Chemical Equilibrium, Class 11, Chemistry Detailed Chapter Notes PDF Download

Types of Equilibrium - Equilibrium, Class 11, Chemistry

Equilibrium

Basically, the term refers to what we might call a "balance of forces". In the case of mechanical equilibrium, this is its literal definition. A book sitting on a table top remains at rest because the downward force exerted by the earth's gravity acting on the book's mass is exactly balanced by the repulsive force between atoms that prevents two objects from simultaneously occupying the same space, acting in this case between the table surface and the book. If you pick up the book and raise it above the table top, the additional upward force exerted by your arm destroys the state of equilibrium as the book moves upward. If you wish to hold the book at rest above the table, you adjust the upward force to exactly balance the weight of the book, thus restoring equilibrium. An object is in a state of mechanical equilibrium when it is either static or in a state of unchanging motion. From the relation f=ma, it is apparent that if the net force on the object is zero, its acceleration must also be zero, so if we can see that an object is not undergoing a change in its motion, we know that it is in mechanical equilibrium.

Another kind of equilibrium we all experience is thermal equilibrium. When two objects are brought into contact, heat will flow from the warmer object to the cooler one until their temperatures become identical. Thermal equilibrium is a "balance of forces" in the sense that temperature is a measure of the tendency of an object to lose thermal energy. A metallic object at room temperature will feel cool to your hand when you first pick it up because the thermal sensors in your skin detect a flow of heat from your hand into the metal, but as the metal approaches the temperature of your hand, this sensation diminishes. The time it takes to achieve thermal equilibrium depends on how readily heat is conducted within and between the objects; thus a wooden object will feel warmer than a metallic object even if both are at room temperature because wood is a relatively poor thermal conductor.

Chemical equilibrium

When a chemical reaction e.g. 2HI(g)  H₂(g) + I₂(g) takes place in a closed container the quantities of components change as some are consumed and others are formed. Eventually this change will come to an end, after which the composition will remain unchanged as long as the system remains undisturbed. The system is then said to be in its equilibrium state, or more simply, "at equilibrium".

It makes no difference whether we start with two moles of HI or one mole each of H₂ and I₂; once the reaction has run to completion, the quantities of these two components will be the same. In general, then, we can say that the composition of a chemical reaction system will tend to change in a direction that brings it closer to its equilibrium composition.

Dynamic equilibrium characteristics:

• The state at which concentrations of reactants or products do not change with time.

• It is attained when rate of forward reaction becomes equal to rate of backward reaction.

• A dynamic equilibrium, attained from either side.

Chemical Equilibrium with Graph

Reversible reaction

A chemical equation of the form A → B represents the transformation of A into B, but it does not imply that all of the reactants will be converted into products, or that the reverse reaction B → A cannot also occur. In general, both processes can be expected to occur, resulting in an equilibrium mixture containing all of the components of the reaction system. If the equilibrium state is one in which significant quantities of both reactants and products are present then the reaction is said to incomplete or reversible. In principle, all chemical reactions are reversible, but this reversibility may not be observable if the fraction of products in the equilibrium mixture is very small, or if the reverse reaction is kinetically inhibited.

Homogeneous equilibrium

The system in which all the reactant and product have same physical state.

CH₃COOC₂H₅(aq) H₂O(aq)  CH₃COOH (aq) C₂H₅OH (aq)

Heterogeneous equilibrium

The system in which atleast one reactant or product have different physical states from others. eg.

CaCO₃(s)  CaO(s) CO₂(g), This is three phase system

Salt H₂O(aq) → Salt (aq), Three phase system.

Law of Mass Action

During the period 1864-1879 Cato Guldberg and Peter Waage showed that an equilibrium can be approached from either direction (see the hydrogen iodide illustration above), implying that any reaction aA  bB → cC  dD is really a competition between a "forward" and a "reverse" reaction. When a reaction is at equilibrium, the rates of these two reactions are identical, so no net (macroscopic) change is not observed, although individual components are actively being transformed at the microscopic level. Guldberg and Waage showed that the rate of the reaction in either direction is proportional to what they called the "active masses" of the various components:

rate of forward reaction = k_f[A]^a[B] ^b

rate of reverse reaction = k_r[C]^c[D]^d

in which the proportionality constants k are called rate constants and the quantities in square brackets represent concentrations. If we combine the two reactants A and B, the forward reaction starts immediately, but the formation of products allows the reverse process to get underway. As the reaction proceeds, the rate of the forward reaction diminishes while that of the reverse reaction increases. Eventually the two processes are proceeding at the same rate, and the reaction is at equilibrium: rate of forward reaction = rate of reverse reaction

k_f[A]^a [B]^b = k_r[C]^c[D]^d If we now change the composition of the system by adding some C or withdrawing some A (thus changing their "active masses"), the reverse rate will exceed the forward rate and a change in composition will occur until a new equilibrium composition is achieved. The Law of Mass Action is thus essentially the statement that the equilibrium composition of a reaction mixture can vary according to the quantities of components that are present.

Equilibrium constant

aA + bB  cC + dD

At equilibrium, R_f = R_b

k_f[A]^a [B]^b = k_r[C]^c [D]^d → 

K_c → equilibrium constant in terms of concentration

 (K_p = eq. constant in terms of partial pressure)

for the reaction, N₂(g) + 3 H₂(g)  2NH₃(g)

, 

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Law of Chemical Equilibrium and Equilibrium Constant - Equilibrium, Class 11, Chemistry 

Law of Chemical Equilibrium

At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the equilibrium Law or Law of Chemical Equilibrium

Units of Equilibrium Constant

The value of equilibrium constant K_C can be calculated by substituting the concentration terms in mol/L and for K_p partial pressure is substituted in Pa, kPa, bar or atm. This results in units of equilibrium constant based on molarity or pressure. unless the exponents of both the numerator and denominator are same.

For the reactions,

H₂(g) + I₂(g) 2HI, K_C and K_P have no unit.

N₂O₄(g)  2NO₂ (g), K_C has unit mol/L and K_P has unit bar or atm

Equilibrium constants can also be expressed as dimensionless quantities if the standard state of reactants and products are specified. For a pure gas, the standard state is 1 bar. Therefore a pressure of 4 bar in standard state can be expressed as 4 bar/1 bar = 4, which is a dimensionless number. Standard state (C₀) for a solute is 1 molar solution and all concentrations can be measured with respect to it. The numberical value of equilibrium constant depends on the standard state chosen. Thus in this system both K_p and K_C are dimensionless quantities and represented as K_p° & K_C° respectively

Relation between K_p and K_c

For the reaction aA bB  cC dD

K_p=K_C(RT)^Dng

where Dn_g = (c d) - (a b) = no. of moles of gaseous products - no. of moles of gaseous Reactants

if Dn = 0 K_p = K_c

significance of equilibrium constant

(a) Using K_eq to Predict Relative Concentrations

The size of the equilibrium constant can give us information about the relative amounts of

reactants and products present at equilibrium.

• When K << 1

The reaction lies to the left (mostly reactants)

• When K >> 1

The reaction lies to the right (mostly products)

• When K = 1

The reaction lies in the middle (mix of reactants and products)

(b) Calculating Equilibrium concentrations

Ex. Phosgene is a poisonous gas that dissociates at high temperature into two other poisonous gases, carbon monoxide and chlorine. The equilibrium constant K_p= 0.0041 atm at 600K. Find the equilibrium composition of the system after 0.124 atm of COCl₂ initially is allowed to reach equilibrium at this temperature.

Sol. 

Substitution of the equilibrium pressures into the equilibrium expression gives

This expression can be rearranged into standard polynomial formx² 0041x-0.00054=0 and

solved by the quadratic formula, but we will simply obtain an approximate solution by iteration.

Because the equilibrium constant is small, we know that x will be rather small compared to

0.124, so the above relation can be approximated by

which gives x=0.025. To see how good this is, substitute this value of x into the denominator of the original equation and solve again:

This time, solving for x gives 0.0204. Iterating once more, we get

and x = 0.0206 which is sufficiently close to the previous to be considered the final result. The

final partial pressures are then 0.104 atm for COCl₂, and 0.0206 atm each for CO and Cl₂.

Note: using the quadratic formula to find the exact solution yields the two roots -0.0247 (which we ignore) and 0.0206, which show that our approximation is quite good. Note: using the quadraticformula to find the exact solution yields the two roots -0.0247 (which we ignore) and 0.0206, which show that our approximation is quite good.

reaction quotient (Q):

At each point in a reaction, we can write a ratio of concentration terms having the same forms as the equilibrium constant expression. The ratio is called the reaction quotient denoted by symbol Q. It helps in predicting the direction of a reaction.

The expresson Q= at any time during rection is called reaction quotient

(i) if Q > K_C reaction will proceed in bakward direction until equilibrium in reached.

(ii) if Q < K_C reaction will proceed in forward direction until equilibrium is established.

(iii) if Q = K_C reaction is at equilibrium

Ex. For the reaction NOBr (g) NO(g) Br₂(g) , K_p=0.15 atm at 90°C. If NOBr, NO and Br₂ are

mixed at this temperature having partial pressures 0.5 atm,0.4 atm & 2.0 respectively, will Br₂ be

consumed or formed?

Sol. 

K_P=0.15

Hence, reaction will shift in backward direction, Therefor Br₂ will be consumed
 

Degree of Dissociation & Vapour Density

nA(g)  A_n(g),

1 0

1 - a , moles of mix = 1 - a  

,

, where V.D. of A = D₀ , V.D. of eq. mix = d

Ex. The Vapour Density of mixture of PCl₅, PCl₃ and Cl₂ is 92. Find the degree of dissociation of PCl₅.

Sol.

   1                      0           0

1 - x                     x           x

 = 104.25 , , a =0.13

Characteristics of equilibrium constant

1. 

After reversing the reaction

After reversing the reaction the equilibrium constant get reversed.

2. 

 and

A B E F  C D G K₃

 → K₃ = K₁ × K₂

when the two reaction are added there equilibrium constant get multiplied.

3. 

After multiplying by n

nA nB  nC nD K₂

 → 

When the reaction is multiplied by any number then eq. constant gets the same number in its power.

Ex. Given the following equilibrium constants:

(1) CaCO₃(s) → Ca²  (aq) CO₃^2-(aq) K₁=10^-8.4

(2) HCO₃^-(aq) → H (aq) CO₃^2-(aq) K₂=10^-10.3

Calculate the value of K for the reaction CaCO₃(s) H (aq)=Ca² (aq) HCO₃^-(aq)

Sol. The net reaction is the sum of reaction 1 and the reverse of reaction 2:

CaCO₃(s) → Ca²  (aq) CO₃^2-(aq) K₁=10^-8.4

H (aq) CO₃^2-(aq)→ HCO₃^-(aq) K_-2=10^-(-10.3)

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CaCO₃(s) H (aq) → Ca² (aq) HCO₃^-(aq) K=K₁/K₂=10^{(-8.4 10.3)} =10 ^1.9

Comment: This net reaction describes the dissolution of limestone by acid; it is responsible for the eroding

effect of acid rain on buildings and statues. This is an example of a reaction that has practically no tendency to take place by itself (the dissolution of calcium carbonate) begin "driven" by a second reaction having a large equilibrium constant. From the standpoint of the LeChâtelier principle, the first reaction is "pulled to the right" by the removal of carbonate by the hydrogen ion. "Coupled" reactions of this type are widely encountered in all areas of chemistry, and especially in biochemistry, in which a dozen or so reactions may be linked in this way.

Ex. The synthesis of HBr from hydrogen and liquid bromine has an equilibrium constant Kp = 4.5×10¹⁵ at 25°C. Given that the vapor pressure of liquid bromine is 0.28 atm, find Kp for the homoge neous gas-phase reaction at the same temperature.

Sol: The net reaction we seek is the sum of the heterogeneous synthesis and the reverse of the vaporiza tion of liquid bromine:

Here are some of the possibilities for the reaction involving the equilibrium between gaseous water and its elements:

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Effect of Temperature Change - Equilibrium, Class 11, Chemistry

Effect of Temperature on Equilibrium constant

VAN'T HOFF EQUATION

where K is the equilibrium constant, T is temperature, DH is the enthalpy of the reaction and R is the gas constant. It provides means to detemine how the equilibrium constant for a reaction or process will vary with temperature.

Ex. H₂O(l)  H (aq) OH^-(aq) , DH= 13.7 kcal/mol , K_eq at 25°C =10^-14. Calculate K_eq at 60°C ?

Sol. 

K₂= 1.26×10^-12

LeChâtelier principle

If a system at equilibrium is subjected to a change of pressure, temperature, or the number of moles of a substance, there will be a tendency for a net reaction in the direction that tends to reduce the effect of this change.

1. EFFECT OF CHANGE IN CONCENTRATION

The change in concentration can affect gaseous systems or liquid solution systems only. However this does not affect the solid and pure liquid systems since their active masses are always taken as unity. By using le Chatelier's principle, the effect of change in concentration on systems at equilibrium can be explained as follows:

1. When the concentration of reactant(s) is increased, the system tries to reduce their concen tration by favoring the forward reaction.

2. When the concentration of product(s) is increased, the system tries to reduce their concentration by favoring the backward reaction.

3. When the concentration of reactant(s) is decreased, the system tries to increase their concentration by favoring the backward reaction.

4. When the concentration of product(s) is decreased, the system tries to increase their concentration by favoring the forward reaction.

When the concentration of reactants is increased, the number of effective collisions between them increases which in turn increases the rate of forward reaction. Thus the forward reaction is more favored over the backward reaction until the new equilibrium is established. At this new equilibrium, the rates of both forward and backward reactions become equal again and the reaction quotient becomes approximately equal to the equilibrium constant.

Remember that small changes in concentration do not affect the equilibrium constant.

For decomposition reaction , PCl₅(g) PCl₃(g) Cl₂(g)

Let the concentration of PCl₅ is doubled to disturb the equilibrium. This will change the reaction quotient, Q to:

After disturbing the equilibrium, the value of Q becomes less than KC. In order to restore the Q value to KC, the concentration of PCl₅ must be decreased while the concentrations of PCl₃ and Cl₂ are to be increased. This is achieved by favoring the forward reaction
 

The forward reaction is also favored by removing the products from the reaction mixture (decrease in the concentration of products). Upon removal of products, the rate of forward reaction becomes greater than that of backward reaction momentarily. This will also decrease the reaction quotient. Hence the system tries to reestablish the equilibrium by converting more reactants to products so as to make the rates of both forward and backward reactions become equal again.

For example, in case of the decomposition of PCl₅, if the concentration of Cl₂ is increased by two times at equilibrium, the Q value becomes greater than the KC value.

Hence the system tries to restore the value of Q to KC again. The backward reaction is favored to decrease the concentration of Cl₂. However the concentration of PCl5 also decreases automatically while the concentration of PCl₅ increases while doing so.

2. EFFECT OF CHANGE IN PRESSURE

The change in pressure only affects the equilibrium of systems involving at least one gas. The le Chatelier's principle can be applied to understand the effect of change in pressure on the systems at equilibrium as follows.

1. When the partial pressure of any of the gaseous reactants or of the products is increased, the position of equilibrium is shifted so as to decrease its partial pressure. This is usually achieved by favoring the reaction in which there is decrease in the number of moles of gaseous components.

2. When the partial pressure of any of the gaseous reactants or of the products is decreased, the position of equilibrium is shifted so as to increase its partial pressure. This can be achieved by favoring that reaction in which there is increase in the number of moles of gaseous components.

However, it is not always correct to say that the equilibrium is shifted whenever there is a change in the total pressure of the system. The equilibrium is not always disturbed upon changing the pressure of the entire system. It is only disturbed whenever there is a change in the partial pressure of any or all of the gaseous reactants or products in the equilibrium for which the Dn_g ¹ 0. Where Dn_g = (no. of moles of gaseous products) - (no.of moles of gaseous reactants) Strictly speaking, the equilibrium is only shifted when the ratio of product of partial pressures of products to the product of partial pressures of reactants i.e., the reaction quotient in terms of partial pressures, Qp is disturbed. The position of equilibrium is shifted so as to make Qp become equal to the value of Kp again.

The Q_p can be changed in the following cases:

1. By adding or removing any gaseous reactant or product at constant volume. The effect is same as changing the concentration as explained above.

2. By changing the volume of the system (or in other words by changing the pressure of entire system) at equilibrium for which the Dng ¹ 0. In this case, however, the pressure of the entire system is also changed.

For the decomposition of PCl₅, the Kp can be written as: 

For this reaction Dng = (1 + 1)-(1) = 1

Hence for this reaction, if the pressure of the system is increased by 2 times by halving the volume, the reaction quotient, Qp is doubled. In order to restore back the Qp value again to Kp, the denominator value i.e., the partial pressure of PCl₅must be increased. This is only achieved by favoring the forward reaction in which less number of gaseous products are formed. i.e., Two moles of products (PCl₃ and Cl₂) are converted to one mole of reactant (PCl₅)

Therefore we can say, if the pressure of the system is increased, the system tries to decrease it by favoring the reaction in the direction of decreasing the number of moles of gaseous components.

3. Addition of Inert Gas

(A) At constant volume :there is no effect of adding inert gas on the state of equilibrium at constant volume.

(B) At constant pressure : On Adding of inert gas At constant pressure, the reaction proceeds in that direction where the sum of stoichiometric coefficient of gaseous components is greater.

4. EFFECT OF CHANGE IN TEMPERATURE

The effect of temperature can be understood by using le Chatelier's principle as follows:

1. Increase in the temperature of the system favors the endothermic reaction. The increase in temperature increases the amount of heat in the system. Hence it tries to remove the excess of heat by favoring that reaction in which heat is absorbed i.e., the endothermic reation.

2. Decrease in the temperature of the system favors the exothermic reaction. In this case, the temperature is decreased by removing the heat content from the system. Hence the system tries to restore the temperature back by favoring the exothermic reaction i.e., the reaction in which the heat is liberated.

It is very important to note that, during the change in temperature, the system establishes a new equilibrium for which the value of equilibrium constant is different from the original constant i.e., the equilibrium constant depends on the temperature.

EFFECT OF CATALYST

A catalyst has no effect on the position of the equilibrium since it increases not only the rate of forward reaction but also the rate of backward reaction. However it does help the system to reach the equilibrium faster.

INDUSTRIAL APPLICATIONS OF LE CHATELIER'S PRINCIPLE

1. HABER PROCESS

In Haber process, the ammonia is synthesized by combining pure nitrogen and hydrogen gases in 1:3 ratio in presence of finely powdered iron catalyst and molybdenum promoter at around 450^oC and at about 250 atm. of pressure.

The le Chatelier's principle helps in choosing these conditions to improve the yields of ammonia as explained as below.

Effect of pressure:

Hence the synthesis of ammonia is favored by increasing the pressure of the system. Industrially, 100 - 250 atm. of pressure is employed. Effect of temperature: Since the forward reaction is exothermic, the increase in temperature favors the backward reaction i.e., the dissociation of ammonia. That means according to le Chatelier's principle, the synthesis of ammonia is favored at lower temperatures. However the reaction will be too slow at lower temperatures (a kinetic restriction). Hence this reaction is carried out at optimal temperatures i.e., at about 450 - 550 oC to overcome the kinetic barrier. Removal of ammonia: The forward reaction can also be favored by removing ammonia from the system from time to time by liquefying it. Catalyst: To increase the speed of the reaction, finely powdered or porous iron is used as catalyst. Its efficiency can be improved by adding molybdenum or oxides of potassium and aluminium.

2. CONTACT PROCESS

In the contact process, sulfuric acid, the king of chemicals, is manufactured on large scale. The major steps involved in the process are:

The crucial step is the oxidation of sulfur dioxide, SO₂ to sulfur trioxide, SO₃. It is a reversible reaction. At normal conditions, the equilibrium lies far to the left and the amount of sulfur trioxide formed is very small. To improve the yield of sulfur trioxide, the reaction is carried out at around 450^oC and 2 atm pressure in presence of V₂O₅ or Pt, which acts as a catalysts.

These conditions are chosen by applying le Chatelier's principle as explained below.

Effect of pressure: In the forward reaction (formation of sulfur trioxide), the number of moles of gaseous components is decreasing.

Hence the forward reaction is favored by increasing the pressure of the system. However, at high pressures, the iron towers used in the contact process are corroded. Hence the process is carried out at optimal pressures like 2 atm.

Effect of temperature: Since the forward reaction is exothermic, at higher temperatures the backward reaction i.e., the dissociation of sulfur dioxide is more favored. However the reaction will be too slow at lower temperatures. Hence this reaction is carried out at optimal temperatures i.e., around 450° C.

Catalyst: To increase the speed of the reaction, V₂O₅ or Pt are used as catalysts

Ex. Under what conditions will the following reactions go in the forward direction?

1. N₂(g) + 3H₂(g)  2NH₃(g) 23 Kcal.

2. 2SO₂(g) + O₂(g)  2SO₃(g) 45 Kcal.

3. N₂(g) + O₂(g)  2NO(g) - 43.2 Kcal

4. 2NO(g) + O₂(g)  2NO₂(g) 27.8 Kcal

5. C(s) + H₂O(g)  CO₂(g) + H₂(g) x k cal.

6. PCl₅(g)  PCl₃(g) + Cl₂(g) - X kcal.

7. N₂O₄(g)  2NO₂(g) - 14 kcal

Sol. 1. Low T, High P, excess of N₂ and H₂ .

2. Low T, High P, excess of SO₂ and O₂

3. High T, any P , excess of N₂ and O₂

4. Low T, High P, excess of NO and O₂

5. Low T, Low P, excess of C and H₂O

6. High T, low P, excess of PCl₅

7. High T, Low P, excess of N₂O₄

Simultaneous Equilibrium

A_(s)  B_(g) + C_(g)

D_(s)  B_(g) + E_(g)

 Applicable only when at least one of the product is common in both the reaction.

Ex. The pressure at equilibrium over solid A is 50 atm and over solid D is 68 atm if both solid A and D are heated simultaneously then find

(i) the total pressure over the solids.

(ii) In the above question find the mole ratio of C and B

(iii) mole fraction of C

Sol. (i)

, 

, 

 ...(i)

 ...(ii)

total pressure = p_B  p_C  p_E = x y x y = 2 (x y)

Also 

→ 

→ Total pressure = 2 (x y) = 

(ii) At constant temp and volume P ≈ n

→ pressure Ratio will their mole ratio

by eq. (i)/(ii) 

(iii) Mole fraction of C

As we know 

→  →  ,, 

→ mole fraction of C = 

Ex.6 A(s)  At eq., pressure = 18 atm

C(s)  At eq., pressure = 36 atm

Calculation

(i) total pressure at new equilibrium when both the solids are heated simultaneously.

(ii) mole ratio of B and D

(iii) mole fraction of B in the mixture.

Sol. A(s)  H₂S(g) + B(g), Kp₁ = (9)² = 81

                            x + y        x

C(s)  H₂S(g) + D(g), Kp₂ = (18)² = 324

                     x + y        y

total pressure = x + y + x + y = 2 (x + y)

Kp₁ = x(x + y)

Kp₂ = y(x + y)

→  = (x + y)² → x + y = 

→ total pressure = 2 (x + y) = 2  = 

(ii) mole ratio of B & D =  =  = 

(iii) mole fraction of B in mixture =  = 

Heterogeneous reactions: the vapor pressure of solid hydrates

Many common inorganic salts form solids which incorporate water molecules into their crystal structures.

These water molecules are usually held rather loosely and can escape as water vapor. Copper(II) sulfate, for example forms a pentahydrate in which four of the water molecules are coordinated to the Cu2 ion while the fifth is hydrogen-bonded to SO₄^--. This latter water is more tightly bound, so that the pentahydrate loses water in two stages on heating: CuSO₄.5H₂OCuSO₄.H₂OCuSO₄

These dehydration steps are carried out at the temperatures indicated above, but at any

temperature, some moisture can escape from a hydrate. For the complete dehydration of the

pentahydrate we can define an equilibrium constant CuSO₄.5H₂O(s) → CuSO₄(s)  + 5 H₂O(g) K_p = 1.14×10^-10

The vapor pressure of the hydrate (for this reaction) is the partial pressure of water vapor at which the two solids can coexist indefinitely; its value is K_p^1/5 atm. If a hydrate is exposed to air in which the partial pressure of water vapor is less than its vapor pressure, the reaction will proceed to the right and the hydrate will lose moisture. Vapor pressures always increase with temperature, so any of these compounds can be dehydrated by heating. Loss of water usually causes a breakdown in the structure of the crystal; this is commonly seen with sodium sulfate, whose vapor pressure is sufficiently large that it can exceed the partial pressure of water vapor in the air when the relative humidity is low. What one sees is that the well-formed crystals of the decahydrate undergo deterioration into a powdery form, a phenomenon known as efflorescence. When a solid is able to take up moisture from

the air, it is described as hygroscopic. A small number of anhydrous solids that have low vapor pressures not only take up atmospheric moisture on even the driest of days, but will become wet as water molecules are adsorbed onto their surfaces; this is most commonly observed with sodium hydroxide and calcium chloride. With these solids, the concentrated solution that results continues to draw in water from the air so that the entire crystal eventually dissolves into a puddle of its own making; solids exhibiting this behavior are said to be deliquescent.

One of the first hydrates to be investigated in detail was calcium sulfate hemihydrate (CaSO₄×1/2 H₂O) which LeChâtelier showed to be what forms when when the form of CaSO₄ known as plaster of Paris hardens; the elongated crystals of the hydrate bind themselves into a cement-like mass.

Ex. At what relative humidity will copper sulfate pentahydrate lose its waters of hydration when the air temperature is 30°C? What is K_p for this process at this temperature?

Sol. From the table, we see that the vapor pressure of the hydrate is 12.5 torr, which corresponds to a relative humidity of 12.5/31.6 = 0.40 or 40%. This is the humidity that will be maintained if the hydrate is placed in a closed container of dry air.

For this hydrate, Kp = , so the partial pressure of water vapor that will be in equilibrium with the hydrate and the dehydrated solid (remember that both solids must be present to have equilibrium!),expressed in atmospheres, will be (12.5/760)⁵= 1.20 ´ 10^-9.

Relationship Between 

Free energy, G, denotes the self intrinsic electrostatic potential energy of a system. This means that in any molecule if we calcualte the total electrostatic potential energy of all the charges due to all the other charges , we get what is called the free energy of the molecule. It tells about the stability of a molecule with respect to another molecule. Lesser the free energy of a molecule more stable it is.Every reaction proceeds with a decrease in free energy.The free energy change in a process is expressed by DG.If it is negative, it means that product have lesser G than reactants, so the reaction goes forward.If it is positive the reaction goes reverse and if it is zero the reaction is at equilibrium.

DG is the free energy change at any given concentration of reactants and products . If all the reactants and products are taken at a concentration of 1 mole per liter, the free energy change of the reaction is called DG^o ( standard free energy change ).

One must understand that DG^o is not the free energy change at equilibrium . It is free energy change when all the reactants and products are at a concentration of 1 mole/L . DG^o is related to K (equilibrium constant ) by the relation , DG^o= -RT ln K.

K may either be kc or Kp.Accordingly we get . The units of DG^o depends only or R.

T is always is Kelvin, and if R is in Joules, DG^o will be in joules, and if R is calories than DG^o will be in calories.

Physical Equilibrium

Phase Transitions

There are number of graphical means to help understand the relationships between the different phases of a particular substance. The first thing we need to do when looking at the transitions to different phases is to establish some definitions.

There are three particular phases between which we will be examining discrete phase changes. These are solids, liquids, and gases. Each transition has a particular name.

For example, the transition from liquid to gas is called vaporization. Vaporization is endothermic, and this transition leads to an increase in entropy. The reverse transition of a gas going to a liquid is condensation. Since condensation is simply the reverse of vaporization, the changes in enthalpy and entropy will be exactly the same, but opposite in sign. So DH_condensation=-DH_vaporization .

Let us consider the following physical equilibrium

H₂O(l)  H₂O(g)

1. Effect of temperature

Since it is an endothermic reaction therefore reaction will proceed in forward direction on increasing temp. (Le Chatelier's principle)

2. Effect of pressure :

for the above equilibrium DH > 0

DV ie V_g - V_l = +ve

→ 

→ on increasing P, T increases

But As reaction is to complete at the initial temp, therefore temperature must be lowered. This will make the reaction go to direction as suggested by Le-chatelior (endothermic reaction)

Ex. A(s) (d = 1.14 gm/cc)

B(s) (d = 1.5 gm/cc)

A(s)  B(s) Heat

If mass of A and B are equal, on increasing the pressure, will for formation of B or ¯ ?

Sol. A(s)  B(s)

As density of solid B is greater than solid A, V_B < V_A → V_B - V_A = - ve

since reaction is exothermic, DH = - ve

→  = , = +ve

→ On increasing pressure, the rate of formation of B will be enhanced.

LE CHATERLIER'S PRINCIPLE AND PHYSICAL EQUILIBRIUM

Consider physical equilibrium

Solid  Liquid  vapour

(i) Effect of pressure on melting

When solid melts there is a decrease in volume for some solids (ice) and increase in volume for some solid (sulphur)When ice melts there is decrease in volume, So at constant volume there is a decrease in pressure . If pressure is increase as ice water system, the equilibrium wil go in forward direction. The above phenomena can also be explained from the following phase quilibrium diagram If we see phase diagram for ice water system.

So on increasing pressure, melting point is increased so equilibrium is shifted in forward direction.

ii. Vapour pressure of liquids.

Consider liquid vapour equilibrium Since evaporation of liquid is endothermic, hence rise in temperature will favour evaporation.

iii. Effect of temperature on solubility

If formation of solution is endothermic process then increase in temperature increases solubility. In formation of solution is exothermic then increase in temperature will lower the solubility.

PHASE DIAGRAM :

Triple Point

One very special point on a phase diagram is the triple point. This is the temperature and pressure at which three phases are in equilibrium.Typically, when we are talking about a triple point it is the solid, liquid, gas triple point. For CO₂this is at a pressure of 5.11 atm and -56.4 °C. Many substances have more than one solid phase, and therefore they can have more than one triple point.

Critical Point

If you follow the vaporization curve up to high temperature and pressure, you notice it simply stops. It is because this is the critical point (the on the diagram) for the substance. At temperatures and pressures greater than this point, the definition of a liquid and a gas disappear and the substance exists as a super-critical fluid (SCF). What is a super-critical fluid? It is a fluid (it takes the shape of its container) like a liquid and a gas, but it has a density that can vary between the two extremes of the liquid and the gas. This makes SCF extremely interesting. Typically, the molecules are either very close together or very far apart. In a SCF, they can essentially be anywhere in between.

What happens when you go from a gas to a SCF Essentially nothing. That is, the boundary between the liquid and the SCF and the gas and the SCF is an imaginary line defined by the critical point. This is not a phase transtion. I repeat. Going from a gas to a SCF or a liquid to a SCF is not a phase transition. Going form a liquid to a gas, there is a discrete change in enthalpy, entropy, volume, density. This is a phase transition. When you "cross the line" from a gas to an SCF, there are no discrete changes.It is a continuous change. Thus, if the diagram shows this boundary, it is usually marked with dotted lines to note it is different.

Note: It is fairly easy to make CO₂ a SCF (relatively low temperature and pressure). As a result, it has wide industrial use as a SCF solvent.