Chapters 5 and 19

Energy = capacity to do work Kinetic = energy of motion Potential = energy of position relative to

other objects Work = energy used to cause an object

with mass to move against a force Force = push or pull exerted on an object

Heat = energy used to cause temp. to increase

1 22

Q Q r

kF

F = the electric force between 2 charged objects

k = Coulomb’s constant = 8.99 x 109 J-m/C2

Q1 and Q2 are the electrical charges

r is the distance of separation between the two objects

System = portion singled out for study Open Closed Isolated

Surroundings = everything else

Work is energy used to cause an object to move against a force w = F x d [Eq. 5.3]

Heat is energy transferred from a hotter object to a colder one

Energy is conserved. Internal energy

Sum of all kinetic and potential energies of the system’s components

Change in energy is what we can know E = Efinal – Einitial Magnitude of the change is important Direction of change is important (+ or -)

Shows the direction and magnitude of the energy change for a chemical reaction

E is positive when energy is added to the system and negative when energy exits

E = q + w (q is heat and w is work) Endothermic

System absorbs heat Exothermic

System releases heat

Depends only on the present state of the system, and not the path it took to get there

Energy of a system

Final

P-V work = pressure-volume work, which is involved in the expansion or compression of gases

w = -PV Enthalpy = heat flow in processes @

constant P where no other work is done except P-V work Internal energy + P*V H = E + PV

Change in enthalpy = heat gained/lost @ constant P

Enthalpy change that accompanies a reaction

Heat of reaction (Hrxn) H = Hproducts – Hreactants

1. Enthalpy is an extensive property.1. Directly proportional to the amount of

reactant consumed in the reaction.2. Enthalpy change for a rxn is equal in

magnitude but opposite in sign to H for the reverse rxn.

3. The enthalpy change for a rxn depends on the state of the reactants and products.

H2O (s) H2O (l) H = 6.01 kJ

• The stoichiometric coefficients always refer to the number of moles of a substance

Thermochemical Equations: show enthalpy changes as well as the mass relationships

• If you reverse a reaction, the sign of H changesH2O (l) H2O (s) H = -6.01 kJ

• If you multiply both sides of the equation by a factor n, then H must change by the same factor n.

2H2O (s) 2H2O (l)H = 2 x 6.01 = 12.0 kJ

H2O (s) H2O (l) H = 6.01 kJ

• The physical states of all reactants and products must be specified in thermochemical equations.

Thermochemical Equations

H2O (l) H2O (g) H = 44.0 kJ

How much heat is evolved when 266 g of white phosphorus (P4) burn in air?P4 (s) + 5O2 (g) P4O10 (s) H = -3013 kJ

266 g P41 mol P4

123.9 g P4x -3013 kJ

1 mol P4x = -6470 kJ

Hydrogen peroxide can decompose to water and oxygen by the following reaction: 2H2O2 (l) → 2H2O (l) + O2 (g) H = -196 kJCalculate the value of q when 5.00 g of

hydrogen peroxide decomposes at constant pressure. kJ 14.4- )

OH mol 2kJ 196(*)

OH g 34.02OH mol 1

( * OH g 5.00 Heat 2222

2222

Measurement of heat flow Calorimeter = device used to measure

magnitude of temp. change that the heat flow produces

Heat capacity = C = amount of heat required to raise the temp. by 1K (or 1°C). Extensive property; units are J/K or J/°C

Molar heat capacity (Cm)= heat capacity of one mole of a substance

Specific heat (Cs)= heat capacity of one gram of a substance Intensive property

q = m * Cs * T

T*mq

change) re(Temperatu * substance) of (gramsferred)heat trans of(quantity Cs

How much heat is given off when an 869 g iron bar cools from 940C to 50C?

Cs of Fe = 0.444 J/g • 0C

t = tfinal – tinitial = 50C – 940C = -890C

q = m*Cs*t= 869 g *0.444 J/g • 0C * –890C= -34,000 J

Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g-K. Calculate the temperature change 50.0 kg of rocks would undergo if they emitted 450. kJ of heat.

Coffee cup calorimeter Pressure of the system = atmospheric

pressure Assume that the calorimeter contains

all heat generated in the reaction Use q = m*Cs * T

H = qrxn

No heat enters or leaves!

When 50.0 mL of 0.100 M silver nitrate and 50.0 mL of 0.100 M hydrochloric acid are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30°C to 23.11°C. The temperature increase is caused by the following reaction:AgNO3 (aq) + HCl (aq) → AgCl (s) + HNO3 (aq)

Calculate H for this reaction in kJ/mol AgNO3, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g-°C.

Bomb calorimetry Bomb calorimeter = used to study

combustion reactions qrxn = -Ccal * T

Constant-Volume Calorimetry

No heat enters or leaves!

qsys = qwater + qbomb + qrxn

qsys = 0qrxn = - (qwater + qbomb)qwater = m x Cs x tqbomb = Cbomb x t

Reaction at Constant V

H ~ qrxn

H = qrxn

Measured in a constant-volume bomb calorimeter

A 0.5865-g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is 4.812 kJ/°C. The temperature increases from 23.10°C to 24.95°C. Calculate the heat of combustion of lactic acid (a) per gram and (b) per mole.

kJ/mol 1370- acid lactic mol 1acid lactic g 09.90*

acid lactic g 1kJ 15.2-

kJ/g 15.2- g 0.5865kJ 8.9022-

kJ 8.9022- C)C)(1.85kJ/ (4.812- T*C- q

C1.85 23.10) - (24.95 T

calrxn