thermochemistry: chemical energy
DESCRIPTION
Thermochemistry: Chemical Energy. Thermodynamics01. Thermodynamics: study of energy and it’s transformations Energy: capacity to do work, or supply heat Energy = Work + Heat Kinetic Energy: energy of motion E K = 1 / 2 mv 2 (1 Joule = 1 kg m 2 /s 2 ) (1 calorie = 4.184 J) - PowerPoint PPT PresentationTRANSCRIPT
Chapter 8Chapter 8Chapter 8Chapter 8
Thermochemistry: Chemical Energy
Thermochemistry: Chemical Energy
Thermodynamics 01
• Thermodynamics: study of energy and it’s transformations
• Energy: capacity to do work, or supply heat
Energy = Work + Heat
• Kinetic Energy: energy of motion
EK = 1/2 mv2 (1 Joule = 1 kgm2/s2)
(1 calorie = 4.184 J)
• Potential Energy: stored energy
Thermodynamics 02
• Conservation of energy law: Energy cannot be created or destroyed; it can only be converted from one form into another.
Thermodynamics 03
• Thermal Energy: kinetic energy of molecular motion (translational, rotational, and vibrational)
• Heat: the amount of thermal energy transferred between two objects at different temperatures
• Chemical Energy: potential energy stored in chemical bonds released in the form of heat or light
Thermodynamics 04
• First Law of Thermodynamics: energy of an
isolated system must be kept constant
Thermodynamics 05
• System reactants + products• Surroundings everything else
• Energy changes are measured from the point of view of the system!
• ∆E is negative energy flows out of the system • ∆E is positive energy flows into the system
Thermodynamics 06
Work 07
w = –PV
QuickTime™ and aSorenson Video decompressorare needed to see this picture.
Sign of w08
negative positive
positive negative
w = -PV expansion
w = -PV contraction
Work Units 09
w = -PV (J or kJ)
1L x 1000mL x 1cm3 x 1m3
1L 1mL (100cm)3
1000
101 x 103 kg
ms2
= 101 kgm2 = 101J
s2
m2w = L x atm =
Energy and Heat 10
Energy = Work + Heat
E = w + q = q - PV
q = E + PV
When a person does work, energy diminishes
w = negative
E = negative
Heat and Enthalpy 11
• 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: qp = E + PV = H
enthalpy
State Functions 12
State Function:
• value depends only on the present state of the system
• path independent
• when returned to its original position, overall change is zero
State Functions 13
• State and Nonstate Properties: The two paths below give the same final state:
N2H4(g) + H2(g) 2 NH3(g) + heat (188 kJ)
N2(g) + 3 H2(g) 2 NH3(g) + heat (92 kJ)
• temperature, total energy, pressure, density, volume, and enthalpy (∆H) state properties
• nonstate properties include heat and work
Enthalpy 14
• Enthalpy or heat of reaction:
•H = H(products) - H(reactants)
• States of the reactants and products are important! (g, l, s, aq)
• Thermodynamic standard state: P = 1atm, [ ] = 1M,
T = 298.15K (25ºC)
Standard Enthalpy of Reaction 15
Thermodynamic standard state: P = 1atm, [ ] = 1M,
T = 298.15K (25ºC)
Standard enthalpy of reaction (Hº)
N2(g) + 3H2(g) 2NH3(g) Hº = -92.2kJ
Enthalpy Changes 16
• Most changes in a system involve a gain or loss in enthalpy
• Physical (melting of ice in a cooler)
• Chemical (burning of gas in your car)
Physical Changes 17
• Enthalpies of Physical Change:
QuickTime™ and aSorenson Video decompressorare needed to see this picture.
Chemical Changes 18
• Enthalpies of Chemical Change: Often called heats of reaction (Hreaction).
Endothermic: Heat flows into the system from the
surroundings H is positive
Exothermic: Heat flows out of the system into the
surroundings H is negative
Enthalpy Changes 19
• 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 x [C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) ]
H =(-2219kJ x 3) = –6657 kJ
Example 20
• How much work is done (in kilojoules), and in which direction, as a result of the following reaction?
w = -0.25kJ
Expansion, system loses -0.25kJ
Example 21
• The following reaction has E = –186 kJ/mol.• Is the sign of PV positive or negative?
• What is the sign and approximate magnitude of H?
Contraction, PV is negative, w is positive
H = E + PV
H = (-186kJ) + (1atm) (-1mole)
H = negative (slightly more than E)
Example 22
The reaction between hydrogen and oxygen to yield water vapor has H° = –484 kJ. How much PV work is done, and what is the value of E (kJ) for the reaction of 0.50 mol of H2 with 0.25 mol of O2 at atmospheric pressure if the volume change is
–5.6 L?
PV = -0.57kJ
Contraction, so w is positive
E = -120.43kJ
Example 23
The explosion of 2.00 mol of solid TNT with a volume of approximately 0.274 L produces gases with a volume of 448 L at room temperature. How much PV (kJ) work is done during the explosion? Assume P = 1 atm, T = 25°C.
2 C7H5N3O6(s) 12 CO(g) + 5 H2(g) + 3 N2(g) + 2 C(s)
PV = 45.2kJ
Expansion, so w = -45.2kJ
Example 24
How much heat (kJ) is evolved or absorbed in each of the following reactions?
1.) Burning of 15.5 g of propane:
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) Hº = –2219 kJ
2.) 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)
Hº = +80.3 kJ
-780kJ (exothermic)
+1.24kJ (endothermic)
Hess’s Law 25
• Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction.
3 H2(g) + N2(g) 2 NH3(g) H° = –92.2 kJ
Hess’s Law 26
(a) 2 H2(g) + N2(g) N2H4(g) H°1 = ?
(b) N2H4(g) + H2(g) 2 NH3(g) H°2 = –187.6 kJ
(c) 3 H2(g) + N2(g) 2 NH3(g) H°3 = – 92.2 kJ
H°1 = H°3 – H°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ
Standard Heats of Formation 27
Where do H° values come from?
• Standard Heats of Formation (H°f): enthalpy
change for the formation of 1 mole of substance in its standard state
• H°f = 0 for an element in its standard state!
Standard Heats of Formation 28
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 29
• Calculating H° for a reaction:
H° = H°f (products) – H°f (reactants)
Heat of formation must be multiplied by the coefficient of the reaction
C6H12O6 (s) 2C2H5OH (l) + 2CO2 (g)
H° = [2H°f(ethanol) + 2H°f(CO2)] - H°f (glucose)
Standard Heats of Formation 30
-1131Na2CO3(s)49C6H6(l)-92HCl(g)
-127AgCl(s)-235C2H5OH(g)95.4N2H4(g)
-167Cl-(aq)-201CH3OH(g)-46NH3(g)
-207NO3-(aq)-85C2H6(g)-286H2O(l)
-240Na+(aq)52C2H4(g)-394CO2(g)
106Ag+(aq)227C2H2(g)-111CO(g)
Some Heats of Formation, Some Heats of Formation, HHff° ° (kJ/mol)(kJ/mol)
Bond Dissociation Energy 31
• Bond Dissociation Energy (D): Amount of energy needed to break a chemical bond in gaseous state
D = Approximate Hº
H° = D(reactant bonds broken) – D(product bonds formed)
H2 + Cl2 2HCl
H° = (DCl-Cl + DH-H) - (2 D H-Cl)
= [(1 mol)(243 kJ/mol) + (1)(436 kJ/mol] - (2)(432 kJ/mol)
= -185 kJ
Bond Dissociation Energy 32
Calorimetry and Heat Capacity 33
• Calorimetry: measurement of heat changes (q) for chemical reactions
• Constant Pressure Calorimetry: measures the
heat change at constant pressure q = H
• Bomb Calorimetry: measures the heat change
at constant volume such that q = E
Calorimetry and Heat Capacity 34
Constant Pressure Bomb
Calorimetry and Heat Capacity 35
• Heat capacity {C}: amount of heat required to raise the temperature of an object or substance a given amount
Specific Heat: amount of heat required to raise the
temperature of 1.00 g of substance by 1.00°C
Molar Heat: amount of heat required to raise the
temperature of 1.00 mole of substance by 1.00°C
C =
q
T
Calorimetry and Heat Capacity 36
Example 37
The industrial degreasing solvent methylene chloride (CH2Cl2, dichloromethane) is prepared from methane by reaction with chlorine:
CH4(g) + 2 Cl2(g) CH2Cl2(g) + 2 HCl(g)
Calculate H° (kJ)CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) H° = –98.3 kJ
CH3Cl(g) + Cl2(g) CH2Cl2(g) + HCl(g) H° = –104 kJ
H° = -98.3 + -104 = -202kJ
Example 38
Calculate H° (kJ) 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.
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
= [(4)(90.2kJ/mol) + (6)(-241.8)] - [(4)(-46.1) + (5)(0)]
= -905.6kJ
Example 39
Calculate H° (kJ) for the photosynthesis of glucose from
CO2 and liquid water, a reaction carried out by all green
plants.
6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)
= [(1 mole)(-1260kJ/mol) + (6)(0)] - [(6)(-393.5) + (6)(-285.8)]
= 2816kJ
Example 40
• Calculate an approximate H° (kJ) for the synthesis
of ethyl alcohol from ethylene:
C2H4(g) + H2O(g) C2H5OH(g)
• Calculate an approximate H° (kJ) for the synthesis
of hydrazine from ammonia:
2 NH3(g) + Cl2(g) N2H4(g) + 2 HCl(g)
Introduction to Entropy 42
• Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings.
• 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.
Introduction to Entropy 43
• 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.
Negative value of S indicates decreased disorder.
Introduction to Entropy 44
Introduction to Entropy 45
• 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).
Introduction to Entropy 39
• 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)
Introduction to Free Energy 40
• 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
G = 0 Process is at equilibrium
G > 0 Process is nonspontaneous
Introduction to Free Energy 41
• Situations leading to G < 0:H is negative and TS is positiveH is very negative and TS is slightly negativeH is slightly positive and TS is very positive
• Situations leading to G = 0:H and TS are equally negativeH and TS are equally positive
• Situations leading to G > 0:H is positive and TS is negative H is slightly negative and TS is very negativeH is very positive and TS is slightly positive
Introduction to Free Energy 42
• 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
Introduction to Free Energy 43
• 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
Introduction to Free Energy 44
• 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?