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Energy Changes in Chemical Energy Changes in Chemical ReactionsReactions
11
Most reactions give off or absorb energy Energy is the capacity to do work or supply
heat.◦ Heat: transfer of thermal (kinetic) energy
between two systems at different temperatures (from hot to cold)
Metal bar in water Metal bar drilled
Work (w): energy transfer when forces are applied to a system
Heat (q): energy transferred from a hot object to a cold one◦ Radiant energy heat from the sun◦ Thermal energy associated with motion of
particles◦ Potential energy energy associated with
object’s position or substance’s chemical bonds◦ Kinetic energy energy associated with object’s
motion
Describe the difference between the two. SI unit of energy: J
1 watt = 1 J/s, so a 100 Watt bulb uses 100 J each second
We often use the unit of kJ to refer to chemical heat exchanges in a reaction. 1 kJ = 1000 J
Energy is also reported in calories:◦ Amount of energy needed to raise 1 gram of water by
1oC◦ 1 cal = 4.184 J; 1 Cal = 4184 J ◦ Cal (or kcal) is used on food labels
Molecular heat transfer2
2
s
m kg 1 J 1
Heat: form of energy transferred from object at higher temperature to one at lower temperature (from hot object to cold object)
Thermochemistry: study of heat changes in chemical reactions, in part to predict whether or not a reaction will occur
Thermodynamics: study of heat and its transformations
First Law of Thermodynamics: Energy can be converted from one form to another but cannot be created or destroyed
System loses heat (negative); gains heat (positive)
Endothermic reaction: q is positive (q > 0)◦ Reaction (system) absorbs heat◦ Surroundings feel cooler
Exothermic reaction: q is negative (q < 0)◦ Reaction releases heat◦ Surroundings feel warmer
Determine if the following processes are endothermic or exothermic…◦ Combustion of methane◦ Reacting Ba(OH)2 with NH4Cl◦ Neutralization of HCl◦ Melting◦ CaCO3 (s) CaO (s) + CO2 (g)
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◦ Combustion of methane exothermic◦ Reacting Ba(OH)2 with NH4Cl endothermic◦ Neutralization of HCl exothermic◦ Melting endothermic◦ CaCO3 (s) CaO (s) + CO2 (g) endothermic
◦ Combustion, neutralization, and combination reactions tend to be exothermic
◦ Decomposition reactions tend to be endothermic◦ Melting, boiling, and sublimation are endothermic
99
1010
Specific heat (sp. ht.): amount of heat required to raise 1 gram of substance by 1oC
Use mass, specific heat, and T to calculate the amount of heat gained or lost:
q = msT ms = C q = CT◦ Heat capacity (C): amount of heat required to
raise the temperature of a given quantity of a substance by 1oC; C = q / T = J / oC
◦ Molar heat capacity (Cm): amount of heat that can be absorbed by 1 mole of material when temperature increases 1oC; q = (Cm) x (moles of substance) x (T) = J / mol • oC
Calculate the amount of heat transferred when 250 g of H2O (with a specific heat of 4.184 J/g·oC) is heated from 22oC to 98oC.
q = msT Is heat being put into the system or given off by
the system?
If a piece of hot metal is placed in cold water, what gains heat and what loses heat? Which one will have a positive q value and which will have a negative q value?
34.8 g of an unknown metal at 25.2oC is mixed with 60.1 g of H2O at 96.2 oC (sp. ht. = 4.184 J/g·oC). The final temperature of the system comes to 88.4oC. Identify the unknown metal.
Specific heats of metals:◦ Al 0.897 J/g·oC◦ Fe 0.449 J/g·oC◦ Cu 0.386 J/g·oC◦ Sn 0.228J/g·oC
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Heat changes in a reaction can be determined by measuring the heat flow at constant pressure
Apparatus to do this is called a calorimeter.
Heat evolved by a reaction is absorbed by water; heat capacity of calorimeter is the heat capacity of water.
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A 28.2 gram sample of nickel is heated to 99.8oC and placed in a coffee cup calorimeter containing 150.0 grams of water at 23.5oC. After the metal cools, the final temperature of the metal and water is 25.0oC.
qabsorbed + qreleased = 0 Which substance absorbed heat? Which substance released heat? Calculate the heat absorbed by the
substance you indicated above.
A hot piece of copper (at 98.7oC, specific heat = 0.385 J/g•oC) weighs 34.6486 g. When placed in room temperature water, it is calculated that 915.1 J of heat are released by the metal.
What gains heat? What loses heat? What is the final temperature of the metal? Watch signs!!!!
Enthalpy (H) describes heat flow into and out of a system under constant pressure
Enthalpy (a measure of energy) is heat transferred per mole of substance.
At constant pressure, ◦ qpH = Hproducts – Hreactants
◦ H > 0 endothermic (net absorption of energy from environment; products have more internal energy)
◦ H < 0 exothermic (net loss of energy to environment; reactants have more internal energy)
Why does T become constant during melting and evaporating?
Melting, vaporization, and sublimation are endothermic
We can calculate total heat needed to convert a 15 gram piece of ice at -20oC to steam at 120oC.
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2.09 J/g2.09 J/gooCC
4.184 J/g4.184 J/gooCC
2.080 J/g2.080 J/gooCC
334 J/g334 J/g
2250 J/g2250 J/g
Heat of fusion (Hfus): Amount of heat required to melt (solid liquid)
Heat of vaporization (Hvap): Amount of heat required to evaporate (liquid gas)
Heat of sublimation (Hsub): Amount of heat required to sublime (solid gas)
Why are there no values for Hfreezing, Hcondendsation, or Hdeposition?
Shows both mass and enthalpy relationships
2Al (s) + Fe2O3 (s) 2Fe (s) + Al2O3 (s) Ho = -852 kJ
Amount of heat given off depends on amount of material:◦ 852 kJ of heat are released for every 2 mol Al, 1
mol Fe2O3, 2 mol Fe, and 1 mol Al2O3
2Al (s) + Fe2O3 (s) 2Fe (s) + Al2O3 (s) Ho = -852 kJ
How much heat is released if 10.0 grams of Fe2O3 reacts with excess Al?
What if we reversed the reaction? Heat would have to be put in to make the
reaction proceed:◦ 2Fe (s) + Al2O3 (s) 2Al (s) + Fe2O3 (s) Ho = +852 kJ
If a compound cannot be directly synthesized from its elements, we can add the enthalpies of multiple reactions to calculate the enthalpy of reaction in question.
Hess’s Law: change in enthalpy is the same whether the reaction occurs in one step or in a series of steps
Look at direction of reaction and amount of reactants/products
Value changes sign with direction
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Figure 8.5
Values of enthalpy change◦ For a reaction in the reverse direction, enthalpy is
numerically equal but opposite in sign Reverse direction, heat flow changes; endothermic
becomes exothermic (and vice versa); sign of H changes
◦ Proportional to the amount of reactant consumed Twice as many moles = twice as much heat; half as
many moles = half as much heat
HT = H1 + H2 + H3 + ….
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Thermochemical equation:◦H2(g) + I2(s) 2HI(g) H =
+53.00 kJ Two possible changes: Reverse the equation:
◦2HI(g) H2(g) + I2(s) H = -53.00 kJ
Double the amount of material:◦2H2(g) + 2I2(s) 4HI(g) H =
+106.00 kJ2626
Calculate Ho for 2NO (g) + O2 (g) N2O4 (g) Ho = ?
N2O4 (g) 2NO2 (g) Ho = 57.2 kJ
NO (g) + ½ O2 (g) NO2 (g) Ho = -57.0 kJ
We can use known values of Ho to calculate unknown values for other reactions
P4 (s) + 3 O2 (g) P4O6 (s) H = -1640.1 kJ
P4 (s) + 5 O2 (g) P4O10 (s) H = -2940.1 kJ
What is Ho for the following reaction?P4O6 (s) + 2 O2 (g) P4O10 (s) H = ?
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Given: 2NH3(g) N2H4(l) + H2(g) H = 54 kJ N2(g) + H2(g) NH3(g) H = -69 kJ CH4O(l) CH2O(g) + H2(g) H = -195 kJ
Find the enthalpy for the following reaction: N2H4(l) + CH4O(l) CH2O (g) + N2(g) + 3H2(g)
H = ? kJ
2
1
2
3
Given the following equations: 2CO2 (g) O2 (g) + 2CO (g) H = 566.0
kJ ½ N2 (g) + ½ O2 (g) NO (g) H = 90.3 kJ
Calculate the enthalpy change for: 2CO (g) + 2NO (g) 2CO2 (g) + N2 (g) H =
?
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Standard heat of formation Hof): heat
needed to make 1 mole of a substance from its stable elements in their standard states
Hof = 0 for a stable (naturally occurring)
element Which of these have Ho
f = 0?◦ CO(g), Cu(s), Br2(l), Cl(g), O2(g), O3(g), O2(s), P4(s)
Do the following equations represent standard enthalpies of formation? Why or why not?◦ 2Ag (l) + Cl2 (g) 2AgCl (s)
◦ Ca (s) + F2 (g) CaF2 (s)
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Can use measured enthalpies of formation to determine the enthalpy of a reaction (use Appendix B in back of book)
Horxn = nHo
f (products) – nHof (reactants)
◦ sum; n = number of moles (coefficients) Direct calculation of enthalpy of reaction if
the reactants are all in elemental form◦ Sr (s) + Cl2 (g) SrCl2 (g)
◦ Horxn = [Ho
f (SrCl2)] – [Hof (Sr) + Ho
f (Cl2)] = -828.4 kJ/mol
Some Common Substances (25oC)
Horxn = Ho
f,products - Hof,reactants
Calculate values of Ho for the following rxns: 1) CaCO3 (s) CaO (s) + CO2 (g) 2) 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Hof values:
CaCO3: -1207.1 kJ/mol; CaO: -635.5 kJ/mol; CO2: -393.5 kJ/mol; C6H6: 49.0 kJ/mol; H2O(l): -285.8 kJ/mol
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Use Standard Heat of Formation values to calculate the enthalpy of reaction for:
C6H12O6(s) C2H5OH(l) + CO2(g) Hint: Is the equation balanced?
Hof (C6H12O6(s)) = -1260.0 kJ/mol
Hof (C2H5OH(l)) = -277.7 kJ/mol
Hof (CO2(g)) = -393.5 kJ/mol
Bond Dissociation Energy (or Bond Energy, BE): energy required to break a bond in 1 mole of a gaseous molecule
Reactions generally proceed to form compounds with more stable (stronger) bonds (greater bond energy)
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H2 Bond Energy
Bond energies vary somewhat from one mole- cule to another so we use average bond dissociation energy (D)
H-OH 502 kJ/mol Avg O-H = 453
H-O 427 kJ/mol kJ/mol H-OOH 431 kJ/mol
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Horxn = BE (reactants) + - BE (products)
endothermic exothermic energy input energy
released
BE(react) > BE(prod) endothermic BE(react) < BE(prod) exothermic
Use only when heats of formation are not available, since bond energies are average values for gaseous molecules
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Use bond energies to calculate the enthalpy change for the following reaction:
N2(g) + 3H2(g) 2NH3(g) Hrxn = [BEN N + 3BEH-H] + [-6BEN-H] Hrxn = [945 + 3(436)] – [6(390)] = -87 kJ measured value = -92.2 kJ Why are the calculated and measured
values different?
4141
Use bond energies to calculate the enthalpy change for the decomposition of nitrogen trichloride: NCl3 (g) N2 (g) + Cl2 (g)
How many distinct bond types are there in each molecule?
How many of each bond type do we need to calculate Hrxn? ◦ BE(N-Cl) = 200 kJ/mol
◦ BE(N≡N) = 945 kJ/mol
◦ BE(Cl-Cl) = 243 kJ/mol4242
6(N-Cl) + -1(N N) + -3(Cl-Cl) 6(200) + -(945) + -3(243) = -474 kJ
≡
Use q = msT (s = J/g·oC) If given mass of reactant, convert to moles
and multiply by enthalpy to find total heat transferred
If given multiple equations with enthalpies, use Hess’s Law
If given Hof values: products – reactants
If given bond energy (BE) values: +reactants + -products
Identify how to set up the following problems:
Calculate the Ho of reaction for:◦ C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l)
◦ Hof C3H8(g): -103.95 kJ/mol; Ho
f CO2(g): -393.5 kJ/mol; Ho
f H2O(l): -285.8 kJ/mol 8750 J of heat are applied to a 170 g sample of
metal, causing a 56oC increase in its temperature. What is the specific heat of the metal? Which metal is it?
C2H4(g ) + 6F2(g) 2CF4(g) + 4HF(g) Ho = ?◦ H2 (g) + F2 (g) 2HF (g) Ho = -537 kJ
◦ C (s) + 2F2 (g) CF4 (g) Ho = -680 kJ
◦ 2C (s) + 2H2 (g) C2H4 (g) Ho = 52.3 kJ
Use average bond energies to determine the enthalpy of the following reaction◦ CH4 (g) + Cl2 (g) CH3Cl (g) + HCl (g)
(BEC-Cl = 328 kJ/mol)