Transcript Chapter 8

Chapter 8
Thermochemistry: Chemical
Energy
Energy
• Energy – capacity to supply heat or do
work
Energy = Heat + Work
E=q+w
– 2 types of Energy
• Potential Energy
• Kinetic Energy
Energy
• Two fundamental kinds of
energy.
– Potential energy is stored
energy.
– Kinetic energy is the
energy of motion.
•
Law of Conservation of
Energy
– Energy can be converted
from one kind to another
but never destroyed
Energy
• Units
– SI Unit – Joule (J)
– Additional units
• Calorie (Cal) – food calorie
• calorie (cal) – scientific calorie
• Conversions
– 1 cal = 4.184 J
– 1000 cal = 1 Cal
Energy and Chemical Bonds
• Chapter 6
– Kept a careful
accounting of atoms
as they rearranged
themselves
• Reactions also
involve a transfer of
energy
Energy and Chemical Bonds
• A chemical
– Potential - attractive forces in an ionic compound or
sharing of electrons covalent compound
– Kinetic – (often in form of heat) occurs when bonds
are broken and particles allowed to move
– To determine the energy of a reaction it is necessary
to keep track of the energy changes that occur during
the reaction
Internal Energy and State
Functions
• In an experiment: Reactants
and products are the system;
everything else is the
surroundings.
• Energy flow from the system
to the surroundings has a
negative sign (loss of energy).
• Energy flow from the
surroundings to the system
has a positive sign (gain of
energy).
Internal Energy and State
Functions
• Tracking energy changes
– Energy changes are measured from the point
of view of the system (Internal Energy - IE)
• Change in Energy of the system – ΔE
– ΔE = Efinal - Einitial
Internal Energy and State
Functions
• IE depends on
– Chemical identity, sample size, temperature,
etc.
– Does not depend on the system’s history
• Internal Energy is a state function
– A function or property whose value depends
only on the present state (condition) of the
system, not on the path used to arrive at that
condition
Expansion Work
•
E=q+w
– In physics w = force (F) x distance (d)
•
Force – energy that produces movement of an
object
– In chemistry w = expansion work
•
•
•
Force - the pressure that the reaction exerts on
its container against atmospheric pressure
hence it is negative
Distance – change in volume of the reaction
w = -PΔV
Energy and Enthalpy
• ΔE = q – PΔV
• The amount of heat exchanged between the system and
the surroundings is given the symbol q.
q = DE + PDV
– At constant volume (DV = 0): qv = DE
– At constant pressure: Energy due to heat and work but work
minimal compared to heat energy
• qp = DE + PDV = DH
– Enthalpy change (heat of reaction): DH = Hproducts – Hreactants
The Thermodynamic Standard
State
• ΔH = amount of energy absorbed or released in
the form of heat
 DH = Hproducts – Hreactants
• Important factors
– States of matter
– Thermodynamic standard state – most stable form of a
substance at 1 atm and at a specified temperature, usually
25oC; and 1 M concentration for all substances in solution
─ DH – valid for the reaction as written including exact # of
moles of substances
» N2H4(g) + H2(g)  2 NH3(g) + heat (188 kJ)
Enthalpies of Physical and
Chemical Change
Enthalpies of Physical and
Chemical Changes
• Enthalpies of Chemical Change: Often called heats of
reaction (DHreaction).
– Endothermic: Heat flows into the system from the surroundings
and DH has a positive sign. Unfavorable Process
– Exothermic: Heat flows out of the system into the surroundings
and DH has a negative sign. Favorable process
Enthalpies of Physical and
Chemical Changes
• Reversing a reaction changes the sign of DH for
a reaction.
– C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DH = –2219 kJ
– 3 CO2(g) + 4 H2O(l)  C3H8(g) + 5 O2(g) DH = +2219 kJ
• Multiplying a reaction increases DH by the same
factor.
– 3 C3H8(g) + 15 O2(g)  9 CO2(g) + 12 H2O(l) DH = –6657
kJ
Problems
• How much heat (in kilojoules) is evolved or
absorbed in each of the following reactions?
• Burning of 15.5 g of propane:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
DH = –2219 kJ
• Reaction of 4.88 g of barium hydroxide octahydrate with
ammonium chloride:
Ba(OH)2·8 H2O(s) + 2 NH4Cl(s)  BaCl2(aq) + 2 NH3(aq) + 10 H2O(l)
DH = +80.3 kJ
Determination of Heats of Reaction
•
•
•
•
Experimentally – calorimetry
Hess’s Law
Standard Heat’s of Formation
Bond Dissociation Energies
Calorimetry and Heat Capacity
• Calorimetry is the science of measuring heat
changes (q) for chemical reactions. There are
two types of calorimeters:
• Bomb Calorimetry: A bomb calorimeter measures the heat
change at constant volume such that q = DE.
• Constant Pressure Calorimetry: A constant pressure
calorimeter measures the heat change at constant
pressure such that q = DH.
Calorimetry and Heat Capacity
Calorimetry and Heat Capacity
• Heat capacity (C) is the amount of heat required to
raise the temperature of an object or substance a
given amount.
q
C =
DT
–Specific Heat: The amount of heat required to raise the
temperature of 1.00 g of substance by 1.00°C.
–Molar Heat: The amount of heat required to raise the
temperature of 1.00 mole of substance by 1.00°C.
Problems
• What is the specific heat of lead if it takes 96 J to
raise the temperature of a 75 g block by 10.0°C?
• When 25.0 mL of 1.0 M H2SO4 is added to 50.0 mL
of 1.0 M NaOH at 25.0°C in a calorimeter, the
temperature of the solution increases to 33.9°C.
Assume specific heat of solution is 4.184 J/(g–1·°C–1),
and the density is 1.00 g/mL–1, calculate the heat
absorbed or released for this reaction.
Hess’s Law
• Allows the enthalpy to be determined for:
– Reactions that occur too quickly or take too
long to use calorimetry
– Reactions that are too dangerous
• Works like the Haber process in chapter 6
– Take reactions for which the heat is known
and manipulate them to give the desired
reaction
Standard Heats of Formation
• Standard Heats of Formation (DH°f): The
enthalpy change for the formation of 1 mole of
substance in its standard state from its
constituent elements in their standard states.
• The standard heat of formation for any element
in its standard state is defined as being ZERO.
 DH°f = 0 for an element in its standard state
Standard Heats of Formation
H2(g) + 1/2 O2(g)  H2O(l)
3/ H (g)
2 2
+ 1/2 N2(g)  NH3(g)
2 C(s) + H2(g)  C2H2(g)
DH°f = –286 kJ/mol
DH°f = –46 kJ/mol
DH°f = +227 kJ/mol
2 C(s) + 3 H2(g) + 1/2 O2(g)  C2H5OH(g)
DH°f = –235 kJ/mol
Standard Heats of Formation
• Calculating DH° for a reaction:
DH° = Σ[DH°f (products) x moles] – Σ[DH°f (Reactants) x moles]
• For a balanced equation, each heat of formation must be
multiplied by the stoichiometric coefficient.
– aA + bB
cC + dD
 DH° = [cDH°f (C) + dDH°f (D)] – [aDH°f (A) + bDH°f (B)]
Problems
• Calculate DH° (in kilojoules) for the reaction of
ammonia with O2 to yield nitric oxide (NO) and
H2O(g), a step in the Ostwald process for the
commercial production of nitric acid.
• Calculate DH° (in kilojoules) for the
photosynthesis of glucose from CO2 and liquid
water, a reaction carried out by all green plants.
Energy Calculations
• Other methods for calculating enthalpies
– Bond dissociation energies – measures the
energy given off by the formation of bonds in
the products and substracts the energy
required to break bonds in the reactants
Why do chemical reactions occur?
• A chemical reaction will move from less
stability to greater stability.
– Achieved by giving off more energy than is
absorbed by the reactants
• This indicates that exothermic reactions occur by
why do endothermic reactions occur?
• Gibb’s Free Energy
 DG = DH – TDS
 DH – enthalpy, T – temperature, DS - entropy
An Introduction to Entropy
• Second Law of Thermodynamics: Reactions proceed
in the direction that increases the entropy of the system
plus surroundings. (increases the degree of disorder)
• A spontaneous process is one that proceeds on its own
without any continuous external influence.
• A nonspontaneous process takes place only in the
presence of a continuous external influence.
An Introduction to Entropy
An Introduction to Entropy
An Introduction to Entropy
• The measure of molecular disorder in a system is called
the system’s entropy; this is denoted S.
• Entropy has units of J/K (Joules per Kelvin).
 DS = Sfinal – Sinitial
– Positive value of DS indicates increased disorder (favorable).
– Negative value of DS indicates decreased disorder
(unfavorable).
Problems
• Predict whether DS° is likely to be positive or
negative for each of the following reactions.
Using tabulated values, calculate DS° for each:
– a. 2 CO(g) + O2(g)  2 CO2(g)
b. 2 NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)
c. C2H4(g) + Br2(g)  CH2BrCH2Br(l)
d. 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g)
An Introduction to Free Energy
• To decide whether a process is spontaneous, both
enthalpy and entropy changes must be considered:
• Spontaneous process:
Decrease in enthalpy (–DH).
Increase in entropy (+DS).
• Nonspontaneous process: Increase in enthalpy (+DH).
Decrease in entropy (–DS).
An Introduction to Free Energy
• Gibbs Free Energy Change (DG): Weighs the
relative contributions of enthalpy and entropy to
the overall spontaneity of a process.
DG = DH – TDS
 DG < 0 Process is spontaneous (favorable)
 DG = 0 Process is at equilibrium
 DG > 0 Process is nonspontaneous (unfavorable)
Problems
• Which of the following reactions are
spontaneous under standard conditions at
25°C?
– a. AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
DG° = –55.7 kJ
– b. 2 C(s) + 2 H2(g)  C2H4(g)
DG° = 68.1 kJ
– c. N2(g) + 3 H2(g)  2 NH3(g)
DH° = –92 kJ; DS° = –199 J/K
An Introduction to Free Energy
• Equilibrium (DG° = 0): Estimate the
temperature at which the following reaction will
be at equilibrium. Is the reaction spontaneous at
room temperature?
–
N2(g) + 3 H2(g)  2 NH3(g)
DH° = –92.0 kJ
DS° = –199 J/K
– Equilibrium is the point where DG° = DH° – TDS° = 0
Problem
• Benzene, C6H6, has an enthalpy of
vaporization, DHvap, equal to 30.8 kJ/mol
and boils at 80.1°C. What is the entropy
of vaporization, DSvap, for benzene?
Optional Homework
• Text - 8.28, 8.32, 8.50, 8.52, 8.56, 8.58,
8.66, 8.70, 8.74, 8.82, 8.88, 8.90
• Chapter 8 Homework from website
Required Homework
• Chapter 8 Assignment