Chapter 8 Thermochemistry: Chemical Energy
Jan 12, 2016
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 = E + PV
– At constant volume (V = 0): qv = E
– At constant pressure: Energy due to heat and work but work minimal compared to heat energy
• qp = E + PV = H
– Enthalpy change (heat of reaction): H = Hproducts – Hreactants
The Thermodynamic Standard State
• ΔH = amount of energy absorbed or released in the form of heat H = 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
─ H – 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 (Hreaction).
– Endothermic: Heat flows into the system from the surroundings
and H has a positive sign. Unfavorable Process
– Exothermic: Heat flows out of the system into the surroundings
and H has a negative sign. Favorable process
Enthalpies of Physical and Chemical Changes
• Reversing a reaction changes the sign of H for a reaction.
– C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) H = –2219 kJ
– 3 CO2(g) + 4 H2O(l) C3H8(g) + 5 O2(g) H = +2219 kJ
• Multiplying a reaction increases H by the same factor.
– 3 C3H8(g) + 15 O2(g) 9 CO2(g) + 12 H2O(l) H = –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)
H = –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)
H = +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 = E.
• Constant Pressure Calorimetry: A constant pressure
calorimeter measures the heat change at constant
pressure such that q = H.
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.
–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.
C =
q
T
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 (H°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.
H°f = 0 for an element in its standard state
Standard Heats of Formation
H2(g) + 1/2 O2(g) H2O(l) H°f = –286 kJ/mol
3/2 H2(g) + 1/2 N2(g) NH3(g) H°f = –46 kJ/mol
2 C(s) + H2(g) C2H2(g) H°f = +227 kJ/mol
2 C(s) + 3 H2(g) + 1/2 O2(g) C2H5OH(g) H°f = –235 kJ/mol
Standard Heats of Formation
• Calculating H° for a reaction:
H° = Σ[H°f (products) x moles] – Σ[H°f (Reactants) x moles]
• For a balanced equation, each heat of formation must be multiplied by the stoichiometric coefficient.
– aA + bB cC + dD
H° = [cH°f (C) + dH°f (D)] – [aH°f (A) + bH°f (B)]
Problems
• Calculate H° (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 H° (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 G = H – TS
H – enthalpy, T – temperature, S - 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).
S = Sfinal – Sinitial
– Positive value of S indicates increased disorder (favorable).
– Negative value of S indicates decreased disorder
(unfavorable).
Problems
• Predict whether S° is likely to be positive or negative for each of the following reactions. Using tabulated values, calculate S° 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 (–H).
Increase in entropy (+S).
• Nonspontaneous process: Increase in enthalpy (+H).
Decrease in entropy (–S).
An Introduction to Free Energy
• Gibbs Free Energy Change (G): Weighs the relative contributions of enthalpy and entropy to the overall spontaneity of a process.
G = H – TS
G < 0 Process is spontaneous (favorable)
G = 0 Process is at equilibrium
G > 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)
G° = –55.7 kJ
– b. 2 C(s) + 2 H2(g) C2H4(g)
G° = 68.1 kJ
– c. N2(g) + 3 H2(g) 2 NH3(g)
H° = –92 kJ; S° = –199 J/K
An Introduction to Free Energy
• Equilibrium (G° = 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)
H° = –92.0 kJ S° = –199 J/K
– Equilibrium is the point where G° = H° – TS° = 0
Problem
• Benzene, C6H6, has an enthalpy of
vaporization, Hvap, equal to 30.8 kJ/mol
and boils at 80.1°C. What is the entropy
of vaporization, Svap, 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