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Chemistry 5

Chapter-7

Thermochemistry

Part-2

21 October 2002

Great Job on Exam!

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Three ways to increase energy of system?

Work = force x distance

Consider expansion of gas how much work is done?

Demonstration- Observations?

Heat & Work

adding material to it

heating it

doing work on it -- for example, winding up a spring or pushinga weight.-- it takes energy to do work

If pressure inside can exceeds external pressure, then

lid flys off! How much work is done???

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Pressure-Volume Work:

Work ( w ) = force (F) x distance (h)

==

w = - P ext x V

significance of negative sign?

significance of P ext ?

Units

WORK

(remember: P = F/A)

P x A x hP x V

When gas expands ( V > 0), work is done on surroundings; thenegative sign signifies energy leaves (as work) the system.

External pressure against which system expands or thatcompressing the system.

The pressure-volume work (L-atm) can be expressedin more familiar energy units of joule (J) using idealgas constants: 101.33 J/L-atm

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Internal Energy, U

Changing the internal energy The internal energy of a system can bechanged by heating or doing work.

cold water hot water

Energy, Heat & Work

Is the total kinetic + potential energy of the system

Includes energy associated with molecular motions:

A system only containsinternal energy; it does not contain energy in the form of heat or work.

Ucold Uhot

U = U hot U cold , change in internal energy

= U final - U initial

heat

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Change in internal energy is the sum of theenergy transferred as heat and thattransferred as work.

Heat and work are equivalent ways of changing the energy of a system.

Changing the Internal Energy

I N T E R N A L E N E R G Y

, U

initial state, U

final state, U f

heat, q

work, w U = U

f U

i= q + w

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First Law

isolated system

system not isolated

Some conventions:

energy entering system

energy leaving system

First Law of Thermodynamics

The internal energy of an isolated system isconstant: U isolated sys = 0

For a system that is not isolated: U = q + w

Energy entering the system has positive sign:

heat absorbed by system, q > 0work done on system, w > 0

Energy leaving the system hasnegative sign:

heat released by system, q < 0work done by system, w < 0

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Changing Internal Energy

U = q + w

C 8H 18(l) + (25/2) O 2(g)

8 CO 2(g) + 9H 2O(g) E lost asHEAT

and WORK

U initial

U finalE lost as

HEAT

E N E R G Y

Extract all the energy change as heatExtract energy as work and heat (a car)

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Changing Internal Energy

U = q + w

E N E R G Y Initial State

Final State

We can have different combinations of q and w, toreach same final state.

U depends only on initial and final states!

q < 0w = 0

q < 0w < 0

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What is a state function

Internal energy is a state function

Path Dependent Functions

heat & work

Example consider expansion of ideal gas onevs. two steps.

State Functions

A state function is a property with a value that depends only on the current state of system and not how it was prepared.

Key to thermochemistry is that the change in a state function is independent about how changeoccurred; that is, independent of pathway.

It is difficult to obtain an absolute value of U ;however, it is generally the change in internal energy, U , that is most important.

values depend on the path followed when a system undergoes change!

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Constant Volume

U = q + w

Constant Pressure

importance?

Enthalpy Change, H

Heats of Reaction Enthalpy

= q P ext V = q 0= q V

Many (most) chemical and biochemical reactionsare carried out under constant pressure (not constant volume); q V q P

At constant pressure: U = q P + w = q P - P V

q P =

U + P

V Since U, P & V are all state functions, q P is also one.We give this a special name enthalpy, H!

H = U + P V

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U = H P V; how important is work term?

Enthalpy vs. Internal Energy Changes

2CO(g) + O 2(g) 2CO 2(g)

q P = -566 kJ = H

Evaluate P-V work: P V = P(V f V i )=RT(n f n i )

U = H P V, constant pressure= -564 kJ

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Changes of State

vaporization

melting

sublimation

Standard States

Endothermic & Exothermic Processes

Enthalpy: Other Key Points

Compound (l) Compound (g), H vap > 0

Compound (s) Compound (l), DH melt > 0

Compound (s) Compound (g), DH sub > 0

Enthalpy changes are precisely defined with respect to state. We can define standard state enthaply, H o ,which corresponds to change for reactants and products in standard state.Temperature usually specified as 298.15 K.

Endothermic heat goesinto system during reaction.

Exothermic hear released by system during reaction.