure 177
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Calorimetry
Measures heat flow OUT of a systemHeat released by system = heat absorbed by
surroundings In aq rxns system = chemicals surroundings = water
Good bc know Cwater and Dwater to calc mwater
CalorimeterEnthalpy (H) Δ = q at constant pressureqsys = ΔH = -qsurr - ΔH = exothermic rxn+ ΔH = endothermic rxn
Lab Chem skip this slideCalorimetry CalculationsPractice
When 500 mL of water containing 050 mol HCl at 225degC is mixed with 500 mL of water containing 050 mol of NaOH at 225 degC in a calorimeter the temperature of the solution increases to 260degC How much heat (in kJ) was released by this reactionWater = surroundings rxn= systemMass water = (500mL + 500mL) x 100gmL
= 1000 g H2OΔT = 260 ndash 225 = 35degCHwater = m x C x ΔT
= (1000g)x( 4184 JgdegC)x(35degC)= 14644 J = 146 kJ (endothermic)
Hsys = - Hwater = -146 kJ (exothermic)
More Cal Calculations A small pebble is heated up and placed in a
foam cup calorimeter containing 250 mL of water at 250degC The water reaches a max temp of 264degC How many joules of heat were released by the pebbleSystem = pebble surroundings = waterqwater = -qpebble
q = mCΔT
= (250mL x 100 gmL) x (4184 JgdegC) x (264 ndash 250 degC)
= (250 g) x (4184 JgdegC) x (14degC) = 14644 J = 146 J
qpebble = - qwater = - (146 J) = -146 J (exothermic)
Thermochemical Equations
In a chemical equation the heat (enthalpy) change for the reaction can be written as either a reactant (endothermic) or product (exothermic)
Thermochemical equation States MUST be listed since different ΔH
Ex H2O (l) requires 44 kJ more than H2O (g) to separate into constituent gases
Heat of reactionThe amount of heat absorbed or released during a
reaction depends on the number of MOLES of reactants involved
Heat of combustion ΔH given in kJ but understood to be kJmol
Figure 177
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Lab Chem skip this slideCalorimetry CalculationsPractice
When 500 mL of water containing 050 mol HCl at 225degC is mixed with 500 mL of water containing 050 mol of NaOH at 225 degC in a calorimeter the temperature of the solution increases to 260degC How much heat (in kJ) was released by this reactionWater = surroundings rxn= systemMass water = (500mL + 500mL) x 100gmL
= 1000 g H2OΔT = 260 ndash 225 = 35degCHwater = m x C x ΔT
= (1000g)x( 4184 JgdegC)x(35degC)= 14644 J = 146 kJ (endothermic)
Hsys = - Hwater = -146 kJ (exothermic)
More Cal Calculations A small pebble is heated up and placed in a
foam cup calorimeter containing 250 mL of water at 250degC The water reaches a max temp of 264degC How many joules of heat were released by the pebbleSystem = pebble surroundings = waterqwater = -qpebble
q = mCΔT
= (250mL x 100 gmL) x (4184 JgdegC) x (264 ndash 250 degC)
= (250 g) x (4184 JgdegC) x (14degC) = 14644 J = 146 J
qpebble = - qwater = - (146 J) = -146 J (exothermic)
Thermochemical Equations
In a chemical equation the heat (enthalpy) change for the reaction can be written as either a reactant (endothermic) or product (exothermic)
Thermochemical equation States MUST be listed since different ΔH
Ex H2O (l) requires 44 kJ more than H2O (g) to separate into constituent gases
Heat of reactionThe amount of heat absorbed or released during a
reaction depends on the number of MOLES of reactants involved
Heat of combustion ΔH given in kJ but understood to be kJmol
Figure 177
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
More Cal Calculations A small pebble is heated up and placed in a
foam cup calorimeter containing 250 mL of water at 250degC The water reaches a max temp of 264degC How many joules of heat were released by the pebbleSystem = pebble surroundings = waterqwater = -qpebble
q = mCΔT
= (250mL x 100 gmL) x (4184 JgdegC) x (264 ndash 250 degC)
= (250 g) x (4184 JgdegC) x (14degC) = 14644 J = 146 J
qpebble = - qwater = - (146 J) = -146 J (exothermic)
Thermochemical Equations
In a chemical equation the heat (enthalpy) change for the reaction can be written as either a reactant (endothermic) or product (exothermic)
Thermochemical equation States MUST be listed since different ΔH
Ex H2O (l) requires 44 kJ more than H2O (g) to separate into constituent gases
Heat of reactionThe amount of heat absorbed or released during a
reaction depends on the number of MOLES of reactants involved
Heat of combustion ΔH given in kJ but understood to be kJmol
Figure 177
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Thermochemical Equations
In a chemical equation the heat (enthalpy) change for the reaction can be written as either a reactant (endothermic) or product (exothermic)
Thermochemical equation States MUST be listed since different ΔH
Ex H2O (l) requires 44 kJ more than H2O (g) to separate into constituent gases
Heat of reactionThe amount of heat absorbed or released during a
reaction depends on the number of MOLES of reactants involved
Heat of combustion ΔH given in kJ but understood to be kJmol
Figure 177
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Figure 177
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Thermochemical calculations
When carbon disulfide is formed from its elements heat is absorbed Calculate the amount of heat (in kJ) absorbed when 566 g of carbon disulfide formed
C(s) + 2S(s) CS2(l) ΔH = 893 kJ
566g CS2 x 1 mol CS2 (12011 + 2x32066) x 893 kJmol
= 6638 kJ = 664 kJ Does this make sense
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
More Thermochemistry Calculations
The production of iron and carbon dioxide from iron(III) oxide and carbon monoxide is an exothermic reaction How many kilojoules of heat are produced when 340 mol Fe2O3 reacts with an excess of CO
Fe2O3 (s) + 3CO (g) 2Fe (s) + 3CO2 (g) + 263 kJ
340 mol Fe2O3 x 263 kJmol
= 894 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Heat in Changes of State
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Phase change calculations
While heating (temp is changing)q=mcΔT
While changing state (temp is constant)q= ΔHfus or vap x (moles)
ΔHfus for water = 601kJmolΔhvap for water = 407kJmol
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
How much total heat is needed to convert 1000 grams of -10degC ice to 120degC steam
q= mciceΔTcice = 21 JgC
ΔHfus = 601 kJmol = 6010 Jmol
q= mcwaterΔTcwater = 4184 JgC
ΔHvap = 407 kJmol = 40700 Jmol
q= mcsteamΔTcsteam = 17 JgC
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Molar Heat of Formation
The enthalpy change occurring during a reaction can be calculated using the molar heats of formation of the reactants and products
Table 174 on page 530 in your text bookhellip
ΔH0 = ΔHf0 (products) ndash ΔHf
0 (reactants)
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Hessrsquos Law of heat Summation
Hessrsquos law adds two or more ldquohalfrdquo reactions and coordinating changes in enthalpy (ΔH) to get finaltotal reaction ΔH
Ex Derive C(graphite) + 2H2 (g) CH4(g)
(1) H2(g) + 12 O2(g) H2O(l) ΔH = -2858 kJ
(2) C(graphite) + O2(g) CO2(g) ΔH = -2935 kJ
(3) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH = -
8904 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
Answer
C(graphite) + 2H2 (g) CH4(g) ΔH = -747 kJ
HOW Multiply reaction 1 by 2 flip reaction 3 and add
all together ΔH = 2(-2858) + (-3935) + (+8904) = -747 kJ
