Thermochemistry - Dr. VanderVeen

download report

Transcript Thermochemistry - Dr. VanderVeen

Unit theme: Energy
Heat capacity
Specific heat
Units of heat
Potential energy diagrams
Endothermic and exothermic reactions
Heats of changes of state
Thermochemical equations
Hess’ Law
Gibb’s Free energy
What evidence can you see
that energy is involved?
 Defined as the capacity to do work or
supply heat
 Chemical potential energy:
 aka chemical energy
 Energy stored in chemicals because of their
 Different substances store different amount of
 A form of energy that flows from a warmer
object to a cooler object
 Q
 Can’t be measured directly
 Can only measure its effect on temperature
 When heat is added to a system, its
temperature rises
 Temperature is not heat!
 Temperature is a measure of average kinetic energy
 The study of heat changes that occur
during chemical reactions and physical
changes of state
Units of Heat
 calorie
 The quantity of heat needed to raise the
temperature of 1 g of pure water by 1 degree
 1 kcal = 1000 calories = 1 Calorie (nutrition)
 joule
 SI unit, named after British physicist
 newtonmeter
 1 cal = 4.186 joules
 1000 J = 1 kJ
 Commonly used because joules are so small
Heat Capacity
 The amount of heat it takes to change an
object’s temperature by 1 Celsius degree
 Depends partly on mass
 More mass  greater heat capacity
 How many calories are required to heat
32.0 g of water from 25.0oC to 80.0oC?
How many joules is this?
 Answer:
 1760 calories
 7360 joules = 7.36 kJ
Specific Heat
 Not all substances
respond the same
way to the input of
 Some get hot much
more quickly than
Specific Heat
 The amount of heat required to raise the
temperature of 1 g of a substance by 1oC
 A measure of how well a substance
stores heat energy
 Substances with low specific heat (ex.
metals) heat quickly, cool quickly
 Substances with high specific heat (ex.
water) take a long time to heat and cool
Using Specific Heat
 C or Cp
 Units: J/goC or cal/goC
 Q = m C T
Q = total heat change
m = mass
T = temperature change
Phase Changes
 Melting, freezing
 Boiling, condensing
 Sublimation, deposition
 Phase changes occur without
temperature changes
Calculating Energy Changes
in Phase Changes
 Can’t use specific heat, because no
temperature change is involved
 Instead, use this formula
Q = m  Heat of fusion (or heat of
 use heat of fusion for freezing, melting
 use heat of vaporization for boiling,
Heating/Cooling Curves
 Regions A, C, E
have temperature
 Q = mCT
 Choose specific heat
to match state of
 Regions B, D are
phase changes
 B: Q = mHfus
 D: Q = mHvap
Enthalpy of the reaction
 The total energy change associated with
a chemical or physical change
 Given the symbol Hrxn
 Hrxn = (energy of products) – (energy of reactants)
Classifying Heat Changes
Thermite reaction
 Produces molten iron
 Formerly used in
welding railroad
tracks, shipbuilding
 Gives off light and
 Highly exothermic
Exothermic reactions
 Energy is released to
the surroundings
 The temperature of
the surroundings
 The products have
less energy than the
Endothermic Reactions
 Energy is absorbed
from the
 The temperature of
the surroundings
 The products have
MORE energy than
the reactants
 Exothermic reactions
 Energy released by system
 Can treat energy as a product
 H is negative
 2 equivalent ways to write equation:
 CaO + H2O  Ca(OH)2 + 65.2 kJ
 CaO + H2O  Ca(OH)2
H = - 65.2 kJ
 Endothermic reactions
 Energy absorbed by system
 Can treat energy as a reactant
 H is positive
 2 equivalent ways to write equation:
 2 NaHCO3 + 129 kJ  Na2CO3 + H2O + CO2
 2 NaHCO3  Na2CO3 + H2O + CO2
H = + 129 kJ
2 ways to manipulate
thermochemical equations
 1) Write equation backwards
 Sign of H must change
 A+BC
H = + 123 kJ
 CA+B
H = - 123 kJ
 2) Multiply everything by a coefficient
 Must multiply H by coefficient, too!
 3 A + 3B  3C H = 3 (+123 kJ) = + 369 kJ
Problems with
thermochemical equations
 Consider the equation:
2 NaHCO3  Na2CO3 + H2O + CO2
H = + 129 kJ
 How many kJ would be released if 4.5
moles NaHCO3 reacted?
Hess’ Law
 If you add two or more thermochemical
equations to give a final equation, then
you can also add the heat changes to
give the final enthalpy of reaction.
 What is the enthalpy change, Hrxn, for
the decomposition of hydrogen peroxide?
 Target: 2 H2O2(l)  2 H2O(l) + O2(g)
 Given:
 H2(g) + O2(g)  H2O2(l)
 H2(g) + ½ O2(g)  H2O(l)
H = -187.9 kJ
H = -285.8 kJ
Standard Heats of
Formation, Hof
The standard heat of formation of a
compound is the change in enthalpy that
accompanies the formation of one mole
of a substance from its elements in their
standard states.
 The heat of formation of elements in their
standard states is arbitrarily set to zero.
Using heats of formation
 Thermodynamic stability: a measure of the
energy required to decompose the compound
 Compounds with large, negative enthalpies of
formation are thermodynamically stable
 Many heats of formation have been measured.
(see Appendix A-6 in textbook)
 Another way to do Hess’ Law!
 Hrxn = SnHf(products) – SmHf(reactants)
 where
 n represents the coefficients for the products
 m represents the coefficients for the reactants
Example Problems
 Compute Hrxn for the following reaction.
(Refer to Appendix A-6)
 2NO(g) + O2(g)  2 NO2(g)
 Compute Hrxn for the following reaction.
(Refer to Appendix A-6)
 4 FeO(cr) + O2(g)  2 Fe2O3(cr)
 Ans: -144.14 kJ, -560.0 kJ
 Symbol: S
 A quantitative measure of the degree of
disorder in a system
 The greater the disorder, the larger the value
of S
 Solids have a high degree of order (low entropy)
 Gases have a low degree of order (high entropy)
 More particles (moles) results in higher entropy
Entropy, cont.
 Systems tend to proceed to higher
 Examples
 Stirring sugar into your coffee
 The neatness of your locker
 The order of cards in a pack of playing cards
after shuffling
J. Willard Gibbs
 1839-1903
 First American to
earn a Ph.D. in
science from a US
 Yale, 1863
 One of nation’s best
Will a reaction occur
 The answer depends on the balance
between enthalpy (heat changes) and
 Gibb’s Free Energy
 The energy available from the system to do
useful work
Gibb’s Free Energy
 G = H – TS
 If G is negative, the reaction will occur
spontaneously and can proceed on its own.
 If G is positive, the reaction is
nonspontaneous and needs a sustained
energy input to proceed.
 If G is zero, the reaction is at equilibrium
(both the forward and reverse reactions take