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Enthalpy

- In Chapter 2, we looked at enthalpies of reaction.

Remember that enthalpy can be positive or

negative. Examples

MgO(s) H2O(l) ? Mg(OH)2(s) DH -37.1 kJ/mol

C2H4 (g) ? C2H2 (g) H2 (g) DH 174.47 kJ/mol

Reactions with a negative enthalpy are more

common, but chemical reactions can happen on

their own (spontaneously) even if the enthalpy is

positive.

Entropy

There must be a second factor, in addition to the

enthalpy, that determines whether or not a

reaction can occur.

This second factor is the entropy. Entropy is a

much more subtle concept than you may have

suspected from General Chemistry. Before we

define entropy, well begin by noticing some

unusual results we get when we apply the

principles from Chapter 2 to a heat engine

containing one mole of gas.

Thermodynamics

P1, V1

1

P2, V2

4

2

P3, V3

P4, V4

3

A diagram like the one we used in the last

chapter will help a lot.

We have a four-step process

1. Isothermal expansion

2. Adiabatic expansion

3. Isothermal compression

4. Adiabatic compression

Thermodynamics

Because of all the work we did in the last

chapter, we already know what the energy, heat,

and work are for each of these steps.

For the energy

1. DU 0

When you add these together, you get zero. Why

does this make sense?

2. DU CV (Tc - Th)

3. DU 0

4. DU CV (Th - Tc)

Thermodynamics

For the heat

1. qrev

We can make a simplification here...

2. qrev 0

3. qrev

4. qrev 0

Adiabatic Expansion

Recall that

for an adiabatic process.

So for step 2

And for step 4

Adiabatic Expansion

Flipping the first equation and setting the two

equal to each other gives

Adiabatic Expansion

Thermodynamics

For the heat

Using ...

1. qrev

2. qrev 0

3. qrev

4. qrev 0

Thermodynamics

For the work

Again using ...

1. wrev

2. wrev CV (Tc - Th)

3. wrev

4. wrev CV (Th - Tc)

Thermodynamics

So to sum up our findings so far, for the entire

cycle

DU 0

The heat and work are equal but opposite in sign,

as they must be, since DU 0.

qrev

wrev

Thermodynamics

wrev

qrev

Since V2 gt V1, q is a positive number (i.e., the

system absorbs heat) and w is a negative number

(the system performs work).

Nicolas Sadi Carnot came up with all these ideas

in 1824. He was trying to develop a theory of

heat in order to design more efficient steam

engines.

And he DID come up with a nice, easy equation for

engine efficiency, based on the equation for work

above...

Thermodynamics

wrev

The efficiency is given by the actual work

performed (given by the equation above) divided

by the maximum possible work, which is the work

that would have been performed if the engine were

always at the high temperature

efficiency

We can also express this using heat instead of

temperature.

Note that the equation with heat features a plus

sign, because qc (unlike Tc) is a negative number.

Efficiency

Suppose we build a steam engine in which the two

reservoirs are at 1C and 150C. What is the

maximum efficiency of this engine?

0.35212 35.212

What would have to happen in order for an engine

to have 100 efficiency?

Thermodynamics

efficiency

If we play with this equation a little, well

arrive at a very important result.

This is actually a strange and surprising result.

Its also very important, so lets look at it

some more...

Thermodynamics

For our engine, DU 0 because energy is a state

function, so U has to return to its initial value

when we go through a cyclic process.

But q is a path function, so we dont expect it

to be zero for a cyclic process (and, as we just

saw, it isnt).

However, we can see that is a state

function. This was very surprising when it was

first discovered, because state functions are

usually fairly obvious (like volume or pressure),

but this one popped up unexpectedly.

Thermodynamics

So the sum of the for all the different

temperatures used in the cycle (there could be

more than two) is equal to zero.

This new state function was given the name

entropy (symbol S). If the steps in our cycle

are infinitesimally small (meaning its a

reversible process), we have

More Fun With Entropy

We now want to get a more intuitive feel for what

entropy is.

Suppose we have an isolated system (i.e., neither

particles nor energy can be transferred to or

from the surroundings).

The system consists of two chambers with

different temperatures. Particles cant be

exchanged between the two, but energy (as heat)

can.

More Fun With Entropy

From the First Law of Thermodynamics, we know

dUA dqA,rev dwA,rev

dUB dqB,rev dwB,rev

dUA dqA,rev

dUB dqB,rev

dUA TAdSA

dUB TBdSB

More Fun With Entropy

dUA TAdSA

dUB TBdSB

So, overall, dS

But wait! Since this is an isolated system, we

know that no energy can escape. Thus, dUA -dUB.

More Fun With Entropy

What does this prove? Plenty! Suppose the

temperature is higher on the left side. What can

we say about dUA and the term in parentheses?

dUA is negative. So is the term in parentheses.

Thus, dS is positive. The entropy increases!

More Fun With Entropy

Now suppose the temperature is higher on the

right side. What can we say about dUA and the

term in parentheses?

dUA is now positive. So is the term in

parentheses.

Thus, once again, dS is positive. The entropy

increases!

More Fun With Entropy

The only time dS isnt positive is when the

temperature of both chambers is equal. What

happens in this case?

dUA and the term in parentheses are both zero,

and so is dS.

In this situation, energy is exchanged equally

between both chambers thus, it is a reversible

(or equilibrium) process.

More Fun With Entropy

In the other situation (where the temperature is

higher on one side), energy flows only in one

direction, so it is an irreversible (or

spontaneous) process.

As you can tell, the entropy change is higher for

a spontaneous process than for a reversible

process.

Thus, .

The Second Law of Thermodynamics

We can also write DS 0

Either of these two equations is better known as

the Second Law of Thermodynamics.

What it means is that, when a spontaneous process

(like a chemical reaction) occurs, the entropy of

a system always increases.

Entropy

So we now have a good mathematical definition of

entropy but what property does actually

represent?

Entropy is the degree of disorder or randomness a

system has. The more disordered a system is, the

higher the entropy.

Entropy

- So what factors can influence the value of S ?

Think about what gases, liquids, and solids look

like

gas

liquid

solid

molecules have random locations

molecules have orderly locations

high entropy

low entropy

Entropy

- So the entropy depends on the phase of the

molecules (gas, liquid, or solid).

It also depends on the number of different

compounds present

all molecules the same less random

different kinds of molecule more random

low entropy

high entropy

Entropy

- We can use this to predict the entropy change of

a chemical reaction.

NaHCO3(s) HCl(aq) ? CO2(g) H2O(l) NaCl(aq)

The entropy goes up how can you tell?

DS 67.3 J/mol K

O2(g) 2H2(g) ? 2H2O(g)

What happens to the entropy, and why?

DS -278.3 J/mol K

Entropy

- The same logic also applies when considering

similar molecules with different numbers of

atoms. For example, consider the molecules

H3C-CH3(g) H2CCH2(g) HCCH(g)

S 229.2 219.3 200.9 J/mol K

Which has the smallest entropy?

In addition, atoms with a greater mass tend to

have higher entropy. Based on this, which of the

following has the smallest entropy?

FeCl3(s) UCl3(s) CrCl3(s)

S 142.3 159.0 123.0 J/mol K

Entropy

- The structure of a molecule can also give us a

clue about the entropy. For example, both of the

following are liquids with the formula C4H8O

S 246.6 J/mol K 204.3 J/mol K

Which has the smallest entropy?

Entropy

So there are several factors that can affect the

entropy

- The phase of the compound

- The number of atoms of different types

- The mass of the atoms

- The structure of the molecules

Entropy

Now were ready to do some calculations with

entropy! But to do that, we need to get DS in a

more useful form.

Recall that for the reversible, isothermal

expansion of the gas

qrev

So

DS

Entropy

Suppose we have 2.50 L of O2 and 1.00 L of H2,

both at 300K and 1 atm, and both acting like

ideal gases. The two gases are in separate

containers.

A valve between the containers is then opened,

allowing the gases to mix. What is the entropy

of this process?

Entropy

2.50 L of O2 and 1.00 L of H2, both at 300K and 1

atm

We need to know the number of moles of each gas.

0.10155 mol

0.04062 mol

Entropy

2.50 L of O2 and 1.00 L of H2, both at 300K and 1

atm

The total entropy is .

0.28408 J/K 0.42308 J/K

0.70716 J/K

Entropy

Lets work with entropy some more, to get a few

more useful equations! From the First Law, we

know dU dqrev dwrev

But we also know that qrev TdS, so dU TdS -

PdV

Entropy

dU TdS - PdV

- In Chapter 2, we saw that
- dH dU d(PV)

dU PdV VdP

TdS VdP

Now, since we know that H is related to heat, it

makes sense to express dH as a function of T

(which is proportional to heat) and P (which

appears in the second term above, showing that

pressure is a variable here).

Entropy

dH TdS VdP

From our discussion of partial differentials in

Chapter 2, we can write the total derivative of H

So TdS VdP

Entropy

TdS VdP

We want to get an expression with entropy by

itself on one side

TdS

dS

Using the definition of the constant-pressure

heat capacity

dS

Entropy

dS

Compare this to the total derivative of S

This tells us

and

We now have an equations for the way entropy

varies with temperature or pressure!

Entropy

This allows us to write a new equation for DS

This will be very useful shortly!

Entropy

So how can we find S (the absolute entropy of a

substance) instead of DS (the entropy change of a

process)?

- This solid has a low entropy, but not zero

entropy. There is still some disorder in the

system. Why?

The atoms still have some kinetic energy they

vibrate as though they were held together by

springs. How can we reduce the amount of kinetic

energy the atoms have?

The Third Law of Thermodynamics

- The kinetic energy will drop to zero if we

decrease the temperature to 0 Kelvin absolute

zero!

When this happens, the atoms stop moving, and we

have a perfect crystal. The crystal is perfectly

ordered, so the entropy is 0.

This is the Third Law of Thermodynamics the

entropy of a perfect crystal is 0 at absolute

zero.

Entropy

This allows us to back to the equation we

recently found for DS

Suppose we set the lower temperature limit (T1)

at 0 Kelvin. From the Third Law, we know that S1

must then be zero, so the entropy at temperature

T is

At last, we have an expression for the absolute

entropy of a substance at temperature T!

Entropy

- S is called the standard molar entropy. It

represents the entropy at 298.15 K (25C), which

is called standard temperature. Remember, the

entropy is different at other temperatures!

The units for S are . There are a couple of

things to notice about this.

What can we do with this? Wed like to know the

change in entropy for a chemical reaction. As we

did with the enthalpy, we can use

DS Sproducts Sreactants

Entropy

DS Sproducts Sreactants

MgO(s) H2O(l) ? Mg(OH)2(s)

DS 63.24 (26.8 69.91)

-33.47

Does this result make sense? Why?

Entropy

- Determine the entropy of the reaction

O2(g) 2H2(g) ? 2H2O(g)

DS Sproducts Sreactants

2 188.83 (205.0 2 130.58)

-88.5

Remember to account for the number of moles of

each compound!

The Second Law

- What will happen if we mix together solutions of

NaCl and AgNO3?

NaCl(aq) AgNO3(aq) ?

NaNO3(aq) AgCl(s)

DS Sproducts Sreactants

207 96.11 (115.5 221.93)

-34.32

But wait! We said that entropy tends to increase

during a chemical reaction, but here it

decreases. Does this reaction break the Second

Law of Thermodynamics?

The Second Law

NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)

DS -34.32

- Remember what the Second Law says

A spontaneous process always results in an

increase of the entropy of a closed system.

The key here is those last four words. Our

reaction is not a closed system! Why not?

The Second Law

NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)

DS -34.32

reaction

environment closed system

The system here includes both the reaction and

its environment (the beaker, the air in the lab,

etc.).

The entropy of the reaction decreases here this

is allowed only if the entropy of the environment

increases even more.

The Second Law

NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)

DS -34.32

What could make the entropy of the environment

increase?

reaction

environment

heat

Simple! The entropy of the environment increases

if it gets hotter. Remember, dS .

In other words, this reaction must give off heat

so that the environment gets warmer.

The Second Law

NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)

DS -34.32

How do we determine whether this reaction really

does give off heat?

DH Hproducts Hreactants

-446.2 -127.0 (-407.1 -100.2)

-65.9

So the reaction does give off heat.

Phase Transitions

- Theres one more subtlety we need to mention

about entropy. At 0 Kelvin, our substance is

guaranteed to be a solid. However, if T is large

enough, the substance will eventually become a

liquid or gas.

gas

liquid

solid

high entropy

low entropy

At the freezing and boiling points, the entropy

suddenly increases dramatically!

Entropy

Therefore, it makes sense to split this integral

into several pieces. Suppose our substance is a

gas at the final temperature, T

The entropy change during freezing

The entropy change during boiling

The entropy change from 0 Kelvin to the freezing

point, Tfus

The entropy change between the freezing and

boiling points

The entropy change from the boiling point, Tvap,

to T

Real Systems

- So far, weve looked at the entropies of isolated

systems. Unfortunately, things are rarely that

simple in a real experiment.

Usually, our experiments are not isolated.

However, with a little care, we can keep a couple

of properties constant. As long as we do that,

the equations arent too complex.

In most experiments, we either keep volume and

temperature or pressure and temperature constant.

Real Systems

- One thing were interested in is whether or not a

chemical reaction will take place spontaneously

under given conditions.

For our first non-isolated system, lets consider

the case where we hold the volume and temperature

constant.

As we often do, we start with the First Law dU

dq dw For this system, what do we know about

the variables in this equation?

Constant Volume Temperature

dU dq dw

dU dq

Because DV 0.

Were trying to determine if the reaction is

spontaneous, so it makes sense to consider the

entropy of this system. We know that, by

definition, entropy is

dS

Constant Volume Temperature

dU dq

dS

Solving the second equation for dq and

substituting into the equation for dU gives

us dU TdS Keep in mind that this equation

applies only to systems with constant V and T!

When does dU TdS? When is dU lt TdS?

Constant Volume Temperature

dU TdS

dU TdS 0

Since T is a constant, we can write this as d(U

TS) 0

What is this equation telling us?

If V and T are constant, a process (like a

chemical reaction) is spontaneous if d(U TS)

0.

Constant Volume Temperature

d(U TS) 0

This result is so important, the term in

parentheses gets its own variable, A, called the

Helmholtz free energy.

A U TS so dA 0

So the value of A decreases during a spontaneous

process. Once the process stops (for example,

when a chemical reaction reaches equilibrium), A

stops changing (dA 0).

Constant Volume Temperature

A U TS

Since we know that T is constant, we can

write DA DU TDS

And since we know that DA 0 DU TDS 0

Note that if DA 0, that doesnt mean that the

process is impossible just that it isnt

spontaneous! What will have to happen in order

to make the process occur?

Constant Volume Temperature

DU TDS 0

What happens as we raise the temperature?

The reaction is more spontaneous. How is this

reflected in the behavior of the system?

Constant Pressure Temperature

Its much more common to have reactions at a

constant pressure rather than a constant volume.

dU dq dw

dU dq PdV

Because dw -PdV.

dS

Constant Pressure Temperature

dS

dU dq PdV

Solving the second equation for dq and

substituting into the equation for dU gives

us dU TdS PdV Keep in mind that this

equation applies only to systems with constant P

and T!

This expression isnt as simple as the one for

constant V and T, but its even more useful.

Constant Pressure Temperature

dU TdS PdV

dU TdS PdV 0

d(U TS PV) 0

This result is so important, the term in

parentheses gets its own variable, G, called the

Gibbs free energy.

G U TS PV so dG 0

So the value of G decreases during a spontaneous

process. Once the process stops (for example,

when a chemical reaction reaches equilibrium), G

stops changing (dG 0).

Constant Pressure Temperature

G U TS PV

You may recall that one equation for energy

is DU DH PDV

Thus, U H PV. Substituting this for U in the

first equation gives

G H TS DG DH TDS

This result should be familiar from General Chem!

Constant Pressure Temperature

DG DH TDS

As we said earlier, DG 0, so that also means

that DH TDS 0

As usual, the two sides of the equation are only

equal for a reversible process.

This equation has some interesting consequences!

Constant Pressure Temperature

DH TDS 0

Consider the reaction NaCl(aq) AgNO3(aq) ?

NaNO3(aq) AgCl(s)

At standard temperature, DH -65.9 and DS

-34.32 .

DH TDS -65900 298.15(-34.32)

-65900 10232.5

-55667.5

So DG lt 0, which means this reaction is

spontaneous.

Constant Pressure Temperature

DH TDS 0

NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)

At 298 K, DH -65.9 , DS -34.32

, DG -55.67

What happens as we raise the temperature?

DS is negative. Since a negative entropy change

makes a reaction less spontaneous, the reaction

will become increasingly unlikely as we raise the

temperature.

Constant Pressure Temperature

NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)

At 298 K, DH -65.9 , DS -34.32

, DG -55.67

Lets calculate DH TDS at 485 K. What else do

we need to know?

CP NaCl 50.5, AgNO3 93.1, NaNO3 92.9, AgCl

50.8

What does this tell you?

If all the compounds in our reaction have the

same phase, we can ignore the effects of the heat

capacity!

Constant Pressure Temperature

NaCl(aq) AgNO3(aq) ? NaNO3(aq) AgCl(s)

At 298 K, DH -65.9 , DS -34.32

, DG -55.67

Lets calculate DH TDS at 485 K.

-65900 485 (-34.32) -49254.8

So the reaction is less spontaneous at this

temperature, as expected.

Constant Pressure Temperature

2 Cu(s) 2 HCl(g) ? 2 CuCl(s) H2(g)

At 298 K, DH -89.8 kJ, DS -137.1 J/K, DG

-48.9 kJ

Now lets calculate DG at 700 K. We do this the

same way we did it in the last example, right?

Wrong! This time, the compounds have different

phases. We must determine the new DH and DS at

this temperature. How do we do that?

Constant Pressure Temperature

2 Cu(s) 2 HCl(g) ? 2 CuCl(s) H2(g)

At 298 K, DH -89.8 kJ, DS -137.1 J/K, DG

-48.9 kJ

First, lets calculate DH. Whats the equation

we want?

-89800 7555

-82245 J

Constant Pressure Temperature

2 Cu(s) 2 HCl(g) ? 2 CuCl(s) H2(g)

At 298 K, DH -89.8 kJ, DS -137.1 J/K, DG

-48.9 kJ

Now lets calculate DS. Whats the equation we

want?

-137.1 16.1

-121.0 J/K

Constant Pressure Temperature

2 Cu(s) 2 HCl(g) ? 2 CuCl(s) H2(g)

At 298 K, DH -89.8 kJ, DS -137.1 J/K, DG

-48.9 kJ

Finally, lets calculate DH TDS at 700 K.

-82245 700 (-121) 2455 J

What does this tell us?

Measuring Entropy

So far, weve determined the entropy by either

looking at a very small system or using the heat

capacities over a wide range of temperatures.

Often, neither of these techniques is practical.

How can we determine the entropy of a process

when not all the heat capacities are known?

The Helmholtz free energy can help, even if were

not at constant V and T!

Measuring Entropy

Remember our equation for the Helmholtz free

energy

A U TS

dA dU TdS SdT

Neither of these equations looks very promising

if we want to know S, because both equations

contain terms that are hard to measure (U, A, and

their derivatives).

However, remember this equation we learned for dU

of a reversible process dU TdS PdV

Measuring Entropy

Combining the two equations gives

dA PdV SdT

Now, this looks more helpful! dV and dT are easy

to measure. The only problem is dA. How can we

get rid of it?

The work we did with partial derivatives will

help! Whats the total derivative of A? dA

Measuring Entropy

dA PdV SdT

dA

-P

-S

We now have an equation for S! Unfortunately, we

still have A in the equation. We can get rid of

it if we differentiate the first equation by T

and the second one by V.

Recall that the terms on the left are called

cross derivatives. What do we know about them?

Measuring Entropy

An equation like this one, which we get by

setting cross derivatives equal to each other, is

called a Maxwell Relation. This particular

equation is very useful!

dS

DS

This is very handy! Its the first easy-to-use

equation for entropy weve had that works even

for real gases!

Measuring Entropy

DS

Suppose we used this equation on an ideal gas.

What would the term in parentheses look like?

P

DS

This is exactly what we found earlier for ideal

gases!

Measuring Entropy

But wait! Why stop with the Helmholtz free

energy? We can get another helpful expression

for entropy if we start with the Gibbs free

energy

G U TS PV

dG dU TdS SdT PdV VdP

Once again, we substitute our equation for dU of

a reversible process dU TdS PdV

dG SdT VdP

Measuring Entropy

dG SdT VdP

Once again, this looks helpful! dT and dP are

easy to measure, but how can we get rid of dG?

Use the total derivative of G. dG

-S

V

Measuring Entropy

-S

V

Well now get rid of the G by differentiating the

first equation by P and the second one by T.

So...

This is another Maxwell Relation!

Measuring Entropy

dS

DS

Measuring Entropy

DS

Suppose we used this equation on an ideal gas.

What would the term in parentheses look like?

V

DS

Measuring Free Energy

There are a couple more useful equations to

notice. As we saw earlier, if we know the

enthalpy and entropy of a process, the Gibbs free

energy is easy to calculate. But what if we

dont?

We use an equation we got in passing earlier

V

DG

Notice that this only applies to an isothermal

process!

Measuring Free Energy

DG

For an ideal gas, this becomes

DG

Compare this to the equation we just got for

entropy

DS

Why does the similarity make sense?

Measuring Free Energy

We just derived a useful equation by starting

with the expression for the variation with Gibbs

energy with pressure. We can get another

important equation if we start with the

relationship between G and temperature.

-S

This relationship would apply to a static

(unchanging) system. For a system that does

undergo change (like a chemical reaction), wed

have

-DS

Measuring Free Energy

-DS

Lets think about this for a second. Another

expression we have that involves DG, DS, and T is

DG DH TDS

DS

If we combine this equation with the one at the

top of the slide, we get

Measuring Free Energy

We can get a very important result by taking the

derivative of the second term

Multiplying through by T gives

Measuring Free Energy

But at the top of the previous slide, we also had

So we can set the right side of the top equation

equal to the left side of the lower one

The Gibbs-Helmholtz Equation

This is an important result! This is the

Gibbs-Helmholtz Equation, and is very helpful

when we look at how temperature can affect the

spontaneity of a chemical reaction.

Activity

A very useful quantity we can measure for any

solution is its activity. The activity is simply

the ratio of the vapor pressure of a component in

a solution versus the vapor pressure of the pure

compound

For an ideal solution, the activity is equal to

the mole fraction of A. For a nonideal solution,

its not.

Activity

Why is the activity useful? If we take the ratio

of the activity and the mole fraction, we get a

number called the activity coefficient

If we have an ideal solution, gA will be equal to

1. Thus, the activity coefficient can be used to

give us an idea of how nonideal a solution is

the farther gA is from 1, the less ideal it is.

Activity

The benzene in a benzene/water solution has a

vapor pressure of 608 mmHg when the mole fraction

of benzene is a 0.191. At the same temperature,

pure benzene would have a vapor pressure of 225

mmHg. Calculate the activity and the activity

coefficient.

2.70

Unsurprisingly, the activity is not equal to the

mole fraction, as it would be for an ideal

solution.

Activity Coefficients

14.15

So this solution is very far from ideality! Can

you think of a reason why?

Water and benzene are not miscible! Weve been

working with an example in which the two

components are barely soluble in one another.

Miscible liquids yield activity coefficients much

closer to 1.

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You can use PowerShow.com to find and download example online PowerPoint ppt presentations on just about any topic you can imagine so you can learn how to improve your own slides and presentations for free. Or use it to find and download high-quality how-to PowerPoint ppt presentations with illustrated or animated slides that will teach you how to do something new, also for free. Or use it to upload your own PowerPoint slides so you can share them with your teachers, class, students, bosses, employees, customers, potential investors or the world. Or use it to create really cool photo slideshows - with 2D and 3D transitions, animation, and your choice of music - that you can share with your Facebook friends or Google+ circles. That's all free as well!

For a small fee you can get the industry's best online privacy or publicly promote your presentations and slide shows with top rankings. But aside from that it's free. We'll even convert your presentations and slide shows into the universal Flash format with all their original multimedia glory, including animation, 2D and 3D transition effects, embedded music or other audio, or even video embedded in slides. All for free. Most of the presentations and slideshows on PowerShow.com are free to view, many are even free to download. (You can choose whether to allow people to download your original PowerPoint presentations and photo slideshows for a fee or free or not at all.) Check out PowerShow.com today - for FREE. There is truly something for everyone!

You can use PowerShow.com to find and download example online PowerPoint ppt presentations on just about any topic you can imagine so you can learn how to improve your own slides and presentations for free. Or use it to find and download high-quality how-to PowerPoint ppt presentations with illustrated or animated slides that will teach you how to do something new, also for free. Or use it to upload your own PowerPoint slides so you can share them with your teachers, class, students, bosses, employees, customers, potential investors or the world. Or use it to create really cool photo slideshows - with 2D and 3D transitions, animation, and your choice of music - that you can share with your Facebook friends or Google+ circles. That's all free as well!

For a small fee you can get the industry's best online privacy or publicly promote your presentations and slide shows with top rankings. But aside from that it's free. We'll even convert your presentations and slide shows into the universal Flash format with all their original multimedia glory, including animation, 2D and 3D transition effects, embedded music or other audio, or even video embedded in slides. All for free. Most of the presentations and slideshows on PowerShow.com are free to view, many are even free to download. (You can choose whether to allow people to download your original PowerPoint presentations and photo slideshows for a fee or free or not at all.) Check out PowerShow.com today - for FREE. There is truly something for everyone!

presentations for free. Or use it to find and download high-quality how-to PowerPoint ppt presentations with illustrated or animated slides that will teach you how to do something new, also for free. Or use it to upload your own PowerPoint slides so you can share them with your teachers, class, students, bosses, employees, customers, potential investors or the world. Or use it to create really cool photo slideshows - with 2D and 3D transitions, animation, and your choice of music - that you can share with your Facebook friends or Google+ circles. That's all free as well!

For a small fee you can get the industry's best online privacy or publicly promote your presentations and slide shows with top rankings. But aside from that it's free. We'll even convert your presentations and slide shows into the universal Flash format with all their original multimedia glory, including animation, 2D and 3D transition effects, embedded music or other audio, or even video embedded in slides. All for free. Most of the presentations and slideshows on PowerShow.com are free to view, many are even free to download. (You can choose whether to allow people to download your original PowerPoint presentations and photo slideshows for a fee or free or not at all.) Check out PowerShow.com today - for FREE. There is truly something for everyone!

For a small fee you can get the industry's best online privacy or publicly promote your presentations and slide shows with top rankings. But aside from that it's free. We'll even convert your presentations and slide shows into the universal Flash format with all their original multimedia glory, including animation, 2D and 3D transition effects, embedded music or other audio, or even video embedded in slides. All for free. Most of the presentations and slideshows on PowerShow.com are free to view, many are even free to download. (You can choose whether to allow people to download your original PowerPoint presentations and photo slideshows for a fee or free or not at all.) Check out PowerShow.com today - for FREE. There is truly something for everyone!

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Copyright 2015 CrystalGraphics, Inc. — All rights Reserved. PowerShow.com is a trademark of CrystalGraphics, Inc.

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