Date post: | 05-Jan-2016 |
Category: |
Documents |
Upload: | mervyn-sanders |
View: | 223 times |
Download: | 2 times |
1
Calorimetry
Calorimetry
• Calorimetry - the accurate and precise measurement of heat change for chemical and physical processes.
• The device used to measure the absorption or release of heat in chemical or physical processes is called a Calorimeter
Calorimetry
• Foam cups are excellent heat insulators, and are commonly used as simple calorimeters
A Cheap Calorimeter
• For systems at constant pressure, the heat content is the same as a property called Enthalpy (H) of the system
Calorimetry
• Changes in enthalpy = H• q = H These terms will be used
interchangeably in this textbook• Thus, q = H = m x C x TH is negative for an exothermic
reactionH is positive for an endothermic
reaction
Calorimetry
• Calorimetry experiments can be performed at a constant volume using a device called a “bomb calorimeter” - a closed system
9
In terms of bonds
COO C
O
O
Breaking this bond will require energy.
CO
OOO C
Making these bonds gives you energy.In this case making the bonds gives you more energy than breaking them.
10
Exothermic
• The products are lower in energy than the reactants
• Releases energy
2Al (s) + 3Cl2 (g) --> 2 AlCl3 (s) + 1408 kJ
∆H=1408 kJ
11
C + O2 CO2E
nerg
y
Reactants Products
C + O2
C O2
395kJ
+ 395 kJ
12
Endothermic
• The products are higher in energy than the reactants
• Absorbs energy
2 H2O + 575 kJ ------> 2 H2 + 1 O2 (g)
∆H = + 572 kJ
13
CaCO3 CaO + CO2E
nerg
y
Reactants Products
CaCO3
CaO + CO2
176 kJ
CaCO3 + 176 kJ CaO + CO2
14
Chemistry Happens in
MOLES• An equation that includes energy is called
a thermochemical equation
• CH4 + 2O2 CO2 + 2H2O + 802.2 kJ
• 1 mole of CH4 releases 802.2 kJ of energy.
• When you make 802.2 kJ you also make 2 moles of water
• What is the molar enthalpy of CO2 (g) in the reaction for the burning of butane below?
2 C4H10 +13 O2 8 CO2 +10 H2O
∆H=-5315 kJ
Answer: Molar enthalpy is the enthalpy change in equation divided by the balance of CO2
• Molar enthalpy, ∆H substance = 5315 kJ ÷ 8 mol
• = 664 kJ / mol.
• For each of the following rewrite the equation in " H " notation, for one mole of the underlined substance.
Fe2O3 (s)+3CO(g)→3CO2(g)+2Fe(s)+25kJ
• Answer:1/3 Fe2O3 (s)+CO(g)CO2(g)+2/3 Fe(s) ∆H = - 8.3 KJ
4 NH3(g)+5O2 (g)→4 NO(g)+6H2O(l)+1170kJ
2 HCl (g)+96 KJ → H2 (g)+Cl2 (g)
N2 (g)+3 H2 (g) → 2 NH3 (g)+92 KJ
2 CO2 (g)+566 KJ →2 CO (g)+ O2 (g)
4 Al (s) +3 O2 (g) →2 Al2O3 (s)+3360 KJ
18
Thermochemical Equations
• A heat of reaction is the heat change for the equation, exactly as written– The physical state of reactants and
products must also be given.
– Standard conditions for the reaction is 101.3 kPa (1 atm.) and 25 oC
19
CH4 + 2 O2 CO2 + 2 H2O + 802.2 kJ
• If 10. 3 grams of CH4 are burned completely, how much heat will be produced?
10. 3 g CH4
16.05 g CH4
1 mol CH4
1 mol CH4
802.2 kJ
= 514 kJ
20
CH4 + 2 O2 CO2 + 2 H2O + 802.2 kJ
• How many liters of O2 at STP would be required to produce 23 kJ of heat?
• How many grams of water would be produced with 506 kJ of heat?
• How much heat will be released if 65 grams of butane is burned in a lighter according the equation:
2 C4H10 +13 O2 8 CO2 +10 H2O
∆H=-5315 kJ
= 2976.4 kJ
= 3.0 MJ
104
104104 2
5315
14.58
165
HmolC
kJ
g
HmolCHgC
• Calculate the heat released when 120 grams of Iron (III) oxide is formed by the following equation
2 Fe2O3 (s) → 4 Fe(s)+3 O2 (g)
∆H=1625 kJ
mol
kJ
g
OmolFeOgFe
2
1625
70.159
1120 32
32
= 610.5 kJ= 610 kJ
• Q = n ∆H (substance)
Where n = # of moles
• What mass of carbon dioxide must form to create 1200 kJ of heat when the following reaction occurs?
C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)
∆H=- 2808kJ
Answer: 110 grams
• 3) What mass of oxygen is needed to completely react and release 550 kJ of heat in the following reaction?
• 4Fe (s)+3O2 (g) → 2 Fe2O3 (s)
∆H=- 1625 kJ
Answer: 32 grams
26
Summary, so far...
27
Enthalpy• The heat content a substance has at a
given temperature and pressure• Can’t be measured directly because
there is no set starting point• The reactants start with a heat content• The products end up with a heat content• So we can measure how much enthalpy
changes
28
Enthalpy• Symbol is H• Change in enthalpy is H (delta H)• If heat is released, the heat content of
the products is lowerH is negative (exothermic)
• If heat is absorbed, the heat content of the products is higherH is positive (endothermic)
29
Ene
rgy
Reactants Products
Change is down
H is <0
30
Ene
rgy
Reactants Products
Change is upH is > 0
31
Heat of Reaction• The heat that is released or absorbed in a
chemical reaction• Equivalent to H
• C + O2(g) CO2(g) + 393.5 kJ
• C + O2(g) CO2(g) H = -393.5 kJ
• In thermochemical equation, it is important to indicate the physical state
• H2(g) + 1/2O2 (g) H2O(g) H = -241.8 kJ
• H2(g) + 1/2O2 (g) H2O(l) H = -285.8 kJ
32
Heat of Combustion• The heat from the reaction that
completely burns 1 mole of a substance
33
• OBJECTIVES:– Classify, by type, the heat changes
that occur during melting, freezing, boiling, and condensing.
34
• OBJECTIVES:– Calculate heat changes that occur
during melting, freezing, boiling, and condensing.
35
Heats of Fusion and Solidification
• Molar Heat of Fusion (Hfus) - the heat absorbed by one mole of a substance in melting from a solid to a liquid
• Molar Heat of Solidification (Hsolid) - heat lost when one mole of liquid solidifies
36
Heats of Fusion and Solidification
• Heat absorbed by a melting solid is equal to heat lost when a liquid solidifies– Thus, Hfus = -Hsolid
37
Heats of Vaporization and Condensation
• When liquids absorb heat at their boiling points, they become vapors.
• Molar Heat of Vaporization (Hvap) - the amount of heat necessary to vaporize one mole of a given liquid.
38
Heats of Vaporization and Condensation
• Condensation is the opposite of vaporization.
• Molar Heat of Condensation (Hcond) - amount of heat released when one mole of vapor condenses
• Hvap = - Hcond
39
Heats of Vaporization and Condensation
The large values for Hvap and Hcond are the reason hot vapors such as steam is very dangerous– You can receive a scalding burn from
steam when the heat of condensation is released!
40
Heats of Vaporization and Condensation
• H20(g) H20(l) Hcond = - 40.7kJ/mol
41
Heat of Solution
• Heat changes can also occur when a solute dissolves in a solvent.
• Molar Heat of Solution (Hsoln) - heat change caused by dissolution of one mole of substance
• Sodium hydroxide provides a good example of an exothermic molar heat of solution:
42
Heat of Solution
NaOH(s) Na1+(aq) + OH1-
(aq)
Hsoln = - 445.1 kJ/mol
• The heat is released as the ions separate and interact with water, releasing 445.1 kJ of heat as Hsoln thus becoming so hot it steams
H2O(l)