Transcript Calorimetry Tutorial

Thermochemistry
The study of energy
Chapter 6
1
Kinetic Energy Ek
The energy of an object in
motion
Ek= ½ mv2
Mass need to be in kg
A car moving at 40 mph has
a greater kinetic energy
than a car moving at 20
mph
2
Types of kinetic energy
• Work: energy used to
cause an object with
mass to move
• Heat: energy used to
cause the
temperature of an
object to increase
3
Potential Energy
• Stored energy in an
object by virtue of its
position.
• Units of energy:
Joule J,
1 Calorie Cal = 4.184 J
4
5. 1 The Nature of Energy
• Energy is capacity to do work or to
produce heat.
• Thermochemistry is the study of heat
change in chemical reactions.
5
Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies that
are at different temperatures.
Temperature is a measure of the thermal energy
Said another way temperature is a measure of random motion
(KE) of particles.
Temperature (K)
= Thermal Energy (J)
7
6.2
Physics vs. Chemistry
Chemical energy: is the energy stored within the
bonds of chemical substances (potential)
Thermal energy: is the energy associated with the
random motion of atoms and molecules (Kinetic
molecular theory) (kinetic)
In this chapter we will discuss the transfer of these
types of energy.
8
Energy Systems
System: portion we single out to study typically
the chemical
Ex: reactants products.
Surroundings: everything else
Ex: Reaction vessel, environment
9
10
Types of Systems
11
C
B
A
12
Exothermic
A reaction that results in
the evolution of heat.
Thus heat flows out of
the system
Exo = out = heat loss from
system = energy loss =
-heat
13
An Exothermic Reaction
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
14
15
Endothermic
A reaction that absorbs heat
from their surroundings.
Thus heat flows into a
system.
Endo = In to = heat gain to
system = energy gain = +
heat
16
An Endothermic Reaction
energy + 2HgO (s)
2Hg (l) + O2 (g)
energy + H2O (s)
H2O (l)
17
18
Work, work, work
• Recall that energy
Can be transferred in the
form of motion and heat
Work = force x distance
Units
f = Newton, N
d = meter, m
W = N*m or Joule, J
19
• Force is any kind of push or pull exerted
on an object.
– Ex: gravity, mechanical, electrostatic
• Distance is how far the object moved as a
result of the force applied.
20
Work in Biology
• Think about water
moving from the
ground up the trunk of
a tree. What part of
the system if any
undergoes a change
in potential energy? Is
work done in the
process?
21
Guided Questions
What is changing location ?
Does this change involve potential energy?
22
• Water moves up the trunk against the
force of gravity. Thus the potential energy
of the water does change
• Work is movement of a mass over a
distance against an opposing force.
23
24
Homework
Chang: pg 255
1, 9, 10
****BL:Pg 188
• 1, 2, 3, 6, 9, 11,
25
5.2 First law of thermodynamics
Aka: The law of conservation of energy
Energy is neither created nor destroyed, thus
energy is converted from one form to
another.
Thus the energy of the universe is constant
Universe = System + Surroundings
26
Chemical energy lost by combustion = Energy gained by the surroundings
system
surroundings
27
Internal Energy
• The energy (E) of a system can be
defined as the sum of the kinetic and
potential energies of all of the particles
in a system.
• Internal energy can be changed by a
flow of HEAT, WORK, or BOTH
29
Change in a system energy
E = Efinal - Einitial
E > 0 = system gained energy
E < 0 = system lost energy to the surroundings
Enthalpy can not be negative, think about it like a bank account (EX pg 235)
30
Relating heat to work
E = q + w
E = change in system’s internal
energy
q = heat
w = work
31
Signs signs every where are signs
32
Write a figure
description for this
figure so that a
student reading this
text would
understand.
Identify key terms in
your description.
33
Work (w)
W > 0 = work is done ON the system
( + W)
W < 0 = work is done BY the system on the
surroundings.
(-W)
34
Prove it….
• Write an example that illustrates work
being done by a system
Then write a sentence illustrating work being
done on a system
35
Heat (q)
q > 0 = heat is added to the system.
(+ q endothermic)
q < 0 = heat is released from the
system (-q exothermic)
36
Prove it ….
• Write an example that illustrates heat
being added to a system
Then write a sentence illustrating heat being
released form a system
37
Example:
Calculate E for a system
undergoing an endothermic
process in which 15.6 kJ of heat
flows and where 1.4 kJ of work is
done by the system.
38
Answer
q = + 15.6 (endothermic)
W = - 1.4 kJ (work is done by the system)
E = q + w
E = 15.6 + (- 1.4) = 14.2
39
State Function
A Property of a function that is determined by specifying its
condition or its present state
The value of a state function depends only on the present
state of the system - not how it arrived there
Energy IS A STATE FUNCTION
Enthalpy IS A STATE FUNCTION
Temperature IS A STATE FUNCTION
Heat IS NOT A STATE FUNCTION
Work IS NOT A STATE FUNCTION
40
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
41
Example
A
B
C
It does not mater whether I heated the water (C  B) or cooled that
water (A  B) to get it to the temperature. Internal energy at
point B would be the same regardless.
42
Change in energy ΔE
• ΔE is a state function because it is
independent of pathway.
(all we care about is energy at Ei and energy
at Ef.
Heat (q) and work (w) are not state
functions!!!!!
43
Example
44
5.2 Homework
Chang: pg 255
11,12,17,18
***BL: Pg 189
17, 19, 21, 22, 25, 26
45
5.3 Enthalpy
Enthalpy (H):
Accounts for the heat flow in chemical
changes occurring at constant pressure
when no forms of work other than P, V
work
*** state function****
Results from a change in P (atm) or V (L)
of the system.
46
What is Pressure volume work
Build up of gas that causes a piston to lift
against the force of gravity.
Work = -P ΔV
47
48
Enthalpy : H = E + PV
E = internal energy
Surroundings
Surroundings
HEAT
System
HEAT
System
ΔH >0
ΔH <0
Endothermic
Exothermic
49
Don’t forget you can manipulate
these equations
Enthalpy = H = E + PV
E = H  PV
H = E + PV
50
Example:
• Calculate the work associated with the
expansion of gas from 46 L to
64 L at a constant external pressure of 15
atm.
W = -PΔV
52
Answer
w = -PΔV
P = 15 atm
V = 64L – 46L = 18L
w = -15atm (18L)
= -270 L*atm
To convert to J 101.3 J/1 L*atm
53
5.3 Homework
Chang: 15, 16,
BL: Pg 190
• #’s 27 A and C only, 29 skip A do B, 30
• HINT ON #30 write a balance chemical
reaction first, don’t over think it.
55
5.4 Enthalpy Changes
• Enthalpy: The heat content of a chemical
system is called the enthalpy (symbol: H)
• The enthalpy change (Δ H) is the amount
of heat released or absorbed when a
chemical reaction occurs at constant
pressure.
56
More enthalpy
• Enthalpy, along with the pressure and
volume of a system, is a state function
(property of a system that depends only on
its state, not how it arrived at its present
state).
57
Enthalpy of reactions
To calculate the enthalpy of a reaction we
need to subtract the final enthalpy from the
initial enthalpy or
Hrxn = Hproducts – Hreactants
58
What is the enthalpy of a rxn?
• The heat of the reaction ΔH or ΔHrxn that occurs
from the change of products to reactants.
Balanced equations that show an enthalpy
change are called thermochemical equations.
2H2 (g) + O2 (g)  2H2O (g)
ΔH = - 483.6 kJ
59
Enthalpy Diagrams
• We can use diagrams to illustrate the
change in enthalpy of a reaction.
Exothermic = - ΔH value
Endothermic = + ΔH value
60
Thermochemical Equations
Is H negative or positive?
System absorbs heat
Endothermic
H > 0
6.01 kJ are absorbed for every 1 mole of ice that
melts at 00C and 1 atm.
H2O (s)
H2O (l)
H = 6.01 kJ
61
6.4
Thermochemical Equations
Is H negative or positive?
System gives off heat
Exothermic
H < 0
890.4 kJ are released for every 1 mole of methane
that is combusted at 250C and 1 atm.
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l) H = -890.4 kJ
62
6.4
64
Reaction time
Reaction time
65
Guidelines for thermochemical
reactions
1. Enthalpy is an extensive property
Ex: combustions of 1 mol = ΔH -40 kJ
combustions of 2 mol = ΔH -80 kJ
2. Enthalpy of a forward reaction is equal to (but
opposite in sign) of enthalpy of the reverse
reaction
2H2 (g) + O2 (g)  2H2O (g) ΔH = - 483.6 kJ
2H2O (g)  2H2 (g) + O2 (g) ΔH = + 483.6 kJ
66
3. Enthalpy of the reaction depends on the
state of the reactants and the products.
Producing H2O (l) or H2O (g) would change
the enthalpy value for a thermochemical
equation
H2O (s)
H2O (l)
H2O (l)
H = 6.01 kJ
H2O (g)
H = 44.0 kJ
67
Steps
CH4 (g) + 2O2(g)  CO2 (g) + 2H2O (l) ΔH = -890 kJ
1. Use mole ratios to establish a proportion
for every 1mol CH4 (g) burned -890 kJ of heat is
produced
2. Use dimensional analysis to convert g to mol to
heat. (since we know heat production from 1
mole of CH4)
4.50g CH4 1mol CH4
16.0g CH4
-890kJ
1mol CH4
=
-250kJ
69
Homework
Chang:
BL: Pg: 190
#’s 31, 33, 36, 37, 40
Read 5.5 Calorimetry
70
Calorimetry
• Measures heat flow
• Calorimeter: used to
measure the
exchange of heat that
accompanies
chemical reactions.
71
How it works
• Reaction using known quantities of
reactants is conducted in an insulated
vessel that is submerge in a known qty of
water.
• The heat created from the reaction will
increase the temperature of the water
surrounding the vessel.
• The amount of heat emitted can be
calculated using the total heat capacity of
the calorimeter and its contents.
72
Heat Capacity
• The temperature change experienced by an object when it
absorbs a certain amount of heat energy.
• The heat required to raise the temperature of a substance by
1K or 1ºC
q = CΔT
q = heat energy released or absorbed by the rxn (J)
C = heat capacity (J/K)
ΔT = change in temperature (K)
NOTE: when a sample gains heat (+q) ΔT is positive
when a sample loses heat (-q) ΔT is negative
73
A note on temperature scale
A temperature change in
Kelvin is equal in
magnitude to the
temperature change in
degrees Celsius.
In the following calculations
we are looking at the
value for the quantity of
the ΔT not an actual Temp
reading.
Thus 1K = 1ºC
74
Molar Heat Capacity
• The heat capacity of one mole of a substance is
called its molar heat capacity.
• If the molar heat capacity for 1 mol =20 J/mol*K
Then the molar heat capacity for 10 moles = 20(10)
200 J/mol*K (20 J/mol*K for each mole present in
the reaction)
75
Specific Heat
• The heat capacity of 1 gram of a substance.
• The heat required to raise 1 gram of a
substance by 1K or 1ºC
q = sm ΔT
q= heat energy released or absorbed by the rxn (J)
s = specific heat (J/g*K)
m = mass of solution (g)
ΔT = change in temperature in (K)
76
Enthalpy and Specific Heat
• FYI: When reactions
occur at constant
pressure
ΔH = q
Thus enthalpy change
(ΔH) is equal to heat
(q)
77
How much heat is given off when an 869 g iron bar cools
from 940C to 50C?
s of Fe = 0.444 J/g • 0C
t = tfinal – tinitial = 50C – 940C = -890C
q = mst = 869 g x 0.444 J/g • 0C x –890C = -34,000 J
78
6.5
Specific Heat
• Specific heat of a substance can be
determined by measuring the temperature
change that a known mass of a
substances undergoes when it gains or
loses a specific quantity of heat.
Specific heat = qty of heat transferred
(g of substance) x (temp change)
79
Example
• 209J is required to increase the temp of
50.0 g of water by 1.00K.
• Thus the specific heat of water is
4.18 J/gK
s=
209 J
(50.0g) (1.00 K)
80
Example:
• How much heat in kJ is required to
increase the temperature if 150 g of water
from 25°C to 42°C. Specific heat of water
is 4.18 J/g°C. (Memorize specific heat
water)
81
Answer
q = msΔT
q = 150 (4.18)(42-25)
q = 10.659 kJ
82
Molar Heat Capacity Example
• If you have 1 mol of water with a specific
heat of 4.18 J/g*K, the molar heat capacity
of water is.
• 4.18 J/g*K (18.0g) = 75.2J/mol*K
•
1mol
• What if you have 26 moles?
83
• IF we assume that no heat is
lost. then the heat gained by
the solution is lost from the
heated metal ball.
Heat lost = Heat gained
qsol = (s) x (g of solution) x ΔT = -qrxn
ΔT > 0 means the rxn is exothermic
qrxn < 0
Metal
ball
85
Example
• When a student mixes 50 mL of 1.0M HCl and
50 mL of 1.0 M NaOH in a coffee cup
calorimeter the temperature of the resulting 100
mL solution increases from 21ºC to 27.5 ºC.
Calculate the enthalpy change for the reaction
assuming that the calorimeter loses only a
negligible amount of heat. The density of the
solution is 1.0 g/mL with a specific heat of
4.18 J/g*K
86
50 mL
1.0 M
HCl
+
50 mL
1.0 M
NaOH
100 mL
=
ΔT = 27.5-21.0 = 6.5
S = 4.18 J/g*K
q = smΔT
D = 1.0 g/mL
m = 100mL (1.0 g) = 100 g
1 mL
q = (4.18 J/gK)(100 g)(6.5)
= 2.7 x103 J
Now we must go back to the problem and determine if
the heat (q) is endothermic (+) or exothermic (-)
Because the temperature increases heat is being
released and the reaction is exothermic
FINAL ANSWER q = = - 2.7 x103 J
87
One step further
We solved for how much energy was produced from
100g of solution…
q = = - 2.7 x103 J
What if the question asked for heat produced from 1 mol
HCl?
(0.05L HCl)(1.0 mol HCl) = 0.05 mol HCl in solution
1 L HCl
-2.7 103J/ (0.050 mol HCl) = -54000 J/mol of HCl
88
A coffee cup calorimeter initially contains
125 g of water at 24.2°C. Potassium
bromide (10.5g) also at 24.2°C is added to
the water, and after the KBr dissolves, the
final temp is 21.1°C. Calculate the
enthalpy change for dissolving the salt in
J. Assume that the specific heat capacity
of the solution is 4.184 J/°Cg and that no
heat transferred to the surroundings or to
the calorimeter.
89
q = msΔT
(think m total)
m = 125g H2O + 10.5KBr = 135.5g total
q= 135.5 (4.184)(21.1- 24.2)
q = -1757.4 J
90
*******Example*****
A 46.2 g sample of copper is heated to
95.4 ºC and then placed into a calorimeter
containing 75.0 g of water at 19.6 ºC. The
final temp of the metal and water was
21.8 ºC. Calculate the specific heat of
copper, assuming that all the heat lost by
copper is gained by water.
91
Assume:
Heat lost by copper = heat gained by the water
qlost = 46.2g(s)(95.4-21.8 ºC )
= 3400.32(s)
qgained = 75.0 g(4.18)(21.8-19.6 ºC )
= 689.7
3400.32(s) = 689.7
S = 0.2 J/ºC *g
92
Bomb Calorimetry
• Used to study
combustion reactions
qrxn = -Ccal x ΔT
93
Homework
Brown Lemay Pg 191-2
#’s
Sp heat/ capacity 43, 44, 45, 46, 47,
Calorimetry 49, 50
94
6.3 Hess’s Law
“Heat of summation”
states that if reactions are carried out in a
series of steps, ΔH for the reaction will be
equal to the sum of the enthalpy changes
by the individual steps.
95
Reactants  Products
The change in enthalpy is the same whether the
reaction takes place in one step or a series of
steps.
Goal: manipulate the steps of the equation
(multiple, reverse) to cancel out terms to isolate
the final reaction and calculate the new enthalpy
(ΔH )
96
97
Calculations via Hess’s Law
1. If a reaction is reversed, H is also reversed.
N2(g) + O2(g)  2NO(g)
2NO(g)  N2(g) + O2(g)
2.
H = 180 kJ
H = 180 kJ
If the coefficients of a reaction are multiplied by
an integer, H is multiplied by that same integer.
2NO(g)  N2(g) + O2(g)
6NO(g)  3N2(g) + 3O2(g)
-180 x 3 = -540
H = 180 kJ
H = 540 kJ
98
Solving Hess’s Law Problems
Given the following data, calculate the ΔH for the reaction:
H2S(g) +
2O2(g)  H2SO4(l)
H2SO4(l)  SO3(g) +
H2O(g)
ΔH =-706.5KJ
ΔH =184.5KJ
H2O(g) 
H2O(l)
ΔH=-99KJ
______________________________________
SO3(g) + H2O(l)  H2S(g) + 2O2(g)
ΔH=
99
Solving Hess’s Law Problems
Given the following data, calculate the ΔH for the reaction:
C2H2 (g)+ 5/2 O2 ( g)  2CO2 (g) + H2O (l) ΔH = -1300kJ
C (s) + O2 (g)  CO2 (g)
H2 (g) +
½ O2 (g)  H2O (l)
ΔH = -394 kJ
ΔH = -286 kJ
2C (s) + H2 (g)  C2H2 (g)
100
C2H2 (g)+ 5/2 O2  2CO2 (g) + H2O (l)
C (s) + O2 (g)  CO2 (g)
H2 (g) +
½ O2 (g)  H2O (l)
ΔH = -1300kJ
ΔH = -394 kJ
ΔH = -286 kJ
__________________________________
2C (s) + H2 (g)  C2H2 (g)
ΔH =
101
Homework
• BL packet
• #’s 57, 59, 60, 61, 62
102
5.7 Enthalpy of Formation
• Change in enthalpy due to the formation of
substances from stable forms of its
component elements.
Standard enthalpy of formation (ΔHfº ) is the
change in enthalpy for the reaction that
forms 1 mol of the compound from its
elements in their standard states.
103
Where to Find ΔHfº Values
• Appendix 3 pg A-8
• NOTE THE STATE
OF YOUR
MOLECULE!!!
105
ΔHrxn = H (products) – H (reactants)
multiply any ∆H by its coefficient
106
Example:
Using enthalpies of formation, calculate the
standard change in enthalpy for the following
reaction. Then determine if the reaction is
endothermic or exothermic:
2Al (s) + Fe2O3 (s)  Al2O3 (s) + 2Fe (s)
107
ΔHrxn = H (products) – H (reactants)
2Al (s) + Fe2O3 (s)  Al2O3 (s) + 2Fe (s)
H= 0
+
-822
-1669 + 0
(-1669 + 0) - (0
+
-822)
ΔHrxn = -847 kJ VERY exothermic
108
Homework
• #’s 67, 71, 72
(You will need to look up H in BL table)
109