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