Met Thermo 2

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Met. Thermo for PTPGFirst Law of Thermodynamics

Energy exists in many forms, such as heat, light, chemical energy, and electrical energy. Energy is

the ability to bring about change or to do work. Thermodynamics is the study of energy.

First Law of Thermodynamics: Energy can be changed from one form to another, but it cannot

be created or destroyed. The total amount of energy and matter in the Universe remains constant,merely changing from one form to another. The First Law of Thermodynamics (Conservation)

states that energy is always conserved, it cannot be created or destroyed. In essence, energy can

be converted from one form into another.

According to Einsteins famous formula, E = mc2, mass is also a condensed form of energy.

From the Second Law of Thermodynamics we know that "in all energy exchanges, if no energy

enters or leaves the system, the potential energy of the state will always be less than that of the

initial state." This is also commonly referred to as entropy. A spring-driven watch will run until

the potential energy in the spring is converted, and not again until energy is reapplied to the spring

to rewind it. In the process of energy transfer, some energy will dissipate as heat. Entropy is a

measure of disorder: The flow of energy maintains order and life. Entropy wins when organismscease to take in energy and die.

Potential vs. Kinetic energy

Potential energy, as the name implies, is energy that has not yet been used, thus the term potential.

Kinetic energy is energy in use (or motion). A tank of gasoline has a certain potential energy that

is converted into kinetic energy by the engine. Batteries, when new or recharged, have a certain

potential. When placed into a tape recorder and played , the potential in the batteries is

transformed into kinetic energy to drive the speakers. When the potential energy is all used up, the

batteries are dead. In the case of rechargeable batteries, their potential is reelevated or restored.

Chemicals may also be considered from a potential energy or kinetic energy standpoint. Onepound of sugar has a certain potential energy. If that pound of sugar is burned the energy is

released all at once. The energy released is kinetic energy (heat). So much is released that

organisms would burn up if all the energy was released at once.

Heat and Work

Heat and work are both forms of energy. They are also related forms, in that one can be

transformed into the other. Heat energy (such as steam engines) can be used to do work (such as

pushing a train down the track). Work can be transformed into heat, such as might be experienced

by rubbing your hands together to warm them up.

Work and heat can both be described using the same unit of measure. Sometimes the calorie is

the unit of measure, and refers to the amount of heat required to raise one (1) gram of water one(1) degree Celsius. Heat energy is measured in kilocalories, or 1000 calories. Typically, we use

the SI units of Joules (J) and kilojoules (kJ). One calorie of heat is equivalent to 4.187 J. You will

also encounter the term specific heat, the heat required to raise one (1) gram of a material one (1)

degree Celsius. Specific heat, given by the symbol "C", is generally defined as:

Where:

C = specific heat in calories/gram-degrees Celsius , q = heat added in calories, M = mass in

grams , T = rise in temperature of the material in degrees Celsius.The value of C for water is 1.00 calories/gram-degrees Celsius.
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The values for specific heat that are reported in the literature are usually listed at a specific

pressure and/or volume, and you need to pay attention to these settings when using values from

textbooks in problems or computer models.

Example Problem: If a 2.34 g substance at 22 degrees celsius with a specific heat of 3.88 cal/gC

is heated with 124 cal of energy, what is the new temperature of the substance?

Two other common heat variables are the Heat of fusion and the Heat of vaporization. Heat of

fusion is the heat required to melt a substance at is normal melting temperature, while the heat of

vaporization is the heat required to evaporate the substance at its normal boiling point.

Chemical work is primarily related to that of expansion. In physics, work is defined as:w = distance X (opposing force)

Where: w = work, in joules (N*m) (or calories, but we are using primarily SI units)

distance is in meters ; opposing force is in newtons (kg*m/s2)

In chemical reactions, work (w) is generally defined as : w = distance X (area X pressure)

The value of distance x area is actually the volume. If we imagine a reaction taking place in a

container of some volume, we measure work by pressure times the change in volume.

w = dV x P

Where: dV is the change in volume, in litres ; If dV=0, then no work is done.

Enthalpy

Enthalpy is an interesting concept: it is defined by its change rather than a single entity. A state

property, the word enthalpy comes from the Greek "heat inside". If you have a chemica l system

that undergoes some kind of change but has a fixed volume, the heat output is equal to the change

in internal energy (q = dE). We will define the enthalpy change, dH or H, of a system asbeing equal to its heat output at constant pressure:

H = q at constant pressure

Where: H = change in enthalpy ; We define Enthalpy itself as: H = E + PV ..[1]Where: H = enthalpy ; E = energy of the system ; PV = pressure in atm. x volume in litres

In practice, we are only interested in the change in enthalpy, or. H

. H = Hfinal - Hinitial or

. H = H(products) - H(reactants)

Tables of enthalpies are generally given as . H values.

Example Problem: Calculate the dH value of the reaction:

HCl + NH3 --> NH4Cl

(dH values for HCl is -92.30; NH3 is -80.29; NH4Cl is -314.4)


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We can also represent enthalpy change with the equation:

dH = dE + P DV .[2]

Where: dV is the change in volume, in liters ; P is the constant pressure

If you recall, work is defined as P.dV, so enthalpy changes are simply a reflection of the amount

of energy change (energy going in or out, endothermic orexothermic), and the amount of work

being done by the reaction. For example, if dE = -100 kJ in a certain combustion reaction, but 10

kJ of work needs to be done to make room for the products, the change in enthalpy is:

dH = -100 kJ + 10 kJ = - 90 kJ

This is an exothermic reaction (which is expected with combustion), and 90 kJ of energy isreleased to the environment. Basically, you get warmer. Notice the convention used here -- a

negative value represents energy coming out of the system.

SECOND LAW

The second law of thermodynamics is an expression of the universal principle of increasing

entropy, stating that the entropy ofan isolated system which is not in equilibrium will tend to

increase over time, approaching a maximum value at equilibrium, and that the entropy change dS

of a system undergoing any infinitesimal reversible process is given by q / T, where q is the

heat supplied to the system and T is the absolute temperature of the system

Second Law of Thermodynamics - Increased Entropy- The Second Law of Thermodynamics is

commonly known as the Law of Increased Entropy. While quantity remains the same (First Law),

the quality of matter/energy deteriorates gradually over time. How so? Usable energy is inevitably

used for productivity, growth and repair. In the process, usable energy is converted into unusable

energy. Thus, usable energy is irretrievably lost in the form of unusable energy.

"Entropy" is defined as a measure of unusable energy within a closed or isolated system (the

universe for example). As usable energy decreases and unusable energy increases, "entropy"

increases. Entropy is also a gauge of randomness or chaos within a closed system. As usable

energy is irretrievably lost, disorganization, randomness and chaos increase.

The first law allows us to convert heat into work, or work into heat. It also allows us to change the

internal energy of a system by transferring either heat or work between the system and its

surroundings. But it doesn't tell us whether one of these changes is more easy to achieve than

another.

Many chemical and physical processes are reversible and yet tend to proceed in a direction in

which they are said to be spontaneous. This raises an obvious question: What makes a reaction

spontaneous? What drives the reaction in one direction and not the other?

Many spontaneous reactions are exothermic that it is tempting to assume that one of the drivingforces that determines whether a reaction is spontaneous is a tendency to give off energy. The

following are all examples of spontaneous chemical reactions that are exothermic.
http://www.shodor.org/UNChem/glossary.html#endothermichttp://www.shodor.org/UNChem/glossary.html#exothermichttp://www.shodor.org/UNChem/glossary.html#combustionhttp://www.shodor.org/UNChem/glossary.html#endothermichttp://www.shodor.org/UNChem/glossary.html#exothermichttp://www.shodor.org/UNChem/glossary.html#combustion


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2 H2(g) + O2(g) 2 H2O(g) Ho = -241.82 kJ/mol H2O

P4(s) + 5 O2(g) P4O10(s) Ho = -2984 kJ/mol P4O10

There are also spontaneous reactions, however, that absorb energy from their surroundings. At

100oC, water boils spontaneously even though the reaction is endothermic.

H2O(l) H2O(g) Ho = 40.88 kJ/mol

Ammonium nitrate dissolves spontaneously in water, even though energy is absorbed when thisreaction takes place.

H2O

NH4NO3(s) NH4+(aq) + NO3

-(aq) Ho = 28.05 kJ/mol

Thus, the tendency of a spontaneous reaction to give off energy can't be the only driving force

behind a chemical reaction. There must be another factor that helps determine whether a reaction

is spontaneous. This factor, known as entropy, is a measure of the disorder of the system

Entropy and the Second Law of Thermodynamics

The second law of thermodynamics describes the relationship between entropy and the

spontaneity of natural processes.

Second Law: In an isolated system, natural processes are spontaneous when they lead to an

increase in disorder, or entropy.

Spontaneous change is that which, once initiated, proceeds on its own until some state

of equilibrium (mechanical, thermal, chemical, etc.) is attained

This statement is restricted to isolated systems to avoid having to worry about whether

the reaction is exothermic or endothermic. By definition, neither heat nor work can be transferred

between an isolated system and its surroundings.

We can apply the second law of thermodynamics to chemical reactions by noting that the entropyof a system is a state function that is directly proportional to the disorder of the system.

Ssys > 0 implies that the system becomes more disordered during the reaction.

Ssys < 0 implies that the system becomes less disordered during the reaction.

For an isolated system, any process that leads to an increase in the disorder of the system will

be spontaneous. The following generalizations can help us decide when a chemical reaction leads

to an increase in the disorder of the system.

Solids have a much more regular structure than liquids. Liquids are therefore more disordered

than solids.The particles in a gas are in a state of constant, random motion. Gases are therefore

more disordered than the corresponding liquids.

Standard-State Entropies of Reaction

Because entropy is a state function, the change in the entropy of the system that accompanies any

process can be calculated by subtracting the initial value of the entropy of the system from the

final value.

S = Sf - Si

S for a chemical reaction is therefore equal to the difference between the sum of the entropies of

the reactants and the products of the reaction.

S = S(products) - S(reactants)

When this difference is measured under standard-state conditions, the result is the standard-state

entropy of reaction, So.


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So = So(products) - So(reactants)By convention, the standard state for thermodynamic measurements is characterized by the

following conditions.

o All solutions have concentrations of 1M.

o All gases have partial pressures of 0.1 MPa (0.9869 atm)

o Temperature is 298 k (25 0C

.Any process that increases the number of particles in the system increases the amount of

disorder

The sign of S for a reaction can also determine the direction of the reaction.In an isolated system, chemical reactions occur in the direction that leads to an increase in the

disorder of the system.

2 NO2(g) = N2O4(g)

Using a standard-state entropy data table, we find the following information:

Compound S(J/mol-K)

NO2(g) 240.06

N2O4(g) 304.29

In this equation, 1 mole of N2O4 is formed for every 2 moles of NO2 consumed, and the value of

So is calculated as follows: So = So(products) - So(reactants)

= [1 mol N2O4(g) x 304.29 J/mol-K] - [2 mol NO2(l) x 240.06 J/mol-K]

= -175.83 J/K

The sign of So is negative because two molecules combine in this reaction to form a larger, moreordered product.

Second Law of Thermodynamics can also be stated as: It is impossible to extract an amount

of heat QH from a hot reservoir and use it all to do work W . Some amount of heat QC must

be exhausted to a cold reservoir. This means that a perfect heat engine is impossible

1. Heat will not flow spontaneously from a cold object to a hot object.

2. Any system which is free of external influences becomes more disordered with time. This

disorder can be expressed in terms of the quantity called entropy.

3. You cannot create a heat engine which extracts heat and converts it all to useful work.

4. Entropy: a state variable whose change is defined for a reversible process at T where Q is

the heat absorbed.

Entropy: a measure of the amount of energy which is unavailable to do work. Entropy: a measure of the disorder of a system.

From Heat engine concept of a reversible process, S = (heat absorbed at temperature T, Q) /T

Since entropy gives information about the evolution of an isolated system with time, it is said to

give us the direction of "time's arrow". If snapshots of a system at two different times show one

state which is more disordered, then it could be implied that this state came later in time. For anisolated system, the natural course of events takes the system to a more disordered (higher

entropy) state.


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---------------------The Gibbs free energy (also known as the Gibbs function) is defined as

G=HT S. [1]

in which Srefers to the entropy of thesystem. SinceH, Tand Sare all state functions, so is G.

Thus for any change in state, we can write the extremely important relation

G= HT S .[2]

In order to make use of free energies to predict chemical changes, we need to know the free

energies of the individual components of the reaction. For this purpose we can combine the

standard enthalpy of formation and the standard entropy of a substance to get itsstandard free

energy of formation

G=H TS .[3]

and then determine the standard Gibbs free energy of the reaction according to

G = Gf(products) Gf(reactants)

As with standard heats of formation, the standard free energy of a substance represents the free

energy change associated with the formation of the substance from the elements in their most

stable forms as they exist under the standard conditions of 1 atm pressure and 298K.tandard Gibbs

free energies of formation are normally found directly from tables. Once the values for all the

reactants and products are known, the standard Gibbs free energy change for the reaction is found

in the normal way.

The interpretation of G for a chemical change is very simple. For a reaction A B, one of the

following three situations will always apply:

Entropy

Entropy is a measure of the disorder of a system. Take your room as an example. Left to itself,

your room will increase in entropy (i.e., get messier) if no work (cleaning up) is done to contain

the disorder. Work must be done to keep the entropy of the system low. Entropy comes from the

second law of thermodynamics, which states that all systems tend to reach a state of equilibrium.

The significance of entropy is that when a spontaneous change occurs in a system, it will always

be found that if the total entropy change for everything involved is calculated, a positive value

will be obtained. Simply, all spontaneous changes in an isolated chemical system occur with an

increase in entro py. Entropy, like temperature, pressure, and enthalpy, is also a state property and

is represented in the literature by the symbol "S". Like enthalpy, you can calculate the change of S(dS or delta S).

dS = Sfinal - S initial or dS = S(products) - S(reactants) .[3]

dS (or S) is change in entropy ; Sfinal and Sinitial are the final and initial entropies.The following table shows the relationship between the state of a substance and its entropy:

State of substance Relative Entropy (S)

gas highest S

aqueous high S

liquid medium Ssolid lowest S


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Free Energy

The Free energy of a system, represented by the letter "G", is defined as the energy of a system

that is free to do work at constant temperature and pressure. Mathematically, it is

defined as:G = H-TS .[4]

The change in G ( G) can be calculated in the same way as the change H and S G = G(products) - G(reactants) ..[5]

A table relating all of the state properties summarized above is shown below.A spontaneous

reaction is one that occurs without any outside intervention. Processes that are spontaneous in one

direction are non-spontaneous in the reverse direction.

Enthalpy Change Entropy Change Spontaneous Reaction?

Exothermic (dH < 0) Increase (dS > 0) Yes, dG < 0

Exothermic (dH < 0) Decrease (dS < 0) Only at low temps, if T dS < dH

Endothermic (dH > 0) Increase (dS > 0) Only at high temps, if T dS > dH

Endothermic (dH > 0) Decrease (dS < 0) No, dG > 0

Entropy Practice Problem: Given the following entropy values (Al2O3(s) is 51.00; Al(s) is

28.32; H2O(g) is 188.7; H2(g) is 130.6), determine dS for the reaction:

Al2O3(s) + 3H2(g) --> 2Al(s) + 3H2O(g)

THERMODYNAMIC EQUILIBRIUM

Definition of Chemical Equilibrium

Chemical equilibrium applies to reactions that can occur in both directions. For a reaction such

as:

CH4(g) + H2O(g)CO(g) + 3H2(g) .. [1]2 Cu (s) + O2

(g)Cu2O (s) . [2]In general, aA + bB cC + dD [3],where A,B are reactants, C,D are products and a,b,c,d, are the coefficients in the balanced

chemical equation.


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Chemical equilibrium means that, at a given temperature, the rate of formation of the products in

the forward reaction is equal to that of the backward reaction of producing the reactants.

The reaction can happen both ways. So after some of the products are created, the products begin

to react to form the reactants. At the beginning of the reaction, the rate that the reactants are

changing into the products is higher than the rate that the products are changing into the reactants.

Therefore, the net change is a higher number of products.

Even though the reactants are constantly forming products and vice-versa, the amount ofreactants and products does become steady. When the net change of the products and reactants is

zero the reaction has reached equilibrium. The equilibrium is a dynamic equilibrium. The

definition for a dynamic equilibrium is when the amount of products and reactants are constant.

(They are not equal but constant

The concept ofchemical equilibrium is extremely important in metallurgical thermodynamics,

because we can understand , from a given situation, whether a desired change is possible or not.

At a given temperature, there is a constant, called Equilibrium constant (K), which

expresses the chemical equilibrium in a quantitative manner.

(Activity or concentration of products)

----------------------------------------------

(Activity or concentration of Reactants)

If there is gas phase in the equilibrium reaction, then its concentration is replaced by partial

pressure.

For the above reaction 1,

pCO .p3

H2O

K1 = --------------

PCH4 . pH2O

aCuFor oxidation of copper, K2 = = (1/pO2)

pO2

By convention, activities of pure elements and pure compounds are taken as unity. In general,

for eq. [3],

Le Chatelier's Principle

Le Chatelier's principle states that when a system in chemical equilibrium is disturbed by a

change of temperature, pressure, or a concentration, the system shifts in equilibrium composition

in a way that tends to counteract this change of variable. You can affect the outcome of the

equilibrium in a given three ways in a reaction vessel according to Le Chatelier's principle:

Changing concentrations by adding or removing products or reactants Changing partial pressure of gaseous reactants and products. Changing the temperature.

These actions change each equilibrium differently, therefore you must determine what needs tohappen for the reaction to get back in equilibrium.

a) Example involving change of concentration: In the equation : 2NO(g) + O2(g) 2NO2(g) ..(1)


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If you add more NO(g) the equilibrium shifts to the right producing more NO2(g) If you add more O2(g) the equilibrium shifts to the right producing more NO2(g) If you add more NO2(g) the equilibrium shifts to the left producing more NO(g) and O2(g)

b) Example involving pressure change: In the equation : 2SO2(g) + O2(g) 2SO3(g), ..[2]

an increase in pressure will cause the reaction to shift in the direction that reduces pressure, that is

the side with the fewer number of gas molecules. Therefore an increase in pressure will cause ashift to the right, producing more product. (A decrease in volume is one way of increasing

pressure.)

c) Example involving temperature change: In the equation :

N2(g) + 3H2(g) 2NH3 + 91.8 kJ, .[3]

An increase in temperature will cause ashift to the leftbecause the reverse reaction uses the

excess heat. An increase in forward reaction would produce even more heat since the forward

reaction is exothermic. Therefore the shift caused by a change in temperature depends upon

whether the reaction is exothermic or endothermic.

The magnitude of the equilibrium constant gives one a general idea of whether the

equilibrium favours products or reactants.

If the products are favoured over the reactants, then the numerator term will bemuch larger than the denominator, and the equilibrium constant will be much

greater than 1.

If the reactants are favoured, then the denominator term of the mass actionexpression will be larger than the numerator term, and the equilibrium constant will

be less than one

Quantitative Problems

1) By examining the amounts of reactants and products at equilibrium, one may numericallysee how an equilibrium favours either reactants or products. Suppose we are given the following

equilibrium at 500 K:

CO(g) + 2 H2 (g) CH3OH(g)

The equilibrium concentrations are: [CO] = 0.0911 M, [H2] = 0.0822 M, [CH3OH] = 0.00892 M,

what is the value of the equilibrium constant? Does the equilibrium favour reactants or

products?

First, we need to write the mass action expression:

Next, substitute the equilibrium concentrations into the mass action expression, and calculate for

the equilibrium constant:

Since the value of the equilibrium constant is greater than one, (Keq >1), the equilibrium favours

the products.

Example 1 Calculate the amount of heat required to just melt 1 kg pure lead at its melting point

of 600 K by heating from 288 K.Data : Latent heat of fusion) = 1.15 kcal / mole;

CP for solid lead . = 5.63 +2.32 x T cal/ mole


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Atomic weight of lead = 207

[Hint : First calculate the enthalpy change, as Kcal. per mole, for heating solid lead through the

given temp. range. Then add the latent heat, obtaining the total H. Now convert 1 kg lead tomole of lead, which should be multiplied with total H to obtain the heat required, in kcal.]

Example 2 : Calculate the enthalpy change for the reaction :

Al2O3 (s) + 3 SO3 (g) = Al2 (SO4)3

Given : Heat of formation of Al2 (SO4)3 = -822.8 kcal/mole

Al2O3 (s) = - 400 kcal/mole

SO3 (g) = -95 kcal/mole

From the result, explain whether the above reaction os possible thermodynamically.

Example 3 . Derive the general equation for H for the following reaction:

CO (g) + H2O (g) = CO2 (g) + H2 (g)

Given : CP values : CO = 6.3 +1.8 x 10-3 T cal/mole

H2O (g) = 7.2 +2.3 x 10-3 T

CO2 (g) = -6.4 +10.1 x 10-3 T

H2 (g) = 7.0 -0.2 x 10-3 T

H0 at 298 K =-9840 cal

[Hint : CP = CP (products) - CP (reactants)

Then calculate the value of H from the above by integration,

HT =

H

0

+

CP. dT]

Example 4 : Given the relations 2 Cu + O2 = Cu2O (s) ; G0 = -40503 +30 T

and 4 Cu + O2 = 2 Cu2O (s) ; G0 = -81000 +60 T

predict whether both the reactions can take place at 870C.

Example 5 : Determine the temperature above which decomposition of Cu2O can take place at

i) standard pressure (initiallypO2= 1) and ii) at 10-3Hg.

Cu2O (s) 2 Cu(s) + O2 (g); G0 = 40503 - 30 T ..(1)

[Hint : G0 = - RT ln K = - 4.575 T.log K = 4.575 T.[log pO2]1/2 from eq. (1).(2)

Answer for (ii) For low pressure, 760 mm Hg =1 atmos., hence pO2 of 10-3 Hg = how much

atmos.?

pO2 = 0.2 x atmosphere. Put this value of pO2, in atmosphere, in equation (2), right hand side.

Equate this relation with eqn. (1), with T unknown on both sides. Solve for T.

Example 6 : Given MnO(s) = Mn + O2, G0 = 91950 -17.47 T and

C (s)+ O2 (g) = CO (g), G0 = -26700 -21 T ,

Determine the minimum temperature of reduction of MnO(s) by solid carbon, i) at standardpressure and ii) at 10-3Hg.

Met Thermo 2

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