Thermodynamics from Greek thermo dy’namis (heat and power)

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Thermodynamics from Greek thermo dy’namis (heat and power). Studies energy changes and the direction of flow of energy usually in a well-defined part of the universe (the system) Definitions: System: part of the universe in which we are interested

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### Thermodynamicsfrom Greek thermo dy’namis (heat and power)

Studies energy changes and the direction of flow of energy usually in a well-defined part of the universe (the system)

Definitions:

System: part of the universe in which we are interested

Surroundings: where we make our observations (the universe)

Boundary: separates above two

Heat and Work

Heat: transfer of energy that changes motions of atoms in the surroundings in a chaotic manner

Work: transfer of energy that changes motions of atoms in the surroundings in a uniform manner

= F x d

Energy
• Definition: the capacity to do WORK
• Units are Joules (J) = kg.m2/s2

(from KE=1/2mv2)

Work done on a system - system gains energy (w +ve)

Work done by the system - system loses energy (w -ve)

Heat absorbed by the system (endothermic) - system gains energy (q +ve)

Heat released by the system (exothermic) - system loses energy (q +ve)

SYSTEM TOTAL ENERGY (kinetic plus potential) is the

INTERNAL ENERGY (U sometimes E)

Usually measure CHANGE in internal energy ( U )

U=Ufinal – Uinitial

U is a STATE FUNCTION (independent of path)

1st LAW of Thermodynamics

Internal energy of an isolated system is constant (energy can neither be created nor destroyed)

U = q+w

Pressure-Volume work

Against constant external pressure

w = -F.dz but Pex=F/A therefore w= -Pex.dV

Free expansion

w = 0

Calorimetry

Can measure internal energy changes in a “bomb” calorimeter

U=q-P V, but in a constant volume “bomb”, V=0

Thus U=q

Heat Capacity

Amount of energy required to raise the temperature of a substance by 1C (extensive property)

For 1 mol of substance: molar heat capacity (intensive property)

For 1g of substance: specific heat capacity (intensive property)

If heat capacity is independent of Temperature over the range of interest

Most reactions we investigate occur under conditions of constant PRESSURE (not Volume)

Enthalpy

Heat of reaction at constant pressure!

Use a “coffee-cup” calorimeter

to measure it

Heat capacity

Excercise: When 50mL of 1M HCl is mixed with 50mL of 1M NaOH in a coffee-cup calorimeter, the temperature increases from 21oC to 27.5oC. What is the enthalpy change, if the density is 1g/mL and specific heat 4.18 J/g.K?

Problem: Heat Capacities & Temperature Changes

How much heat is required to raise the temperature of 10 g of water and 10g of lead from 0 to 50oC?

specific heat of H2O = 4.18 J/g-oC

specific heat of Pb = 0.128 J/g-oC

q = m×c×∆T

q(H2O) = 10g×4.18 J/g-oC×50oC

= 2090 J

q(Pb) = 10g×0.128 J/g-oC×50oC

= 64 J

Problem: Heats of Chemical Reaction

100 ml solutions of 1.00 M NaCl and 1.00 M AgNO3 at 22.4 oC are mixed in coffee cup calorimeter and the resulting temperature rises to 30.2 oC.

What is the heat per mole of product? Assume the solution density and specific heat are the same as pure water.

Write balanced chemical reaction:

Net ionic: Ag+(aq) + Cl-(aq) → AgCl(s)

Determine heat of reaction:

qrxn= -qcal = -m×c×∆T

m = 200 ml × 1.0g/ml = 200g

c = cH2O = 4.18 J/g-oC

= -200g × 4.18 J/g-oC × (30.2-22.4)

= -6,520 J

Determine heat per mole of product:

stoichiometric reactants, 0.1 mol in 100 ml

qrxn/mol = -6.52 kJ/0.1 mol

= -65.2 kJ/mol

See student activities

U and H (see chapter 10)

Only differ significantly when gases are involved

Standard Enthalpy Changes, Ho

H for a process in which the initial and final species are in their standard states.

Can be reported for any T. Use 298K unless otherwise indicated

Hvapo:1 mole pure liquid vapourises to a gas at 1bar

(+40.66 kJmol-1 at 373K for water) endothermic

Hfuso:1mole pure solid melts to a pure liquid at 1bar

(+6.01 kJmol-1 at 273K for ice) endothermic

Standard Reaction Enthalpy Changes

CaO(s) + CO2(g)  CaCO3(s) rxnHo = -178.3kJmol-1

Thermochemical equations: standard heats of reaction, rxnHo

Represent by an Enthalpy Diagram

Hess’s Law

If a reaction is the sum of two separate reactions then the enthalpy change during that reaction is also the sum of the enthalpy changes in the component reactions.

Hess’s Law

rxnHo = Hoproducts - Horeactants

Standard Heats of Formation

If one mole of the compound is formed under standard conditions from its elements in their standard state then the resulting enthalpy change is said to be the standard molar enthalpy (Heat) of formation, fHo where the subscript indicates this.

By definition the enthalpies of formation of the elements in their standard states are zero.

H2 (g) + 1/2O2 (g)  H2O (l) fHo = -285.8kJmol-1

2C (s) + 3H2(g) + 1/2O2(g)  C2H5OH (l) fHo = -277.7kJmol-1

Hess’s Law

Hess‘s Law is particularly useful for calculating fHo which would not be easy to measure experimentally. fHo for CO cannot be measured as CO2 is also formed when graphite is burned

C(s) + 1/2O2 CO fHo = x

CO + 1/2O2 CO2rxnHo = -283 kJmol-1

_______________________________________

C(s) + O2 CO2fHo = -393.5 kJmol-1

From looking at these equations it is fairly obvious that the sum of the first two enthalpies is equal to the third by Hess‘s Law.

i.e. x - 283 = -393.5 or x = -110.5 kJmol-1.

Enthalpy Changes and Bond Energies

Energy is absorbed when bonds break. The energy required to break the bonds is absorbed from the surroundings.

If there was some way to figure out how much energy a single bond absorbed when broken, the enthalpy of reaction could be estimated by subtracting the bond energies for bonds formed from the total bond energies for bonds broken.

O2(g) 2O(g) H°=490.4 kJ H2(g) 2H(g) H° =431.2 kJ

H2O(g)2H(g) + O(g) H°=915.6 kJ

We can estimate the bond enthalpies of O=O, H-H, and O-H as 490.4 kJ/mol, 431.2 kJ/mol, and 457.7 kJ/mol, respectively.

2H2(g) + O2(g)  2H2O(g) H°= ?

2H2(g) + O2(g)  2H2O(g)

moles of bonds formed

moles of bonds broken

Energy absorbed

Energy released

2 H-H @ 431.2 kJ each 862.4kJ 4 O-H @ 457.7 kJ each 1830.9kJ

1 O=O @ 490.4 kJ each 490.4kJ

_____________________________________________

1352.7kJ 1830.9kJ

H°= 1352.7 - 1830.9 kJ = -478.2 kJ.

(Remember that the minus sign means "energy released", so you add the bond energies for broken bonds and subtract energies for bonds formed to get the total energy.)

A calculation based on enthalpies of formation gave H° = -483.7 kJ

Bonds in a molecule influence each other, which means that bond energies aren't really additive. An O-H bond in a water molecule has a slightly different energy than an O-H bond in H2O2, because it's in a slightly different environment.

Reaction enthalpies calculated from bond energies are very rough approximations!

Foods and Fuels

Enthalpies (heats) of combustion: complete reaction of compounds with oxygen. Measure using a bomb calorimeter.

Most chemical reactions used for the production of heat are combustion reactions. The energy released when 1g of material is combusted is its Fuel Value. Since all heats of combustion are exothermic, fuel values are reported as positive.

Most of the energy our body needs comes from fats and carbohydrates.

Carbohydrates are broken down in the intestines to glucose. Glucose is transported in the blood to cells where it is oxidized to produce CO2, H2O and energy:

C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l) DH°rxn=-2816 kJ

The breakdown of fats also produces CO2 and H2O

Any excess energy in the body is stored as fats

About 100 kJ per kilogram of body weight per day is required to keep the body functioning at a minimum level

Fuels

Energy comes primarily from the combustion of fossil fuels

Coal represents 90% of the fossil fuels on earth. However, it typically contains sulfur, which when combusted can lead to environmental pollution (acid rain)

Solar energy: on a clear day the sun's energy which strikes the earth equals 1kJ per square meter per second.

Hydrogen: clean burning (produces only water) and high fuel value. Hydrogen can be made from coal as well as methane

C(coal) + H2O(g)  CO(g)+H2(g)

CH4(g) + H2O(g)  CO(g) + 3H2(g)

The greater the percentage of carbon and hydrogen in the fuel the higher the fuel value

Fuel cells
• Biofuel cell research in NUIG
• Biomednano website
• Combustion chemistry