Chapter 6 ­ Thermochemistry Part II
Recap ­ Same Rules for Heat and Work
E = q + w
+1.3 x 108J
W = ­p V ­ (1.0 atm)(4.50 x 106 ­ 4.00 x 106L) = ­5.00 x 105 L*atm
­5.00 x 105 L*atm x 101.3J
=
• Heat given off is negative.
• Heat absorbed is positive.
• Work done by system on surroundings is negative.
• Work done on system by surroundings is positive.
• Thermodynamics­ The study of energy and the changes it undergoes.
­5.1 x 107 J
L*atm
E = q + w
= (+1.3 x 108J) + (­5.1 x 107J) = 8 x 107 J
Since more energy is being added, the gas expands, so there is a net increase in internal energy. Hence Change in energy is positive.
6.2 Enthalpy and Calorimetry
Enthalpy and Chemical Reactions
In systems at constant pressure, where the only work is PV, the
Enthalpy (H) is the heat energy exchange that takes place during chemical reactions.
H = E + PV
E = the internal energy of the system
P = the pressure of the system
V = the volume of the system
• Enthalpy is a state function.
change in enthalpy is due only to energy flow as heat (q) ( H = heat of rxn)
H = Hproducts ­ Hreactants
In a case in which the pdts of a rxn have a greater enthaly that the reactants H will be positive . Thus, heat will be absorbed by the system and the rxn is endothermic.
On the other hand, if the enthalpy of the products is less than that of the reactants, H will be negative. Heat is gnerated and given off, the rxn is exothermic.
• Most rxns occur in a constant P situation
H = Hproducts ­ Hreactants
Remember in the system
Enthalpy and Gases
If a rxn produces more gas molecules than reacted = EXPANSION (system does work on surroundings)
If a rxn produces less gas molecules than reacted = compression (surroundings did work on system)
No work is done if:
# gas molecules reacted = # gas molecules produced
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Chapter 6 ­ Thermochemistry Part II
CH4
+
O2
CO2
+
H 2O
1 mole of CH4 = ­890kJ (exothermic)
5.8g CH4 x
1 mol CH4
890 kJ
16.05 g CH4
1 mol CH4
= ­322 kJ
CH4 + 4Cl2 CCl4 + 4HCl, ∆H = ­434kJ
Based on the above reaction, what energy change occurs when 1.2 moles of methane reacts?
A) 5.2 x 105 J are released
B) 5.2 x 105 J are absorbed
C) 3.6 x 105 J are released
D) 3.6 x 105 J are absorbed
E) 4.4 x 105 J are released
A
Specific Heat Capacity
Calorimetry
Calorimetry ­ the science of measuring heat.
Calorimeter ­ a device used to measure heat transfer.
Heat capacity ­ the amount of heat required to change the temperature by one degree (J/ 0C).
• physical property of materials
C = heat absorbed
mass of substance x increase in temperature
Fact: it takes more E to raise the temperature of 2 g object than a 1 g object (by 1 C)
Specific Heat Capacity incorporates mass with heat capacity.
Units: J/g C or J/g K
Molar Heat Capacity ­ heat is given in per mole of a substance.
Units: J/mol C or J/mol K
These are specific to a compound (you have to look these up)
Constant Pressure Calorimetry
(since constant P assume H = qp)
q = m C T
or (or s)
q = moles molar heat T
• If a rxn done in a solution, use the specific heat of water.
• If one substance loses heat to another then ­q1 = q2
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Chapter 6 ­ Thermochemistry Part II
How much heat must be added to a 8.21 g sample of gold to increase its temperature by 6.2 oC? The specific heat of gold is 0.13 J/goC.
q = m x c x T If 40.5 J of heat is added to a 15.4 g sample of silver, how much will the temperature increase by? The specific heat of silver is 0.235 J/
goC.
q = m x c x T q = (8.21g)(0.13 J/g C)(6.2 C) = 6.62 J
0
0
40.5J = (15.4g)(0.235 J/g0C)(x)
x = 11.20C So, water absorbed 6.62 J of heat, that means gold lost ­6.62J of heat!
A lead pellet having a mass of 26.47 g at 89.98 C was placed in a constant­pressure calorimeter of negligible heat capacity containing 100.0 mL of water. The water temperature rose form 22.50 C to 23.17 C. What is the specific heat of the lead pellet?
Plan: Find heat gained by water, Then heat lost by metal. Finally, use the equation C = heat absorbed/ mass of metal x T qwater = (100g)(4.18 J/g 0C)(23.17 ­ 22.500C) = 280.06 J = heat gained by water
A sheet of gold weighting 10.0 g and at a temperature of 18.0 C is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6 C. What is the final temperature of the combined metals? Assume that no heat was lost to the surroundings. (Hint: heat gained by Au = heat lost by Fe. CAu = 0.129 J/g C; CFe =0.444 J/g C)
qcold = ­(qhot)
(10.0g)(0.129 J/g 0C)(Tfinal ­ 18.00C)= ­((20.0g)(0.444 J/g 0C)(Tfinal ­ 55.60C))
1.29 Tf ­ 23.22
Heat lost by metal = m x T ( 89.98­23.17)
= (26.47g)(66.810C) = 1768. g 0C =
10.18 Tf
=
­ 8.88 Tf
+ 493.73
516.95
Tf = 50.78 0C
C = 280.06 J
= 0.158 J/g 0C
1768 g 0C
50.0 mL of pure water at 280. K is mixed with 70.0 mL of pure water at 330. K. What is the final temperature of the mixture?
(50g)(4.18 J/ g 0C)(Tfinal ­ 280K) = ­((70g)(4.18 J/g 0C)(Tfinal ­ 330K))
209Tf ­ 58520
=
501.6Tf
=
Tf = ­292.6Tf
155078
309K or 360C
+
96558
In the lab, you mix two solutions (each originally at the same temp.) and the T of the resulting sol'n decreases. Which of the following is true?
A) the chemical reaction is releasing energy
B) the energy released is equal to m x C x ∆T
C) the chemical reaction is absorbing energy
D) the chemical reaction is exothermic
E) more than one of these
ANS: C
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Chapter 6 ­ Thermochemistry Part II
Constant­Volume Calorimetry
If constant volume, there is no work (w = ­P V [V=0 so w = 0])
q = C T (for constant­volume cal.)
Hess's Law
Enthalpy is a state function: H is independent of the pathway.
Hess's Law ­ going from reactants to products, the change in enthalpy is the same whether the reaction occurs in one step or several steps.
Two rules for calculating enthalpy changes using Hess's Law
1) If a rxn is reversed, then reverse the sign.
2) If coefficients are multiplied by an integer, so is H.
enthalpy is an extensive property ­ more matter reacting produces/releases more heat
When in doubt...trial & error
CH4 + 2O2 CO2 + 2H2O H1 = ­890 kJ
CH4 + 2O2 CO + 2H2O + 1/2O2 H2 = ­607 kJ
CO + 2H2O + 1/2O2 CO2 + 2H2O H3 = ­283 kJ
Consider the following processes
2A 1/2B + C ∆H = 5 kJ
3/2B + 4C 2A + C + 3D ∆H = ­15 kJ
E + 4A C
∆H = 10 kJ
Calculate ∆H for: C E + 3D
A) 0 kJ
B) 10 kJ
C) ­10 kJ
D) ­2 kJ
E) 20 kJ
C
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Chapter 6 ­ Thermochemistry Part II
Standard Enthalpies of Formation
Some reactions occur too slow or too fast to study H.
Hf
Standard Enthalpy of Formation: the change in enthalpy that accompanies the formation of one mole of a compound from its elements with substances in their standard states.
Definitions of Standard State ­ H (p 246)
Compounds:
• Standard state for a gas = 1 atm
• Substance in sol'n, the concentration is 1 M
• Condensed states (solid/liquid) are standard
Elements:
• 1 atm, 25 C
Standard State ­ H
Enthalpy of a reaction
Write the standard enthalpies of formation equations for:
NO2
CO2
Since we can manipulate heats of formation we can use this to find the heat (enthalpy) of the reaction.
Hreaction =
np Hf (products) ­ nr Hf
(reactants)
Elements are not included ( H f = 0 kJ)
H2O
CH3OH
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