Surroundings: Every Thing Else Other Than the System

Chapter 18

Thermodynamics

System: reaction under study.

Surroundings: Every thing else other than the system.

State function: Property that depends only on the current state and not on the path.

Isothermal: Any change at constant temperature.

Enthalpy: Heat of the reaction at constant pressure.

The 1st Law of Thermodynamics: Energy can be converted from one form to another, but it cannot be created or destroyed.

q = heatw = work

E = q + w = heat + work

Everything that is not part of the system (sys) is the surroundings (surr), and vice versa:

Euniv = Esys + Esurr

The total energy of the universe is constant and, therefore, energy cannot be created or destroyed.

Thermodynamics: is the science that studies energy changes that accompany chemical and physical processes.

One of the main objectives in studying thermodynamics is to predict whether or not a reaction will occur when reactants are brought together under specific conditions (for example, at a certain temperature, pressure, and concentration).

A spontaneous reaction: A reaction that does occur under the given set of conditions.

Non-spontaneous reaction: If a reaction does not occur under specified conditions.

We observe spontaneous physical and chemical processes every day, including many of these examples:

A waterfall runs downhill
Water freezes below 0oC and ice melts above 0oC (at 1 atm)
Heat flows from a hotter object to a colder one.
A gas expands in an evacuated bulb
Iron exposed to water and oxygen forms rust.

These examples show that processes that occur spontaneously in one direction cannot, under the same conditions, also take place spontaneously in the opposite direction.

Similarly, a large number of exothermic reactions are spontaneous.

A combustion of methane:

CH4(g) + 2O2(g)  CO2(g) + H2O(l) Ho = -890.4 KJ

Acid-base neutralization reaction

OH-(aq) + H+(aq)  H2O(l)Ho = -56.2 KJ

It is possible for an endothermic reaction to be spontaneous.

Consider a solid-to-liquid phase transition such as

H2O(s)  H2O(l)Ho = 6.01 KJ

We can conclude that, sign of Ho cannot predict spontaneous change

In order to predict spontaneity of a process, we need to know two things about the system:

Enthalpy, H.
Entropy, S,

Entropy: is a measure of the randomness ordisorder of a system.

The greater the disorder of a system, the greater its entropy.

For any substance, the particles in the solid state are more ordered than those in the liquid state, which in turn are more ordered than those in the gaseous state.

Ssolid < Sliquid < Sgas

A system with relatively few equivalent ways to arrange its components, such as a crystalline solid, has relatively small disorder and low entropy.
A system with many equivalent ways to arrange its components, such as a gas, has relatively large disorder and high entropy.

If the change from initial to final results in an increase in randomness,Sf > Si

S = Sf - SiS > 0

Processes that lead to an increase in

entropy of the system (S > 0).

melting: Ssolid < Sliquid
vaporization: Sliquid < Sgas
dissolving.
heating.

Example 18.1

The Second Law of Thermodynamics: the entropy of the universe (system + surroundings) increases in a spontaneous process and remains unchanged in an equilibrium process.

For a spontaneous process:Suniv = Ssys + Ssurr> 0

For an equilibrium process:Suniv = Ssys + Ssurr= 0

If Sunivis negative, this means that the process is not spontaneous in the direction described. Rather, it is spontaneous in the opposite direction.

Entropy Changes in the System (Ssys):

Suppose the system is represented by the following reaction:

aA + bB  cC + dD

The standard entropy of reactionSorxnis given by:

Sorxn= [cSo(C) + dSo(D)] - [aSo(A) + bSo(B)]

Sorxn= Σ nSo (products) - Σ mSo (reactants)

Examples 18.2, 18.3.

When gases are produced (or consumed)

If a reaction produces more gas molecules than it consumes, So > 0.
If the total number of gas molecules diminishes, So < 0.
If there is no net change in the total number of gas molecules, then So may be positive or negative BUT So will be a small number.

Entropy Changes in the Surrounding (Ssurr):

An exothermic process transfers heat from the system to the surroundings and results in an increase in the entropy of the surroundings. S > 0
An endothermic process absorbs heat from the surroundings and thereby decreases the entropy of the surroundings. S < 0

For constant-pressure processes the heat change is equal to the enthalpy change of the system, Hsys. There for, the change in entropy of the surroundings, Ssurr, is proportional to Hsys.

Ssurr α Hsys

The minus sign is used because if the process is exothermic, Hsys is negative and Ssurr is a positive quantity, indicating an increase in entropy.

If the temperature of the surroundings is high, the molecules are already quite energetic. Therefor, the absorption of heat from an exothermic process in the system will have relatively little impact on molecular motion and the resulting increase in entropy will be small. (The higher the temperature, the smaller the Ssurrand visa versa).

Ssurr= - Hsys/T

Consider the following reaction at 25oC:

N2(g) + 3H2(g)  2NH3(g)Ho = -92.6 KJ

Given that Ssys= - 199 J/k, Is the reaction spontaneous at 25oC ?

Ssurr = (-92.6 * 1000)J/298K = 311 J/K

Suniv = Ssys + Ssurr= - 199 J/K + 311 J/K = 112 J/K

Because Sunivis positive, we predict that the reaction is spontaneous.

Thermodynamics can tell us whether a reaction will occur spontaneously under specific conditions, but it does not say how fast it will occur.

The Third Law of Thermodynamics:

The entropy of a perfect crystalline substance is zero at the absolute zero of temperature.

As the temperature increases, the freedom of motion also increases. Thus the entropy of any substance at a temperature above 0K is greater than zero.

Entropy of a pure crystalline substance is zero at 0K.

and when the substance is heated to, say, 298K.

The change in entropy, S is given by:S = Sf - Si

S = Sf because Si is zero. The entropy of the substance at 298K, then, is given by S or Sf , which is called the absolute entropy.

Standard entropies of the elements are not zero.

At the melting point, there is a sizable increase in entropy as the more random liquid state is formed.

At the boiling point there is a large increase in entropy as a result of the liquid to gas transition.

Gibbs Free Energy:

From the 2nd law of thermodynamics,

For a spontaneous process:Suniv = Ssys + Ssurr> 0

For an equilibrium process:Suniv = Ssys + Ssurr= 0

Gibbs free energy or free energy (G) is given by the equation G = H – TS.

The change in free energy (G) of a system for a constant-temperature process is

G = H - TS

G is called free energy because G is the maximum amount of energy released in a process occurring at constant temperature and pressure.

Free energy is the energy available to do work.

G < 0 The reaction is spontaneous in the forward direction.

G > 0 The reaction is nonspontaneous. The reaction is spontaneous in the opposite direction.

G = 0 The system is at equilibrium. There is no net change.

Standard Free-Energy Changes:

The standard free-energy of reaction (Gorxn) is the free-energy change for a reaction when it occurs under standard-state conditions, that is, when reactants in their standard states are converted to products in their standard state.

aA + bB  cC + dD

Gorxn= [cGof(C) + dGof(D)] - [aGof(A) + bGof(B)]

Gorxn= Σ nGof(products) - Σ mGof(reactants)

Standard free energy of formation (Gof) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states.

Gof of any element in its stable form is zero.

Example 18.4

Application of Equation G = H - TS

H / S / G / Description
- / + / - / Spontaneous at all T
+ / - / + / Nonspontaneous at all T
+ / + / + or - / Spontaneous only at high T
- / - / + or - / Spontaneous only at low T

An endothermic reaction can be spontaneous only if there is an entropy increase (an increase in disorder).

Temperature and Chemical Reactions:

Consider the following reaction:

CaCO3(s)  CO2(g) + CaO(s)

Ho = 177.8 KJSo = 160.5 J/K

Go = Ho - TSo = 130.0 KJ

Because Go is a large positive, the reaction is not favored at 250C, In order to make Go negative, we first have to find the temperature at which Go is zero.

0 = Ho - TSoT = Ho / So

T = (177.8*1000)/ 160.5 = 1108 K = 8350C

So, at a temperature higher than 8350C, Gobecome negative, indicating that the decomposition is spontaneous.

For example at 8400C, Go = - 0.8KJ

The fact the Gois a positive value at some temperature below 8350C does not mean that no CO2 is produced, but rather that the pressure of the CO2 gas formed at that temperature will be below 1 atm.

If a system is at equilibrium, then ther is no tendency for spontaneous change in either direction.

The condition G = 0 applies to any phase transition.

Phase Transition:

At the melting point and boiling point at which the phase transition occurs, the system is at equilibrium (G = 0). So

0 = H - TSS = H/T

Consider ice-water equilibrium, H is the molar heat of fusion, T melting point of ice. S is therefore Sice → water = (6010J/mol)/273K = 22.0J/K.mol, this mean that when 1 mol of ice melts at 0oC, there is an increase in entropy of 22.0J/K

Swater → ice = (- 6010J/mol)/273K = -22.0J/K.mol, there is a decrease in entropy of -22.0J/K

We can calculate entropy change in each case using the equation S = H/T as long as the temperature remains at 0oC. The same procedure can be applied to water-steam transition. In this case H is the heat of vaporization and T is the boiling point.

SvapSfus.

Example 18.5.

Free Energy and Chemical equilibrium:

Under conditions that are not standard state, we must us G rather than Go to predict the direction of the reaction.

G = Go + RTlnQ

G = actual free energy, Go = standard free energy, R = gas constant (8.314J/K.mol), T = absolute temperature and Q = reaction quotient.

At a given temperature, Go is constant but RTlnQ is not, because Q varies according to the composition of the reaction mixture.

At equilibrium, G = 0, and Q = K.

Thus0 = Go + RTlnKGo = - RTlnK

In this equation Kp is used for gases, and Kc for reactions in solution.

The larger the K is, the more negative Go is.

We use Go rather than G in the relation, because G changes as the reaction proceeds and becomes zero at equilibrium. On the other hand both Go and K are constant for a particular reaction at a given temperature.

Go < 0, the products are favored over reactants at equilibrium.

Go > 0, the reactants are favored over products at equilibrium.

Relation between Go and K as predicted by the equation Go = - RTlnK

K / lnK / Go / Comments
> 1 / Positive / Negative / Products are favored over reactants at equilibrium.
= 1 / 0 / 0 / Products and reactants are equally favored at equilibrium.
< 1 / Negative / Positive / Reactants are favored over products at equilibrium.

To be consistent with units, we express both Go and G in units of KJ/mol.

For example, Go for the reaction

H2(g) + I2(g)  2HI(g)

is 2.6KJ/mol at 298K, meaning that this is standard free energy per mol of H2, per mol of I2 , and per 2 mol of HI.

Examples 18.6, 18.7, 18.8.

Most biological reactions are nonspontaneous. They made to occur by coupling them to the hydrolysis of ATP (adenosine triphosphate), which has a negative Go value.

Selected Problems: 9-11, 13, 14, 17, 19, 20, 23-32, 38.