chapters 5 and 19. energy = capacity to do work kinetic = energy of motion potential = energy of...
TRANSCRIPT
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 changes
H2O (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 P4
1 mol P4
123.9 g P4
x-3013 kJ1 mol P4
x = -6470 kJ
Hydrogen peroxide can decompose to water and oxygen by the following reaction: 2H2O2 (l) → 2H2O (l) + O2 (g) H = -196 kJ
Calculate the value of q when 5.00 g of hydrogen peroxide decomposes at constant pressure.
kJ 14.4- )OH mol 2
kJ 196(*)
OH g 34.02
OH 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*m
q
change) re(Temperatu * substance) of (grams
ferred)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 = 0
qrxn = - (qwater + qbomb)
qwater = m x Cs x t
qbomb = 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 1
acid lactic g 09.90*
acid lactic g 1
kJ 15.2-
kJ/g 15.2- g 0.5865
kJ 8.9022-
kJ 8.9022- C)C)(1.85kJ/ (4.812- T*C- q
C1.85 23.10) - (24.95 T
calrxn