Thermochemistry: Chemical Energy

Dr. Paul CharlesworthMichigan Technological UniversityDr. Paul Charlesworth

Michigan Technological University

C h a p t e rC h a p t e r 88Thermochemistry: Chemical Energy

Thermochemistry: Chemical Energy

Chemistry 4th EditionMcMurry/Fay

Chemistry 4th EditionMcMurry/Fay

Chapter 08 Slide 2Prentice Hall ©2004

Thermodynamics 01Thermodynamics 01

• Energy: is the capacity to do work, or supply heat.

Energy = Work + Heat

• Kinetic Energy: is the energy of motion.

EK = 1/2 mv2 (1 Joule = 1 kgm2/s2)

(1 calorie = 4.184 J)

• Potential Energy: is stored energy.



• 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).



• Closed System: Only energy can be lost or gained.• Isolated System: No matter or energy is exchanged.



• The law of the conservation of energy: Energy cannot be created or destroyed.

• The energy of an isolated system must be constant.

• The energy change in a system equals the work done on the system + the heat added.

E = Efinal – Einitial = E2 – E1 = q + w

q = heat, w = work


State Functions 01State Functions 01

• State Function: A function or property whose value depends only on the present state (condition) of the system.

• The change in a state function is zero when the system returns to its original condition.

• For nonstate functions, the change is not zero if the path returns to the original condition.


State Functions 02State Functions 02

• 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)

• State properties include temperature, total energy, pressure, density, and [NH3].

• Nonstate properties include the heat.


Enthalpy Changes 01Enthalpy Changes 01

• Enthalpies of Physical Change:



• 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.

Exothermic: Heat flows out of the system into the

surroundings and H has a negative sign.



• 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



• 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



• Thermodynamic Standard State: Most stable form

of a substance at 1 atm pressure and 25°C; 1 M

concentration for all substances in solution.

• These are indicated by a superscript ° to the symbol

of the quantity reported.

• Standard enthalpy change is indicated by the symbol

H°.


Hess’s Law 01Hess’s Law 01

• 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



• Reactants and products in individual steps can be added and subtracted to determine the overall equation.

(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



• 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)

• Use the following data to calculate H° (in kilojoules) for the above reaction:

CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)

H° = –98.3 kJ

CH3Cl(g) + Cl2(g) CH2Cl2(g) + HCl(g)

H° = –104 kJ


Standard Heats of Formation 01Standard Heats of Formation 01

• 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



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



• Calculating H° for a reaction:

H° = H°f (Products) – H°f (Reactants)

• 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)]



-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)



• 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.


Bond Dissociation Energy 01Bond Dissociation Energy 01

• Bond Dissociation Energy: Can be used to determine an approximate value for H°f .

H = D (Bonds Broken) – D (Bonds Formed)

• For the reaction between H2 and Cl2 to form HCl:

H = D(H–Cl) – ∑ {D(H–H) + D(O=O)}





• Calculate an approximate H° (in kilojoules) for the

synthesis of ethyl alcohol from ethylene:

C2H4(g) + H2O(g) C2H5OH(g)

• Calculate an approximate H° (in kilojoules) for the

synthesis of hydrazine from ammonia:

2 NH3(g) + Cl2(g) N2H4(g) + 2 HCl(g)


Calorimetry and Heat Capacity 01Calorimetry and Heat Capacity 01

• 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.



Constant Pressure Bomb



• 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



• 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 H for the reaction.




Introduction to Entropy 01Introduction to Entropy 01

• 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.



• 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.





• 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).



• 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 01Introduction to Free Energy 01

• 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



• 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



• 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



• 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



• 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?

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Thermochemistry: Chemical Energy