Transcript AP Thermo I Notes

Chapter 5
Energy (section 5.1)
 Energy is- the capacity to do work or transfer
heat.
 So what are work and heat?
 Work-energy used to cause an object with mass
to move
 Heat-energy used to cause the temperature of
an object to increase.
Energy
 Kinetic (Ek)-energy of motion
 EK = ½mv2
 Magnitude depends on the object’s mass (m)
and its velocity (v).
 Potential-depends on the object’s position
relative to other objects. In chemistry, it is
expressed as electrostatic potential energy
Potential Energy
 Eel= (KQ1Q2) /d, where K is a constant, and Q
is the electrical charge on the two objects, and
d is the distance between them.
 K = 8.99x109 J-m/C2 (C is in coulombs)
 Q will usually be about the size of electron
charge (1.6x10-19 C)
 When both Q’s have the same sign, the charges
will repel one another, etc.
 When Eel is positive, what is the situation?
Units
 SI unit for energy is the Joule (J) 1 kg-m2/s2
 Joules are tiny, so kJ will be appropriate.
 calorie=4.184J
 1000cal=1 Cal=1kcal
 What is the kinetic energy in joules of a 45g
golf ball moving at 61 m/s?
 What is this energy in calories?
 What happens to this energy when the ball
lands in a sand trap?
Work and Heat
 Two players: the system (that which we are
focused upon), and the surroundings
(everything else).
 Work (w)= F*d A force (F) moves an object a
distance (d).
 When we (surroundings) cause an object
(system) to move, we are performing work on
the system, or transferring energy to it.
 Heat, q, can enter or leave the system.
First Law of Thermodynamics section 5.2
 Energy(E) is neither created nor destroyed.
 The internal energy of a system is the sum of
the kinetic & potential energies of all the
components of the system. (We can’t know
these values, but we can measure their change)
 ∆E=Efinal – Einitial
 Positive ∆E means the system gained energy.
 Negative ∆E means the system lost energy.
First Law, cont.
 Positive ∆E denotes an endothermic process.
 Negative ∆E denotes an exothermic process.
 Endothermic-heat flows into the system from
the surroundings.
 Exothermic-heat flows out of the system to the
surroundings.
State Functions
 A property of a system that is determined by
specifying the system’s condition, or state (in temp.,
pressure, location, etc.)
 The value of a state function depends only on the
present state of the system, not the path it took to get
to that state.
 ∆E is a state function, since it is derived from an
initial and final state, not how it got from initial to
final. It describes the amount of change only.
 q (heat) and w (work) are NOT state functions.
 However, ∆E = q + w
Enthalpy H section 5.3
 In chem., much of our work is in open systems (like
a beaker), so work, w, is not readily obvious.
 In a closed system, with a piston, pressure is
constant, and the piston can be moved via a rxn. That
generates a gas. This work would be visible.
 This would be called pressure-volume work.
 H accounts for heat flow in processes where pressure
is constant, and the only work done is pressurevolume work.
 H2 gas is generated and moves the piston. Pressure-volume work.
Enthalpy cont.
 So, H = E + PV (internal energy plus pressure-volume work
done)
 Therefore, when a change occurs at constant
pressure, ∆H = ∆E + P∆V
 If ∆E= q + w, and w = -P∆V, then:
 ∆H = (qP + w) – w, or ∆H = qP
 This simply means that in most cases, ∆H = q
 Positive ∆H means the system has gained heat from
the surroundings (endothermic).
 Negative ∆H means the system has lost heat to the
surroundings (exothermic).
Enthalpies of Reaction section 5.4
 ∆H = Hproducts – Hreactants
 This change is known as the heat of reaction. It is
the heat change that acoompanies a reaction.
 Example:
 CH4 + 2O2 → CO2 + 2H2O ∆H = -890 kJ
 The production of water from its elements is an
exothermic reaction the gives off 483.6 kJ of heat.
 This is a thermochemical equation.
Enthalpies of reaction cont.
 Enthalpy is an extensive property. If you double the
amount reactants consumed, ∆H doubles also.
Enthalpy is stoichiometric.
 The enthalpy change for a reaction is equal in
magnitude, but opposite in sign, for its reverse
reaction.
 The enthalpy change for a reaction depends on the
state (phase) of the reactants and products.
 ∆H of reaction is usually in the units of kJ/mol.
 The value of ∆H for the reaction below is -72 kJ.
H2 (g)  Br2 (g)  2HBr(g)
 __________ kJ of heat are released when 1.0 mol of HBr
is formed in this reaction.

 A) 144
 B) 72
 C) 0.44
 D) 36
 E) -72
d
 The value of ∆H for the reaction below is -126 kJ.
2Na 2O2 (s)  2H2O(l)  4NaOH(s)  O2 (g)
 The amount of heat that is released by the reaction of 25.0 g
of Na2O2 with water is __________ kJ.

 A) 20.2
 B) 40.4
 C) 67.5
 D) 80.8
 E) -126
a
 The value of ∆H for the reaction below is -1107 kJ:
2Ba(s)  O2 (g)  2BaO(s)
 How many kJ of heat are released when 15.75 g of Ba(s)
reacts completely with oxygen to form BaO(s)?





A) 20.8
B) 63.5
C) 114
D) 70.3
E) 35.1
b
Calorimetry
section 5.5
 ∆H can be measured experimentally, and this
measurement of heat flow is called calorimetry.
 The temp. change a substance undergoes when it
absorbs heat is its heat capacity. Every pure
substance has its own.
 Heat capacity, C, is the heat required to raise the
temperature 1°C. Specific heat capacity, s, is the
heat required to raise 1g of that substance 1°C.
Molar heat capacity, Cmolar……1 mole of a
substance 1°C.
Constant-pressure calorimeter
 Really just a styrofoam cup.
Styrofoam is an excellent
insulator, so it holds heat very
well.
 Constant pressure because it’s
open to the atmosphere.
Constant-volume calorimeter
 Called constant-volume
because it is a closed
system, so pressure
varies, but the volume of
the system does not.
 Used to measure the
energy stored in samples
of fuels and foods.
Calorimetry cont.
 Heat, q, = m*s*∆T
 Heat lost by the system is gained by the
surroundings, and vice versa.
 As stated earlier, ∆H and q are the same thing at
constant pressure, so when we measure heat, we are
measuring ∆H directly.
 When mixing solutions,
qSOLN = (spec. heat of solution)*(g of solution)*∆T
 For solutions, the specific heat will be that of water:
4.184 J/g-°C
Calorimetry
 When 50.0mL of 0.100M AgNO3 and 50.0mL of 0.100M HCl are
mixed in a constant-pressure calorimeter, the temperature of the
mixture increases from 22.30°C to 23.11°C. The temperature
increase is caused by the following reaction:
 AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)
 Calculate ∆H for the reaction, assuming 100.0g of solution, and a
specific heat of 4.184 J/g-°C.
 -68000 J/mol or -68 kJ/mol (q= 338.9 J)
 Would the ∆H value be different if the molarity of your reactants
was different?
 Write a net ionic equation for the reaction, just for fun.
Hess’s Law section 5.6
 Allows you to calculate ∆H for a reaction without
measuring anything, using known ∆H values for
other reactions.
 Stated: if a reaction is carried out in a series of
steps, ∆H for the overall reaction will equal the sum
of the enthalpy changes for the individual steps.
 Some examples:
Enthalpy of Formation section 5.7
 Using Hess’s Law, the enthalpy change can be calculated for
many different kinds of reactions:
 ∆Hvap-heat of vaporization-converting liquid to gas
 ∆Hfus-heat of fusion-melting a solid
 ∆Hcomb-heat of combustion-combusting a substance in oxygen
 ∆Hf-heat of formation-heat associated with the formation of a
compound from its elements. This one is kind of important.
 Since the amount of enthalpy change depends on temp.,
pressure, and state (phase), it helps to compare reactions at what
is called standard state, which is defined as 1 atm pressure and
298K or 25°C. From this, we have developed tables of enthalpy
data called standard enthalpy changes.
Standard enthalpy change
 Symbolized as ∆H°, standard enthalpy change is the enthalpy
change that occurs for a reaction when all of the reactants and
products are at their standard state. That is, whatever state they
are in at 1 atm and 298K.
 So, a reaction with Fe(l) or H2O(s) could not be occurring at
standard state conditions.
 Even more specific, ∆H°f, or standard enthalpy of formation,
is the enthalpy change that occurs when 1 mole of a compound
is formed from its elements at standard state conditions.
 These are listed in Appendix C in the back of the textbook.
∆H°f Std. enthalpy of formation
 Let’s try writing a few equations, with their ∆H°f component
alongside (from Appendix C):
 HBr
 AgNO3
 Hg2Cl2
 C2H5OH
 NH3
 SO2
 RbClO3
 NH4NO3
∆H°f Std. enthalpy of formation
 If an element exists in more than one form at standard conditions,
the most stable form of the element is listed. For example Cgraphite
would be listed instead of Cdiamond.
 ∆H°f of elements is zero, because no energy is needed, the
element already would exist in its standard state at standard
conditions.
 ∆H°f equation or not (and if not, why not?):
 2Na(s) + ½O2(g) → Na2O(s)
 2K(l) + Cl2(g) → 2KCl(s)
 C6H12O6(s) → 6C(diamond) + 6H2(g) + 3O2(g)
Using ∆H°f to find ∆Hrxn
 C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ∆Hcomb = ?
 We can “add” these three equations to find ∆Hcomb.
C3H8(g) → 3C(s) + 4H2(g)
∆H1 = -∆H°f[C3H8(g)]
 3C(s) + 3O2(g) → 3CO2(g)
∆H2 = 3∆H°f[CO2(g)]
 4H2 + 2O2(g) → 4H2O(l)
∆H3 = 4∆H°f[H2O(l)]
 _______________________________________________
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ∆Hrxn = ∆H1 + ∆H2 + ∆H3

 You can use Hess’s Law and ∆H°f equations from a table to
calculate ∆Hrxn for any reaction, so long as you choose the
necessary reactions to work with.
 The short version: To find the heat of reaction, ∆Hrxn,total
all of the ∆H°f‘s for the products, total all of the ∆H°f ’s for
the reactants, and subtract the reactant total from the
product total.
 Or:
 ∆H°rxn= Σn∆H°f(products) – Σm∆H°f(reactants)
 Now try # 69 and 71 in the textbook.