EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0 o C to steam at 127.0 o C? q overall = q ice + q fusion + q water + q boil + q steam q = (10.0g 2.09J/g o C 15.0 o C) + (10.0g 333J/g) + (10.0g 4.18J/g o C 100.0 o C) + (10.0g 2260J/g) + (10.0g 2.03J/g o C 27.0 o C) q = (314 + 3.33×10 3 + 4.18×10 3 + 2.26×10 4 + 548)J = 30.9 kJ
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EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0 o C to steam at 127.0 o C? q overall = q ice + q fusion + q water + q boil + q steam.
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EXAMPLE: How much heat is required to heat 10.0 g of ice at -15.0oC to steam at 127.0oC? qoverall = qice + qfusion + qwater + qboil + qsteam
q = (10.0g 2.09J/goC 15.0oC)
+ (10.0g 333J/g)
+ (10.0g 4.18J/goC 100.0oC)
+ (10.0g 2260J/g)
+ (10.0g 2.03J/goC 27.0oC)
q = (314 + 3.33×103 + 4.18×103 + 2.26×104 + 548)J
= 30.9 kJ
Heat Flow in Reactions
exothermic –reaction that gives off energy
q < 0
isolated system E=0 heat released by reaction raises the temperature of the solvent
constant T, heat is released to the surroundings
endothermic – reaction that absorbs energy
q > 0
Expansion Type Work
w = -PV system does work
P
P
Vinitial
V
V = Vfinal - Vinitial
qp = +2kJ
Do 250 J of work to compress a gas, 180 J of heat are released by the gas
1. The magnitude of is directly proportional to the amount of reaction.
H is for 1 mole of reaction as written
2 H2(g) + O2(g) 2 H2O(l) H = -571.6 kJ
H2(g) + ½ O2(g) H2O(l) H = -285.8 kJ
Can have ½ mole O2 just not ½ molecule
Laws of Thermochemistry
2. H for a reaction is equal in magnitude but opposite in sign to H for the reverse reaction.
H2(g) + ½ O2(g) H2O(l) H = -285.8 kJ
H2O(l) H2(g) + ½ O2(g) H = +285.8 kJ
Laws of Thermochemistry
3. The value of H for the reaction is the same whether it occurs directly or in a series of steps.
Hoverall = H1 + H2 + H3 + · · ·
also called Hess’ Law
Enthalpy Diagram
H2(g) + ½ O2(g) H2O(l) H = -285.8 kJ
H2O(l) H2O(g) H = +44.0 kJ
H2(g) + ½ O2(g) H2O(g) H = -241.8 kJ
Given 3 CO + 3/2 O2 3 CO2 H = -849 kJ
What is H for CO2 CO + ½ O2 ?
100
0
130
-283
kJ
+283
kJ
+849
kJ
-254
7 kJ
+254
7 kJ
0% 0% 0%0%0%
1. -283 kJ
2. +283 kJ
3. +849 kJ
4. -2547 kJ
5. +2547 kJ
Energy and Stoichiometry
• Since H is per mole of reaction we can relate heat to amount of reaction
• Given C2H6 + 7/2 O2 2 CO2 + 3 H2O H=-1559.7 kJ
• If 632.5 kJ are released to surroundings what mass of H2O is formed?
• 632.5 kJ released means H = -632.5 kJ for this much H2O
OHg92.21OHmol1
OHg016.18
reactionmol1
OHmol3
kJ7.1559
reactionmol1kJ5.632 2
2
22
Bomb Calorimeter measure qv
qrxn + qcal = 0qrxn = -qcal
qrxn = - ccalTErxn = qrxn/moles rxnErxn ≈ Hrxn
H = E + (PV)H = E + RTngas
@298K RT = 2.5 kJ/mol
“Coffee Cup” Calorimeter qp
Photo by George Lisensky
Measuring H
• When 25.0 mL of 1.0 M H2SO4 are added to 50.0 mL of 1.0 M KOH, both initially at 24.6C the temperature rises to 33.9C. What is H for H2SO4 + 2 KOH K2SO4 + 2 H2O ? (Assume d = 1.00 g/mL, c = 4.18 J/g.C)
• qsoln = mcT
• m = (25.0 + 50.0)mL×1.00g/mL = 75.0 g
Measuring H cont.
• q=mcT
• qsoln = 75.0 g × 4.18 J/g.C × (33.9-24.6)C
• qsoln = 2916 J
• qrxn + qsoln = 0
• qrxn = -2916 J
Hrxn = qrxn/moles rxn
Measuring H cont
• How many moles rxn?
• 1 mol rxn / 1 mol H2SO4
• 1 mol rxn / 2 mol KOH
4242 SOHmol025.0
mL1000
L1
L1
SOHmol00.1mL0.25
KOHmol050.0mL1000
L1
L1
KOHmol00.1mL0.50
Stoichiometric mixture so 0.025 mol rxn
Measuring H cont
Hrxn = qrxn/moles rxn
Hrxn = -2916 J / 0.025 mol rxn
Hrxn = -116622 J / mol rxn
Hrxn = -117 kJ
H is per mole of reaction as written
If excess Al is added to 50 mL of 0.250 M H2SO4 how many moles of
the following reaction occur?2 Al + 3 H2SO4 Al2(SO4)3 + 3 H2
10
0
0
130
0% 0% 0%0%0%
1. 0.0125 mol
2. 0.0375 mol
3. 0.025 mol
4. 0.00625 mol
5. 0.00417 mol
Hess’s Law
• Can find H for an unknown, or hard to measure, reaction by summing measured H values of known reactions.
EXAMPLE
H for formation of CO cannot readily be measured since a mixture of CO and CO2 is always formed.
C (s) + ½ O2 (g) CO (g) H = ?
C (s) + O2 (g) CO2 (g) H1 = -393.5 kJ
CO (g) + ½ O2 (g) CO2 (g) H2 = -283.0 kJ
C (s) + ½ O2 (g) CO (g) H = H1 - H2
H = H1 - H2 = -393.5 – (-283.0) = -110.5 kJ
Standard Enthalpy of Formation
the enthalpy associated with the formation of 1 mol of a substance from its constituent elements under standard state conditions at the specified temperature
For an element this is a null reaction
O2 (g) O2 (g) H = 0
Hf = 0 for all elements in their standard states
For which one of these reactions is ΔHºrxn
= ΔHºf?
0% 0% 0%0%0%
1. N2(g) + 3 H2(g) 2 NH3(g)
2. C(graphite) + 2 H2(g) CH4(g)
3. C(diamond) + O2(g) CO2(g)
4. CO(g) + ½ O2(g) CO2(g)
5. H2(g) + Cl2(g) 2 HCl(g)
100
0
130
Calculation of Ho
Ho = mols Hfoproducts – mols Hf
oreactants
We can always convert products and reactants to the elements.
Hess’s law says H is the same whether we go directly from reactants to products or go via elements
Example What is the value of Hrxn for the reaction:2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
from Appendix J TextC6H6(l) Hf
o = + 49.0 kJ/mol
O2(g) Hfo = 0
CO2(g) Hfo = - 393.5
H2O(g) Hfo = - 241.8
Hrxn mols Hfoproduct
– mols Hforeactants
ExampleWhat is the value of Hrx for the reaction:2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)