Transcript Thermodynamics I

Thermodynamics I
Energy
• Energy is the capacity to do work
• The forms of energy most important in chemistry are:
– thermal energy (kinetic energy) associated with the random
motion of atoms and molecules
– chemical energy (potential energy) stored in the structural units
of chemical substances, i.e. , covalent bonds, ionic bonds,
hydrogen bonds, etc.
• Other forms exist, but they are of less interest to a chemist
Energy
• Energy can not be lost, nor created
• The first law of thermodynamics says that the total amount
of energy in the universe remains constant
– The principle of conservation of energy
• Energy is transferred from one place to another, but it is
never lost nor created
Energy Changes in Chemical Reactions
• Heat is a transfer of thermal energy from one object to a
colder object
• The system is the part of the universe under observation (in
chemistry, it is often just the reactants and the products of
a reaction)
• The environment is the rest of the universe not included
within the system
Energy Changes in Chemical Reactions
• (a) An open system allows for
the exchange of matter and
energy with the environment
• (b) A closed system allows for
the exchange of energy with the
environment, but not matter
• (c) An isolated system does not
allow the exchange of energy or
matter with the environment
Energy Changes in Chemical Reactions
• Any process that generates heat
is called exothermic
– For example, the combustion
of hydrogen is exothermic,
as the heat leaves the system
and enters the environment
Energy Changes in Chemical Reactions
• Any process that requires heat
from the environment to be
brought into ​to the system is
called endothermic
– e.g., the melting of ice is
endothermic, because you
have to provide heat to melt
the ice
Enthalpy
• Most of the physical and chemical transformations of interest
occur at a constant pressure
• Enthalpy, H, is defined as H = E + PV
• During a reaction at constant pressure, the enthalpy of a reaction,
DH = H(products) - H(reactants), is equivalent to the heat
released or absorbed
• If DH > 0, the reaction is endothermic
• If DH < 0, the reaction is exothermic
Thermochemical Equations
• e.g.; CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(l)
DH = -890.4 kJ
is a thermochemical equation
• The above thermochemical equation says that the combustion of
one mole of CH4(g) produces one mole of CO2(g) and two
moles of H2O(l), and releases 890.4 kJ of heat
• For the inverse reaction
CO2(g) + 2 H2O(l)
CH4(g) + 2 O2(g)
DH = +890.4 kJ
We must provide 890.4 kJ of heat to make it happen
Thermochemical Equations
• If we double the amount of reactants and products, the value of
DH doubles as well
2 CH4(g) + 4 O2(g)
2 CO2(g) + 4 H2O(l)
DH = -1780.8 kJ
• It is important to note the phase of each reactant and product
ex.;
but
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(l)
DH = -890.4 kJ
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
DH = -802.4 kJ
because H2O(g) is higher in enthalpy
2 H2O(l)
2 H2O(g)
DH = +88.0 kJ
Specific Heat and Heat Capacity
• The specific heat (s) of a substance is the amount of heat
required to raise the temperature of a one gram substance by one
Kelvin
• The heat capacity (C) of a substance is the amount of heat
required to raise the temperature of a given mass of this
substance by one Kelvin
• An object with a high heat capacity requires a lot of heat to
warm up and releases a lot of heat when cooling down
Specific Heat and Heat Capacity
• The specific heat is an intensive quantity and heat capacity
is an extensive quantity
C=ms
where m is the mass of the substance (in grams)
• The heat capacity relies on changes in temperature, DT,
with the quantity of heat (Q) absorbed or released during
the process
Q = C DT = m s DT
Specific Heat and Heat Capacity
• Example: The specific heat of iron is 0.444 J/(g K). A 869 g
bar of iron goes from 94oC to 5oC. Calculate the amount of
heat released by the metal.
• Solution:
Q  m s ΔT  (869g)(0.4 44 J
gK
)( 89 K)
Q  34.3  10 4 J  34.3 kJ
Exothermic reaction, thus 34.3 kJ of heat is released.
Constant Volume Calorimetry
• The heat of combustion is
measured in a bomb
calorimeter which has a fixed
volume
• The bomb is insulated so that
there is no heat transfer with
the environment
Qsystem = 0
Qwater + Qbomb + Qreaction = 0
Qreaction = - (Qwater + Qbomb)
Constant Volume Calorimetry
• The heat released by the reaction contributes to the
warming of the water and the bomb calorimeter
Qwater = mwater swater DT
Qbomb = Cbomb DT
where Cbomb was determined in advance by the combustion
of a standard within the bomb calorimeter in question
• N.B. Qreaction is not exactly equal to DH because it is not
operating at constant P (It is a constant V)
Constant Volume Calorimetry
• Example: 1.922 g of methanol (CH3OH) is burned in a constantvolume calorimeter. The temperature of the water and the bomb
calorimeter is increased 4.20 oC. If the quantity of water was
exactly 2000 g and the heat capacity of the calorimeter is 2.02
kJ/K, calculate the molar heat of combustion of methanol. The
specific heat of water is 4.184 J/(g K).
Constant Pressure Calorimetry
• For reactions other than combustion
(i.e., endothermic or less exothermic
reactions)
Qreaction = -(Qwater + Qcalorimeter)
where Qcalorimeter was determined in
advance
• Because this is occurring at constant
pressure
Qreaction = DH
Constant Pressure Calorimetry
• Example: We mix 100 mL of a 0.500 mol/L HCl solution
with 100 mL of a 0.500 mol/L NaOH solution in a constantpressure calorimeter whose heat capacity is 335 J/K. The
initial temperature of the solutions is 22.50 oC. The final
temperature is 24.90 oC. Calculate the heat of the
neutralization reaction :
NaOH(aq) + HCl(aq)
NaCl(aq) + H2O(l)
Assume that the densities and heat capacities of the solutions
are the same as those of pure water (1.00 g/mL and 4.184
J/(g K), respectively)
Constant Pressure Calorimetry
• Solution:
• N.B. This heat was released by the reaction of
(0.500mol/L)(0.100L) = 0.0500 moles of acid and base, so if 1.00
mole of HCl reacts with 1.00 mole of NaOH
heat of neutralization = -2.81kJ / 0.0500 mol = -56.2 kJ/mol
The Standard Enthalpy of Formation
• We are only able to measure changes in enthalpy, DH, and not
absolute enthalpies (i.e., H)
• Because we are only interested in changes in enthalpies, we are
free to choose any reference point (i.e., our “zero”)
• Arbitrarily, we chose to say that the standard enthalpy of
formation, DHfo, of any element in its most stable allotropic form
(i.e., their standard state) is zero
• Then, DHfo for any other substance is the amount of heat produced
or absorbed when one mole of the substance is formed from these
elements in their standard states at a pressure of 1 atm
The Standard Enthalpy of Formation
A few examples:
•
•
•
•
•
•
ΔHfo (H2, g) = 0, by definition
ΔHfo (O2, g) = 0, by definition
ΔHfo (O3, g) > 0, since O2 is more stable
ΔHfo (C, graphite) = 0, by definition
ΔHfo (C, diamond) > 0, since graphite is more stable
ΔHfo (H2O, l) = -285.8 kJ, because the reaction
H2(g) + ½ O2(g)  H2O(l) releases 285.8 kJ of heat
The Standard Enthalpy of Reaction
• Using the standard enthalpy of formation, DHfo, we can calculate
the standard enthalpy of reaction, DHoreaction, i.e., the enthalpy of a
reaction that occurs under a pressure of 1 atm
• ex.; For the reaction

aA  bB  cC  dD
 
DH oréaction  c ΔH of (C)  d ΔH of (D)  a ΔH of (A)  b ΔH of (B)
• The general formula is
where m et n are the stoichiometric coefficients of the reactants
and products

The Standard Enthalpy of Reaction
• Sometimes, we can determine DHfo of a compound by directly
reacting its elements in their standard states
• e.g.; For the reaction
C(graphite) + O2(g)
CO2(g)
The reaction is complete and 393.5 kJ of heat is released,
thus by definition, DHfo(CO2, g) = -393.5 kJ
• However, for other compounds, it is more difficult to find DHfo
because the direct synthesis of the compound from its elements
in their standard states is impossible
e.g.; The reaction: C(graphite) + 2 H2(g)
CH4(g)
does not happen no matter what experimental conditions
are used
Hess’ Law
• Hess’ Law: when the reactants are converted to products,
the enthalpy is the same whether the reaction takes place in
one step or in several steps
• In other words, as long as the starting point (reactants) and
end point (products) are the same, the DH value does not
change with the chosen trajectory to get reactants to
products
Hess’ Law
• Example: Using the following thermochemical equations,
calculate the enthalpy of formation of methane
C(graphite) + O2(g)
2 H2(g) + O2(g)
CH4(g) + 2 O2(g)
CO2(g)
DHoreaction = -393.5 kJ
2 H2O(l)
DHoreaction = -571.6 kJ
CO2(g) + 2 H2O(l) DHoreaction = -890.4 kJ
• Solution: Add the following three thermochemical equations:
C(graphite) + O2(g)
2 H2(g) + O2(g)
CO2(g) + 2 H2O(l)
C(graphite) + 2 H2(g)
CO2(g)
DHoreaction = -393.5 kJ
2 H2O(l)
DHoreaction = -571.6 kJ
CH4(g) + 2 O2(g) DHoreaction = +890.4 kJ
CH4(g)
DHoreaction = -74.7 kJ
Thus DHfo(CH4, g) = -74.7 kJ
Thermodynamics
• Thermochemistry: The study of heat transfer in chemical
reactions
• Thermochemistry is a branch of thermodynamics
• Thermodynamics: The study of transformations from the
point of view of the energy exchanges that accompany
these transformations
• We also consider work (not just heat) in thermodynamics
State Functions
• The state of a system is defined
by the values ​of all macroscopic
variables such as composition,
temperature, pressure, and
volume
• A state function is a property of
the system which is determined
by the state of the system,
regardless of the way the system
has reached this state
State Functions
• e.g.; If 1 L of water is cooled from 30 oC to 20 oC at 1 atm
or if 1 L of water is heated from 10 oC to 20 oC at 1 atm,
the two samples of water will have the same pressure,
volume, temperature, mass, energy, enthalpy, specific heat,
etc., as these properties are state functions
• A state equation is an equation that relates state functions
eg.; PV = nRT is a state equation because it sets the
value of the fourth function, if three of the four
(P, V, n, T) are known
The First Law of Thermodynamics
• The First Law of Thermodynamics:
Energy can be converted from one form to another, or
transferred from one place to another, but it can
neither be created nor destroyed.
• The First Law says that the change in the internal energy of
a system (DU) is offset by a variation of the energy of the
environment so that the energy of the universe does not
change
The First Law of Thermodynamics
• We can express the First Law mathematically as:
DU = Q + W
where Q is the heat transferred to the system and W is the
work done on the system
• If Q > 0, we have an endothermic transformation
• If Q < 0, we have an exothermic transformation
• If W > 0, work is done on the system and the energy of the
system increases
• If W < 0, work is done by the system and the energy of the
system decreases
PV Work
• We will consider electrical work
when we study electrochemistry
• For mechanical work, the
expansion work done by a gas
(PV work) will be especially
significant
W = -P DV
• For a compression (we require
calculus to get this formula)
W = - nRT ln (Vf/Vi)
PV Work
• Example: The volume of gas passes from 264 mL to 971
mL at a constant temperature. Calculate the work done (in
joules) by a gas if it expands (a) in a vacuum, and (b)
opposed to a constant pressure of 4.00 atm.
• Solution: We have to work with SI units:
1 atm = 101 325 Pa
and
1 mL = 1 x 10-6 m3
(a) P = 0 so W = 0
101325 Pa
10 6 m3
)  (971 mL  264 mL)  (
)
(b) W  PΔ V  (4.00 atm)  (
1 atm
1 mL
W  287 J
Work and Heat
• In this last example, we saw that the value of W depends on the
path taken and is therefore not a state function
• Because DU = Q + W, and we also know that U is a state
function and W is not, we know that Q, like W, is not a state
function
• the value of W depends on the path chosen, but the value
of DU (Q + W) is independent of the path, so the value of
Q must also depend on the selected path
Work and Heat
• Example: A gas expands and performs PV work equal to
279 J on the external environment. At the same time, it
absorbs 216 J of heat. What is the variation in energy of
the system?
• Solution:
W = -279 J
Q = +216 J
DU = Q + W = +216 J + (-279 J) = -63 J
Enthalpy and the First Law of Thermodynamics
• At constant volume, the PV work is equal to zero because DV = 0,
and thus DU = Q + W = Q
• For all reactions at constant volume, Q = DU
• However, constant-pressure reactions are of greater interest and if a
gas is produced or consumed, DV is not negligible
e.g.; During the reaction,
2 Na(s) + 2 H2O(l)
2 NaOH(aq) + H2(g)
The H2(g) produced must do PV work when it has to push back the
air so as to enter the atmosphere
Enthalpy and the First Law of Thermodynamics
• For any reaction, DU = Q + W
• For a reaction at constant pressure, DH = Q
• So for a reaction at constant pressure, DU = DH + W
• if one considers only the PV work at constant pressure, W = -P DV
DU = DH - P DV
or
DH = DU + P DV
• Because U, P, and V are state functions, H is also a state function
Enthalpy and the First Law of Thermodynamics
• If we say that DV only depends on changes in the number of moles
of gas produced or consumed (i.e., the volumes of solids and liquids
are negligible in comparison)
DH = DU + PDV = DU + D(PV)
• Making the approximation that the gases behave as ideal gases
DH = DU + D(nRT)
• Since the temperature is constant :
DH = DU + RTDn
Where Dn = moles of gas products – moles of gas reactants
and we use R = 8.3145 J/(K mol)
• We want to heat a sample of water via the combustion of
ethane (C2H6) at a temperature of 25.0 ◦C and a pressure of
1.00 atm (N.B. combustion is the reaction of a substance
with O2(g) to produce CO2(g) and H2O(l)). All of the heat
released in this reaction enters 2.000 kg of water so as to
increase the temperature from 25.0 ◦C to 60.0 ◦C. What
volume of ethane (C2H6) must be used?
•
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•
•
∆H◦f (C2H6, g) = −103.9 kJ mol−1
∆H◦f (CO2, g) = −393.5 kJ mol−1
∆H◦f(H2O, l) = −285.8 kJ mol−1
s (H2O, l) = 4.184 J K−1 g−1