Download Thermo chemistry Dealing with

Thermo chemistry Dealing with
ENERGY & HEAT
The Nature of Energy
Kinetic and Potential Energy
From Physics:
Force is a push or pull on an object. •
Work is the product of force applied to an object over a •
distance:
w=Fd
Energy is the work done to move an object against a force. •
Kinetic energy is the energy of motion: •
Ek  12 mv 2
Potential energy is the energy an object possesses by virtue of
its position.
Potential energy can be converted into kinetic energy. •
Example: a ball of clay dropping off a building.
Energy Units
SI Unit for energy is the joule, J:
E k  12 mv 2 
1
2
2 kg 1 m/s 2
 1 kg m 2 / s 2
 1J
We sometimes use the calorie instead of the joule:
1 cal = 4.184 J (exactly)
A nutritional Calorie:
1 Cal = 1000 cal = 1 kcal
•
Temperature & Energy & Heat
Temperature is a measure of hot and cold – an intensive property.
Thermometers in various scales are used to compare temperature.
Energy is the driving force for changes. A change often is associated
with a certain amount of energy and amounts of energy can be
measured according to the quantities changed. Like other quantities,
energy is an extensive property, unlike temperature.
Heat is energy in transfer or energy transferred and heat is often
used as another term for energy. In this aspect, it is the quantity of
energy transferred or involved.
The Nature of Energy
Systems and Surroundings
System: part of the universe we are interested in.
Surroundings: the rest of the universe.
A system for Thermochemistry
For the study of energy, there must be a reference for its amounts
transferred in or out. Such a reference is called a system.
The boundary isolates the system from its surrounding, so that the
amount of energy or heat transferred in or out can be identified.
Picture a system with boundary, surrounding, and heat transfer in your
mind
Surrounding
System
Environment
Measuring Heat (Energy)
Heat = Energy = Work
Heat is energy – the ability to do work. Units for work, heat or
energy is J
1 cal = 4.184 J (defined. The amount of
1J=1Nm
heat to raise 1 g water for 1 K)
Gaining heat increase T, as the kinetic energy of molecules increase.
Remember these from discussion of gases?
Kinetic energy = ½ m u 2
Heat capacity: the amount of heat required to raise T by 1 K
Heat capacity = C = ms
Specific heat (s): heat capacity of 1gram of substance (no unit)
compare to water (1.000 cal K-1 g-1)
Calorimetry
Heat Capacity and Specific Heat
Calorimetry = measurement of heat flow.
Calorimeter = apparatus that measures heat flow.
Heat capacity = the amount of energy required to raise the
temperature of an object by one degree.
Molar heat capacity = heat capacity of 1 mol of a substance.Specific
heat = specific heat capacity = heat capacity of 1 g of a substance.
Amount of heat = q = sm( T)
joul
q = (specific heat)  (mass of substance)  T.
Calorimeter Experiment
q = C * T
Heat gained
Temperature difference
Heat capacity
Heat capacity of water = 1 cal deg-1 g-1
molar heat capacity of water
= 18 * 4.184 J deg-1 mol-1
= 75.3 J deg-1 mol-1
Heat capacity of water = 1 cal deg-1 g-1
molar heat capacity of water
= 18 * 4.184 J deg-1 mol-1
= 75.3 J deg-1 mol-1
Bomb and Coffee-cup Calorimeter
q = C * T
Use a known amount of q and measure T to determine C.
By measuring T , calculate
(corection) C = q / T
Once C is known, and a known amount of reactants, heat of reaction,
Hreaction, can be evaluated
Bomb Calorimetry (Constant-Volume Calorimetry)
Reaction carried out under constant volume.
Use a bomb calorimeter.
Usually study combustion.
qrxn = -CcalorimeterT
First Law of Thermodynamics
Endothermic and Exothermic Processes
Endothermic: absorbs heat from the surroundings.
An endothermic reaction feels cold.
Exothermic: transfers heat to the surroundings.
An exothermic reaction feels hot.
Endothermic and Exothermic Reactions
Use an energy level diagram to help visualize endothermic and
exothermic reactions.
Explain the sign changes in the following:
H = 6.01 kJ at 298 K
H2O (s)  H2O (l)
H = – 6.01 kJ at 298 K
H2O (l)  H2O (s)
H = 44.0 kJ at 273 K H2O (l)  H2O (g)
H = – 44.0 kJ at 273 K H2O (g)  H2O (l)
H = 181 kJ 2 NO (g)  N2 (g) + O2 (g)
H = – 181 kJ N2 (g) + O2 (g)  2 NO (g)
Explain why the sign changes when the process is reversed
Enthalpy (H)
The quantity of heat transferred between the system and
surroundings carried out under constant pressure.
Can only measure the change in enthalpy:
DH = Hfinal - Hinitial = q / no. of moles
for example...
CH4(g) + 2 O2(g) ® CO2(g) + 2 H2O(g)
DH = - 890.3 kJ/mol
reactants ® products + heat
i.e. less Heat content in products than reactants
i.e. heat is released!!!
for example.. HgO(s) ® Hg(l) + ½ O2(g)
DH = + 90.83 kJ/mol
i.e. more Heat content in products than reactants
i.e. heat is absorbed!!.
Hess’s Law
Hess’s Law is based on the principle of conservation of energy.
The enthalpy change for the overall process is the sum of the
enthalpy changes for the individual steps.
The heat evolved or absorbed in a chemical process is the same
whether the process takes place in one or in several steps.
Hess’s Law Illustrated
A + 2B  AB2
H1
A + B  AB
H2
AB + B = AB2
__add___________________
H1 + H2
A + 2 B = AB2
= H1 2
First Law of Thermodynamics
Internal Energy
Internal Energy: total energy of a system. •
Cannot measure absolute internal energy. •
Change in internal •
energy, DE = Efinal - Einitial
Enthalpies of ReactionFor a reaction
DHrxn = H(products) - H (reactants)
Enthalpy is an extensive property (magnitude DH is directly
proportional to amount):
DH = -802 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(g) DH = -1604 kJ
When we reverse a reaction, we change the sign of DH:
DH = +802 kJ
CO2(g) + 2H2O(g)  CH4(g) + 2O2(g)
Change in enthalpy depends on state:
DH = -88 kJ H2O(g)  H2O(l)
Hess’s Law
Hess’s law: if a reaction is carried out in a number of steps, •
H for the overall reaction is the sum of H for each individual
step.
For example: •
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = -802 kJ
2H2O(g)  2H2O(l)
H = -88 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ
In the above enthalpy diagram note that
H1 = H2 + H3
Enthalpies of Formation
If 1 mol of compound is formed from its constituent elements, •
then the enthalpy change for the reaction is called the enthalpy
of formation, Hof .
Standard conditions (standard state): 1 atm and 25 oC (298 K). •
Standard enthalpy, Ho, is the enthalpy measured when •
everything is in its standard state.
Standard enthalpy of formation: 1 mol of compound is formed •
from substances in their standard states.
Foods and FuelFoods
Foods
Fuel value = energy released when 1 g of substance is burned. •
1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal. •
Energy in our bodies comes from carbohydrates and fats •
(mostly).
Intestines: carbohydrates converted into glucose: •
C6H12O6 + 6O2  6CO2 + 6H2O, DH = -2816 kJ
Fats break down as follows: •
2C57H110O6 + 163O2  114CO2 + 110H2O, DH = -75,520 kJ
Fats: contain more energy; are not water soluble, so are good •
for energy storage.
Fuels
U.S.: 1.0 x 106 kJ of fuel per day.
Most from petroleum and natural gas.
Remainder from coal, nuclear, and hydroelectric.
Fossil fuels are not renewable
Foods and Fuels
Fuel value = energy released when 1 g of substance is
burned.Hydrogen has great potential as a fuel with a fuel value of
142 kJ/g.
Heat of solution
The heat generated or absorbed when a certain amount of
solute dissolved in a certain amount of solvent
Hsolution = Hsolution - Hcomponent
EXAMPLE:
Dissolution of NaCl
•
Na+Clsolid
Na+ + Cl- •
1- Lattice energy
Gaseous state
U = 788 KJ/ mol
2- Heat of hydration
Hhyd = - 784 KJ/mol
HEAT OF SOLUTION
 Hsoln = 4 KJ/ mol
Na+(aq) + Cl-(aq)
Hydrated ions
Hsoln = U + Hhyd = 788 + (- 784) = 4 KJ/ mol
Copyright 1999, PRENTICE HALL
Chapter 5
34
Heat of Dilution
Heat change associated with dilution process •
EXAMPLE:
Dilution of H2SO4, by adding the acid to water slowly with stirring
as it is Exothermic process