Chemical Equillibrium

Chemical Equillibrium

INTRODUCTION

It is an experimental fact that most of the process including chemical reactions, when carried out in a closed vessel, do not go to completion. They proceed to some extent leaving considerable amounts of reactants & products. When such stage is reached in a reaction, it is said that the reaction has attained the state of equilibrium. Equilibrium represents the state of a process in which the properties like temperature, pressure, concentration etc of the system do not show any change with passage of time. In all processes which attain equilibrium, two opposing processes are involved. Equilibrium is attained when the rates of the two opposing processes become equal.

If the opposing processes involve only physical changes, the equilibrium is called Physical Equilibrium. If the opposing processes are chemical reactions, the equilibrium is called Chemical Equilibrium.

Some common physical equilibria are:

Solid ⇔ Liquid

Liquid ⇔ Gas

Gas ⇔ Solution

Generally, a chemical equilibrium is represented as

aA + bB ⇔ xX + yY

Where A, B are reactants and X, Y are products.

Note:

The double arrow between the left hand part and right hand part shows that changes is taking place in both the directions.

On the basis of extent of reaction, before equilibrium is attained chemical reactions may be classified into three categories.

(a) Those reactions which proceed to almost completion.

(b) Those reactions which proceed to almost only upto little extent.

(c) Those reactions which proceed to such an extent, that the concentrations of reactants and products at equilibrium are comparable.

EQUILIBRIUM IN PHYSICAL PROCESS

The different types of physical Equilibrium are briefly described below

(a) Solid – liquid Equilibrium

The equilibrium that exist between ice and water is an example of solid – liquid equilibrium. In a close system, at 0oC ice and water attain equilibrium. At that point rate of melting of ice is equal to rate of freezing of water. The equilibrium is represented as

H2O(s) ⇔ H2O(l)

(b) Liquid-Gas Equilibrium

Evaporation of water in a closed vessel is an example of liquid – gas equilibrium. Where rate of evaporation is equal to rate of condensation. The equilibrium is represented as

H2O(l) ⇔ H2O(g)

(c) Solid – solution equilibrium

If you add more and more salt in water taken in a container of a glass and stirred with a glass rod, after dissolving of some amount. You will find out no further salt is going to the solution and it settles down at the bottom. The solution is now said to be saturated and in a state of equilibrium. At this stage, many molecule of salt from the undissolved salt go into the solution (dissolution) and same amount of dissolved salt are deposited back (Precipitation).

Thus, at equilibrium rate of dissolution is equal to rate of precipitation.

Salt(Solid) ⇔ Salt(in solution)

(d) Gas –Solution equilibrium

Dissolution of a gas in a liquid under pressure in a closed vessel established a gas – liquid equilibrium. The best example of this type of equilibrium is cold drink bottles. The equilibrium that exists with in the bottle is

CO2(g) ⇔ CO2(in solution)

Equilibrium in Chemical Process (Reversible and irreversible reactions)

A reaction in which not only the reactants react to form the products under certain conditions but also the products react to form reactants under the same conditions is called a reversible reaction.

Examples are

(i) 3Fe(s) + 4H2O(g) ⇔ Fe3O4(s) + 4H2(g)

(ii) CaCO3(s) ⇔ CaO(s) + CO2(g)

(iii) N2(g) + 3H2(g) ⇔ 2NH3(g)

If a reaction cannot take place in the reverse direction, i.e. the products formed do not react to give back the reactants under the same condition it is called an irreversible reaction.

Examples are:

(i) AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(g)

(ii) 2M(g) + O2(g) → 2MgO(s)

Note:

If any of the product will be removed from the system, reversible reaction will become irreversible one.

CONCEPT OF CHEMICAL EQUILIBRIUM

Let we have a general reversible reaction,

A + B ⇔ C + D

at time t = 0 a0 b0 0 0

as time will pass, A and B will be converted to C and D. As soon as C and D will be formed, reaction will start in back direction also. A time will come when rate of decomposition of A and B will be equal to the rate of formation of A & B, i.e. the rate of forward reaction will be equal to the rate of reverse reaction. At this stage the reaction is said to be in a state of Chemical Equilibrium.

If we see variation of reactants and products with time on graph we will be having following graph.

When the equilibrium is reacted, the concentration of reactant as well as product remains constant. At this stage reaction seems to be stopped but in actual the reaction is going in forward as well as reverse direction with same rate and hence this equilibrium is called as dynamic equilibrium.

Alt text: Dynamic equilibrium

Characteristics of Chemical Equilibrium

The equilibrium is dynamic i.e. the reaction continues in both forward and reverse directions.

The rate of forward reaction equals to the rate of reverse reaction.The observable properties of the system such as pressure, concentration, density

remains invariant with time.The chemical equilibrium can be approached from either side.A catalyst can hasten the approach of equilibrium but does not alter the state of

equilibrium.

Types of Equilibria

There are mainly two types of equilibria:

(a) Homogeneous: Equilibrium is said to be homogeneous if reactants and products are in same phase.

H2(g) + I2(g) ⇔ 2HI(g)

N2(g) + 3H2(g) ⇔ 2NH3(g)

N2O4(g) ⇔ 2NO2(g)

CH3COOH(l) + C2H5OH(l) ⇔ CH3COOC2H5(l) + H2O(l)

(b) Heterogeneous: Equilibrium is said to be heterogeneous if reactants and products are in different phases

CaCO3(s) ⇔ CaO(s) + CO2(g)

NH4HS(s) ⇔ NH3(g) + H2S(g)

NH2CO2NH4(s) ⇔ 2NH3(g) + CO2(g)

Note:

To write the equilibrium constant expression, the concentration of pure liquid and pure solid assumed to be unity, as the concentration of such substances remain constant, i.e.,

concentration = mole/litre ∝ density

Solved example 1. For the gaseous reaction: C2H2 + D2O ⇔ C2D2 + H2O, ΔH is 530 cal. At 25°C, KP = 0.82. Calculate how much C2D2 will be formed if 1 mole of C2H2 and 2 moles of D2O are put together at a total pressure of 1 atm at 100°C?

Sol. Firstly calculate KP at 100°C by using equation

log (Kp)100oC/(Kp)25

oC = ΔH/2.303R (T2-T1/T1T2)

On solving we get, (KP)100°C = 0.98

C2H2 + D2O ⇔ C2D2 + H2O

Initial mole 1 2 0 0 At eq. mole 1–α 2–α α α

∴ 0.98 = α2/(1-α)(2-α)

0.02α + 2.94α – 1.96 = 0

On solving we get, α = 0.75

Solved example 2. An equilibrium mixture at 300K contains N2O4 and NO2 at 0.28 and 1.1 atm respectively. If the volume of container is doubled, calculate the new equilibrium pressure of two gases.

Sol. N2O4(g) ⇔ 2NO2(g)

0.28 1.1

Kp = (1.1)2/0.28 = 4.32 atm

When volume is doubled pressure will reduced to half

N2O4(g) ⇔2NO2(g)

(0.28/2 – p) (1.1/2 + 2p)

⇒ Kp = (1.1/2+2p)2/(0.28/2–p) = 4.32 ⇒ p = 0.045

⇒ p(N2O4) = 0.14 – 0.045 = 0.095

p(NO2) = 0.55 – 0.045 = 0.64 atm

Solved example 3. At 540K, 0.10 mole of PCl5 are heated in a 8 litre flask. The pressure of the equilibrium mixture is found to be 1 atm. Calculate the Kpfor the reaction.

Sol. PCl5 ⇔ PCl3 + Cl2

0.1 – α α α

Total no. of moles = 0.1 – α + α + α = 0.1 + α

Now, P = nRT/V ⇒α 1 = (0.1+α)×0.082×540/8

⇒ (0.1 + α) = 0.180

α = 0.08

So, Kp = (pPCl3 × pCl2)/(pPCl5)

= (αp/0.1+α)(αp/0.1+α)/(0.1–α)/(0.1+α)p = α2p/(0.1+α)(0.1–α)

= 0.0064×1/0.18×0.02 = 1.78 atm

Solved example 4. In a mixture of N2 and H2 initially in a mole ratio of 1:3 at 30 atm and 300oC, the percentage of ammonia by volume under the equilibrium is 17.8. Calculate the equilibrium constant (KP) of the mixture, for the reaction

N2(g) + 3H2(g) ⇔ 2NH3(g)

Sol. Let the initial moles of N2 and H2 be 1 and 3 respectively (this assumption is valid as KP will not depend on the exact no. of moles of N2 and H2. One can even start with x and 3x). Alternatively

N2(g) + 3H2(g) ⇔ 2NH3

Initial 1 3 0

At eqb. 1-x 3-3x 2x

Since % by volume of a NH3 gas is same as % by mole,

∴ 2x/4–2x = 0.178

∴ x = 4×0.178/(2+2×0.178) = 0.302

∴ Mole fraction of H2 at equilibrium = 3–3x/4–2x = 0.6165

Mole fraction of N2 at equilibrium = 1 - 0.6165 - 0.178

= 0.2055

∴ KP = (XNH3×PT)2/(XN2×PT)(XH2×PT)3 = (0.178×30)2/(0.2055×30)(0.6165×30)3

= 7.31 × 10-4 atm–2

Solved example 5. The density of an equilibrium mixture of N2O4 and NO2 at 1 atm. and 348 K is 1.84 g dm-3. Calculate the equilibrium constant of the reaction, N2O4(g) ⇔ 2NO2(g).

Sol. Let us assume that we start with C moles of N2O4(g) initially.

N2O4(g) ⇔ 2NO2(g)

Initial C 0

At equilibrium C(1-α) 2Cα

Where α is the degree of dissociation of N2O4(g)

Since = Total moles at equilibrium/Total moles initially = Vapour density initial/Vapour density at equilibrium

Initial vapour density = 92/2 = 46

C(1+α)/C = 46/d

Since vapour density and actual density are related by the equation,

V.D. = ρRT/2P = 1.84×0.082×348/2

= 26.25

∴ 1 + α = 46/26.25 = 1.752

∴ α = 0.752

∴ Kp = (2Cα/C(1+α) × PT)2/C(1-α)/C(1+α) × PT = (2×0.75/1.752×1)2/0.248/1.752 × 1

= 5.2 atm

Solved example 6. At temperature T, a compound AB2(g) dissociates according to the reaction

2AB2(g) ⇔ 2AB(g) + B2(g) with a degree of dissociation x, which is small compared with unity. Deduce the expression for x in terms of the equilibrium constant, Kp and the total pressure P.

Sol. Let the initial pressure of AB2(g) be P1

2AB2(g) ⇔ 2AB(g) + B2(g)

P1(1-x) P1x P1x/2

The total pressure is = P1 (1+x/2) ≈ P1 (∴ x<< 1) ∴ P1 = P

So, KP = (Px)2×Px/2/[P(1-x)]2 ≈ Px3/2

∴ x = 3√2Kp/P

LAW OF MASS ACTION

Guldberg and Waage established a relationship between rate of chemical reaction and the concentration of the reactants or, with their partial pressure in the form of law of mass action.

According to this law, “The rate at which a substance reacts is directly proportional to its active mass and rate of a chemical reaction is directly proportional to product of active masses of reactants each raised to a power equal to corresponding stoichiometric coefficient appearing in the balanced chemical equation”.

aA + bB → cC + dD

rate of reaction ∝ [A]a.[B]b

rate of reaction = K[A]a[B]b

where K is rate constant or velocity constant of the reaction at that temperature.

Unit of rate constant (K) = [moles/lit]1–n time–1 (where n is order of reaction.)

Note:

For unit concentration of reactants rate of the reaction is equal to rate constant or specific reaction rate.

At equilibrium the rate of both forward and backward reactions become equal and after achieving equilibria, the concentration of reactants and products remains constant as shown in figure.

Note:

Active mass is the molar concentration of the reacting substances actually participating in the reaction.

Hence

Active mass = number of moles/volume in litres

Active mass of solid is taken as unity.

Illustration 1. Calculate the partial pressure of each component in the following equilibria

N2(g) + 3H2(g) ⇔ 2NH3(g)

Solution: N2(g) + 3H2(g) ⇔ 2NH3(g)

at t = 0 a b 0

at equilibrium a – x b – 3x 2x

ntotal = a – x + b – 3x + 2x = (a + b – 2x)

Partial pressure:

PN2 = nN2/nT × PT = (a–x)/(a+b–2x)×PT

PH2 = nH2/nT × PT = (b–3x)/(a+b–2x)×PT

PNH3 = nNH3/nT × PT = 2x/(a+b–2x)×PT

Law of Chemical Equilibrium

According to this law, the ratio of product of concentration of products to the product of concentration of reactants, with each concentration term is raised to the power by its coefficient in overall balanced chemical equation, is a constant quantity at a given temperature and it is called equilibrium constant.

Derivation of law of chemical equilibrium

Let us consider for the following equilibrium

aA + bB ⇔ cC + dD

then, from Law of mass action

rate of forward reaction r1 ∝ [A]a [B]b

or r1 = Ka [A]a [B]b and rate of reverse reaction r2 ∝ [C]c [D]d

r2 = K2 [C]c [D]d

at equilibrium, r1 = r2

⇒ K1 [A]a[B]b = K2 [C]c[D]d

⇒ K1/K2 = [C]c[D]d/[A]a[B]b

Kc = K1/K2, an equilibrium constant in terms of active masses of reacting species.

For the reaction

SO2Cl2 ⇔ SO2 + Cl2

at t = 0 a 0 0 at equilibrium a - x x x

equilibrium conc. a–x/V x/V x/V (V is volume of container)

So, [SO2Cl2] = a–x/V; [SO2] = x/V = [Cl2]

So, Kc = (x/V)×(x/V)/(a–x/V) = x2/(a–x)V ⇒ Kc = x2/(a–x)×V

CHARACTERISTICS OF EQUILIBRIUM CONSTANT (Kc)

(i) Kc for a particular reaction at given temperature has a constant value.

(ii) Value of Kc always depends on nature of reactants and the temperature, but independent of presence of catalyst or, of inert material.

(iii) Its value is always independent of the initial concentration of reactants as well as the products.

(iv) The value of Kc indicates the proportion of products/product formed at equilibrium. Large Kc value means large proportions of product.

(v) When the reaction is reversed, equilibrium constant for reverse reaction will also be inversed.

K'c = 1/Kc

Let us have

A + B ⇔ C + D

Kc = [C][D]/[A][B]

By reversing the reaction,

C + D ⇔ A + B

K'c = [A][B]/[C][D]=1/Kc

(vi) If the coefficients of reactants of products are halved or, doubled then accordingly, value of K'c will change.

A + B ⇔ C + D

Kc = [C][D]/[A][B]

for 2A + 2B ⇔ 2C + 2D

K'c = [C]2[D]2/[A]2[B]2 = {[C][D]/[A][B]}2 = K2c

(vii) When a number of equilibrium reactions are added, the equilibrium constant, for overall reaction is the product of equilibrium constants of respective reactions.

N2(g) + O2(g) ⇔ 2NO; K1 = [NO]2/[N2][O2]

N2O(g) ⇔ N2(g) + 1/2O2(g); K2 = [N2][O2]1/2/[N2O]

N2O(g) + 1/2O2(g) ⇔ 2NO(g); K3 = [NO]2/[N2O][O2]1/2

K3/K2 = [NO]2/[N2O][O2]1/2 × [N2O]/[N2][O2]1/2 = [NO]2/[N2][O2] = K1

i.e. K3/K2 = K1

Equilibrium constant in terms of partial pressures

Let us consider a general reaction

aA + bB ⇔ cC + dD

PA PB PC PD

From law of mass action

r1 ∝ (Pa)a (PB)b ⇒ r1 = K1(PA)a(PB)b

r2 ∝ (Pc)c (PD)d ⇒ r2 = K2(PC)c(PD)d

at equilibrium

r1 = r2 or K1 (PA)a (PB)b = K2(PC)c(PD)d

⇒ K1/K2 = (PC)a(PD)d/(PA)a(PB)b

⇒ Kp = K1/K2 = (PC)a(PD)d/(PA)a(PB)b

RELATION BETWEEN KP AND KC

For a general equation,

aA + bB cC + dD

Where, a , b, c and d are coefficients of the reacting substance

Kp = (PC)c (OD)d / (PA)a (PB)b

From gas equation

PV = nRT => P = n/V RT

So, PA = [A]RT’ PB = [B]RT

PC = [C]RT; PD = [D]RT

Hence, Kp = ([C]RT)c.([D]RT)d / ([A]RT)a / ([B]RT)b

or, Kp = [C]d[D]d / [A]a[B]b × (RT)[(c+d) – (a+b)]

Where Δng = {(c + d) – (a + b)} = change in the numbers of gaseous moles.

Hence, Kp = Kc.(RT)Δng

When Δn = 0; Kp = Kc

Δn > 0; Kp > Kc

Δn < 0; Kp < Kc

Note:

Similarly we can find equilibrium constant (Kx) interms of mole fraction and can find out its relation with Kp and Kc.

Illustration . At 27°C Kp value for reaction is 0.1 atm, calculate its Kc value.

Solution: KP = Kc(RT)Dn

Δn = 1

Kc = KP / RT = 0.1 / 0.82 × 300 = 4 × 10-3

Illustration . Solid NH4I dissociates according to the reaction at 400 K

NH4I(s) NH3(g) + HI(g) ; Kp = 16 atm. In presence of catalyst HI dissociates in H2 and I2 as 2HI H2 + I2. If partial pressure of H2 at this temp is 1 atm in the container when both the equilibrium exist simultaneously, calculate Kp value of second equilibrium (for the dissociation of HI).

Solution: NH4I(s) NH3 (g) + HI (g) KP1 = 16

2HI H2 + I2

P - x x/2 . x/2 KP2 = ?

Here x/2 = 1 atm

x = 2 atm

P(P -2) = 16

P = 5.1

Illustration . A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on addition of graphite. What is the value of Kp if the total pressure at equilibrium is 0.8 atm?

Solution: CO2(g) + C(s) ——> 2CO(g)

0.5 2n

or, 0.5-n 0.5+n = 0.8

n = 0.3

Kp = (2×0.3)2 / (0.5–0.3) = 1.8 atm

Illustration . Given that the equilibrium constant for the reaction, H2(g) + I2(g) 2HI(g) is 50 at 700K. Calculate the equilibrium constant for the reaction.

HI(g) 1/2 H2(g) + 1/2 l2(g)

Solution: We have H2(g) + I2(g) 2HI(g)

Kc = [Hl]2 / [H2][l2] = 50

The new equilibria is

HI(g) [H2]1/2[l2]1/2 / [Hl] = 1/√Kc

= 1/ √50 = √2 / 10 = 0.141

Illustration . One mole of N2 is mixed with 3 moles of H2 in a 4 litre container. If 25% of N2 is converted into NH3 by the following reaction

N2(g) + 3H2(g) 2NH3(g). Calculate Kc and Kp of the reaction. (Temperature = 227°C and R = 0.08231).

Solution: We have N2(g) + 3H2(g) 2NH3(g)

Percentage N2 reacted, 25%

x = 0.25

Now, (a – x) = 1 – 0.25 = 0.75

b – 3x = 3 – 0.75 = 2.25

= 1.48 × 10–5 L2 mol–2

Now, Kp = Kc . (RT)Δn = 1.48 × 10–5 × [0.0821] × (227 + 273)–2

1.48 × 10–5 / {0.082 × (500)}2

= 8.78 × 10–9

RELATION BETWEEN KP AND KC

For a general equation,

aA + bB cC + dD

Where, a , b, c and d are coefficients of the reacting substance

Kp = (PC)c (OD)d / (PA)a (PB)b

From gas equation

PV = nRT => P = n/V RT

So, PA = [A]RT’ PB = [B]RT

PC = [C]RT; PD = [D]RT

Hence, Kp = ([C]RT)c.([D]RT)d / ([A]RT)a / ([B]RT)b

or, Kp = [C]d[D]d / [A]a[B]b × (RT)[(c+d) – (a+b)]

Where Δng = {(c + d) – (a + b)} = change in the numbers of gaseous moles.

Hence, Kp = Kc.(RT)Δng

When Δn = 0; Kp = Kc

Δn > 0; Kp > Kc

Δn < 0; Kp < Kc

Note:

Similarly we can find equilibrium constant (Kx) interms of mole fraction and can find out its relation with Kp and Kc.

Illustration . At 27°C Kp value for reaction is 0.1 atm, calculate its Kc value.

Solution: KP = Kc(RT)Dn

Δn = 1

Kc = KP / RT = 0.1 / 0.82 × 300 = 4 × 10-3

Illustration . Solid NH4I dissociates according to the reaction at 400 K

NH4I(s) NH3(g) + HI(g) ; Kp = 16 atm. In presence of catalyst HI dissociates in H2 and I2 as 2HI H2 + I2. If partial pressure of H2 at this temp is 1 atm in the container when both the equilibrium exist simultaneously, calculate Kp value of second equilibrium (for the dissociation of HI).

Solution: NH4I(s) NH3 (g) + HI (g) KP1 = 16

2HI H2 + I2

P - x x/2 . x/2 KP2 = ?

Here x/2 = 1 atm

x = 2 atm

P(P -2) = 16

P = 5.1

Illustration . A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on addition of graphite. What is the value of Kp if the total pressure at equilibrium is 0.8 atm?

Solution: CO2(g) + C(s) ——> 2CO(g)

0.5 2n

or, 0.5-n 0.5+n = 0.8

n = 0.3

Kp = (2×0.3)2 / (0.5–0.3) = 1.8 atm

Illustration . Given that the equilibrium constant for the reaction, H2(g) + I2(g) 2HI(g) is 50 at 700K. Calculate the equilibrium constant for the reaction.

HI(g) 1/2 H2(g) + 1/2 l2(g)

Solution: We have H2(g) + I2(g) 2HI(g)

Kc = [Hl]2 / [H2][l2] = 50

The new equilibria is

HI(g) [H2]1/2[l2]1/2 / [Hl] = 1/√Kc

= 1/ √50 = √2 / 10 = 0.141

Illustration . One mole of N2 is mixed with 3 moles of H2 in a 4 litre container. If 25% of N2 is converted into NH3 by the following reaction

N2(g) + 3H2(g) 2NH3(g). Calculate Kc and Kp of the reaction. (Temperature = 227°C and R = 0.08231).

Solution: We have N2(g) + 3H2(g) 2NH3(g)

Percentage N2 reacted, 25%

x = 0.25

Now, (a – x) = 1 – 0.25 = 0.75

b – 3x = 3 – 0.75 = 2.25

= 1.48 × 10–5 L2 mol–2

Now, Kp = Kc . (RT)Δn = 1.48 × 10–5 × [0.0821] × (227 + 273)–2

1.48 × 10–5 / {0.082 × (500)}2

= 8.78 × 10–9

SIGNIFICANCE OF THE MAGNITUDE OF EQUILIBRIUM CONSTANT

(i) A very large value of KC or KP signifies that the forward reaction goes to completion or very nearly so.

(ii) A very small value of KC or KP signifies that the forward reaction does not occur to any significant extent.

(iii) A reaction is most likely to reach a state of equilibrium in which both reactants and products are present if the numerical value of Kc or KP is neither very large nor very small.

Units of Equilibrium Constant

As we have learned that numerical value of equilibrium constant is a function of stoichiometric coefficient used for any balanced chemical equation.

But, what will happen in the size of units of equilibrium constant when the stoichiometric coefficient of the reaction changes? (If sum of the exponent for product is not equal to reactant side sum)

So for the sake of simplicity in the units of Kp for Kc the relative molarity or pressure of reactants and products are used with respect to standard condition. (For solution standard state 1 mole/litre and for gas standard pressure = 1 atm)

Now the resulting equilibrium constant becomes unitless by using relative molarity and pressure.

Illustration . The value of Kp for the reaction 2H2O(g) + 2Cl2(g) 4HCl(g) + O2(g) is 0.035 atm at 400oC, when the partial pressures are expressed in atmosphere. Calculate Kc for the reaction,

1/2O2(g) + 2HCl(g) —> Cl2(g) + H2O(g)

Solution: KP = KC (RT)Δn

Δn = moles of product - moles of reactants = 5 - 4 = 1

R = 0.082 L atm/mol K,

T = 400 + 273 = 673 K

0.035 = KC (0.082 ´ 673)

KC = 6.342 ´ 10-4 mol l-1

KC for the reverse reaction would be 1/KC

KC = 1/6.342×10–4 = 1576.8 (mol l-1)-1

When a reaction is multiplied by any number n (integer or a fraction) the K'C or K'pbecomes (KC)n or (KP)n of the original reaction.

KC for 1/2 O2(g) + 2HCl(g) ——> Cl2(g) + H2O(g)

is √1576.8 = 39.7 (mol.l-1)-½

Illustration . Kp for the equilibrium, FeO(s) + CO(g) Fe(s) + CO2(g) at 1000oC is 0.4. If CO(g) at a pressure of 1 atm and excess FeO(s) are placed in a container at 1000oC, what are the pressures of CO(g) and CO2(g) when equilibrium is attained?

Solution: Assuming ideal gas behaviour, partial pressures are proportional to the no. of moles present. Since moles of CO2 formed equals moles of CO consumed, the drop in partial pressure of CO will equal the partial pressure of CO2 produced. Let the partial pressure of CO2 at equilibrium be ‘x’ atm. Then, partial pressure of CO will be (1 - x) atm.

Since K = pCO2 / pco = x / 1 – x = 0.4 => x = 0.286

Hence PCO = 1-x

= 0.714 atm.

Illustration . At 800 K a reaction mixture contained 0.5 mole of SO2, 0.12 mole of O2 and 5 mole of SO3 at equilibrium. Kc for the equilibrium 2SO2 + O2 2SO3 is 833 lit/ mole. If the volume of the container is 1 litre, calculate how much O2 is to be added at this equilibrium in order to get 5.2 moles of SO3 at the same temperature.

Solution: Suppose x mole of O2 is added by which equilibrium shifts to right hand side and y mole of O2 changes to SO3.

The new equilibrium concentration may be

2 SO2 + O2 2SO3

Moles at I equilibrium 0.5 0.12 5

II equilibrium (0.5-2y) (0.12+x-y) (5+2y)

5 + 2y = 5.2

y = 0.1

Kc =

= 833

Substituting y = 0.1 (V = 1 lit)

We get x = 0.34 mole

THE REACTION QUOTIENT ‘Q’

Consider the equilibrium

PCl5 (g) PCl3(g) + Cl2 (g)

At equilibrium [Cl2].[PCl3] / [PCl5] = KC. When the reaction is not at equilibrium this ratio is called ‘QC’ i.e., QC is the general term used for the above given ratio at any instant of time. And at equilibrium QC becomes KC.

Similarly, PCl2 PPCl2 / PPCl2 is called QP and at equilibrium it becomes KP.

If the reaction is at equilibrium, Q = Kc

A net reaction proceeds from left to right (forward direction) if Q < KC.

A net reaction proceeds from right to left (the reverse direction) if Q >Kc

Illustration . For the reaction,

A(g) + B(g) 2C(g) at 25°C, in a 2 litre vessel contains 1, 2, 3 moles of respectively. Predict the direction of the reaction if

(a) Kc for the reaction is 3

(b) Kc for the reaction is 6

(c) Kc for the reaction is 4.5

Solution: A(g) + B(g) 2C(g)

Reaction quotient Q =

(a) Q > Kc,therefore backward reaction will be followed

(b) Q > Kc

The forward reaction is followed

(c) Q = Kc

The reaction is at equilibrium

Illustration . For the reaction : A(aq)+ B(aq) C(aq)+D(aq) , the net rate of consumption of B at 25°C and at any time 't' is as given below

–d/[b]/ dt = {4´10-4[A] [B] – 1.33×10-5 [C] [D]} mol L-1 min-1

Predict whether the reaction will be spontaneous in the direction as written in reaction mixture in which each A, B, C and D is having a concentration of 1 mol L-1?

Solution: K = K1/K2

= 4 × 10–4 / 1.33 × 10–5 = 30

Q = [C][D] / [A][B]

= 1 × 1 / 1 × 1 = 1 < K

Since Q < K, so the above reaction is spontaneous in the forward direction.

LE CHATELIER’S PRINCIPLE

“When an equilibrium is subjected to either a change in concentration, temperature or, in external pressure, the equilibrium will shift in that direction where the effects caused by these changes are nullified”.

This can be understood by the following example. Overall we can also predict the direction of equilibrium by keeping in mind following theoretical assumption.

PCl5 ——> PCl3 + Cl2

Let us assume that we have this reaction at equilibrium and the moles of Cl2, PCl3 and PCl5 at equilibrium are a, b and c respectively, and the total pressure be PT.

Since PT = (a+b+c) RT / V

KP = abRT / cV

Now if d moles of PCl3 is added to the system, the value of Q would be, a(b+d)RT / cV

We can see that this is more than KP. So the system would move reverse to attain equilibrium.

(i) If we increase the volume of the system, the Q becomes abRT / cV' where V' > V.

Q becomes less, and the system would move forward to attain equilibrium.

(ii) If we add a noble gas at constant pressure, it amounts to increasing the volume of the system and therefore the reaction moves forward.

(iii) If we add the noble gas at constant volume, the expression of Q remains as Q = abRT / cV and the system continues to be in equilibrium.

∴ nothing happens.

(iv) Therefore for using Le-Chatlier’s principle, convert the expression of KP and KC into basic terms and then see the effect of various changes.

Effect of Concentration

Let us have a general reaction,

aA + bB cC + dD

at a given temperature, the equilibrium constant,

Kc = [C]c[D]d / [A]a[B]b

again if α, β, γ and δ are the number of mole of A, B, C and D are at equilibrium

then, Kc = [γ]c[δ]d / [α]a[β]b

If any of product will be added, to keep the Kc constant, concentration of reactants will increase i.e. the reaction will move in reverse direction. Similarly if any change or disturbance in reactant side will be done, change in product’s concentration will take place to minimise the effect.

Temperature Effect

The effect of change in temperature on an equilibrium cannot be immediately seen because on changing temperature the equilibrium constant itself changes. So first we must find out as to how the equilibrium constant changes with temperature.

For the forward reaction, according to the Arrhenius equation,

Kr = Af e–eα/RT

And for the reverse reaction,

It can be seen that

For any reaction, ΔH = Eaf – Ear

From the equation,

logK2/K1 = ΔHo / 2.303R (1/T1 – 1/T2) it is clear that

R = Reactant, P = Product

(a) If ΔHo is +ve (endothermic), an increase in temperature (T2 > T1) will make K2 > K1, i.e., the reaction goes more towards the forward direction and vice-versa.

(b) If ΔHo is -ve (exothermic), an increase in temperature (T2 > T1), will make K2 < K1 i.e., the reaction goes in the reverse direction.

(i) Increase in temperature will shift the reaction towards left in case of exothermic reactions and right in endothermic reactions.

(ii) Increase of pressure (decrease in volume) will shift the reaction to the side having fewer moles of the gas; while decreases of pressure (increase in volume) will shift the reaction to the side having more moles of the gas.

(iii) If no gases are involved in the reaction higher pressure favours the reaction to shift towards higher density solid or liquid.

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

(i) N2(g)+ 3H2(g) 2NH3(g) + 23 k cal.

(ii) 2SO2(g) + O2(g) 2SO3(g) + 45 k cal.

(iii) N2(g) + O2(g) 2NO(g) - 43.2 k cal.

(iv) 2NO(g) + O2(g) 2NO2(g) + 27.8 k cal.

(v) C(s) + H2O(g) CO2(g) + H2(g) + X k cal.

(vi) PCl5(g) PCl3(g) + Cl2(g)- X k cal.

(vii) N2O4(g) 2NO2(g) - 14 k cal.

Solution: (i) Low T, High P, excess of N2 and H2.

(ii) Low T, High P, excess of SO2 and O2.

(iii) High T, any P, excess of N2 and O2

(iv) Low T, High P, excess of NO and O2

(v) Low T, Low P, excess of C and H2O

(vi) High T, Low P, excess of PCl5

(vii) High T, Low P, excess of N2O4.

Solved example . The equilibrium constant KP for the reaction N2(g) + 3H2(g) 2NH3(g) is 1.6 × 10-4 atm-2 at 400oC. What will be the equilibrium constant at 500oC if heat of the reaction in this temperature range is-25.14 k cal?

Solution: Equilibrium constants at different temperature and heat of the reaction are related by the

equation,

log KP2 = –25140 / 2.303 × 2 [773 – 673 / 773 × 673] + log 1.64 ´ 10-4

log KP2 = –4.835

KP2 = 1.462 × 10–5 atm–2

Effect of change of Pressure

The effect of change of pressure on chemical equilibrium can be done by the formula.

Kp = QP × P(Δn)

Case A: When Δn = 0

Kp = Qp

And equilibria is independent of pressure.

Case B: When Δn = – ve

Kp = QP × P–Δn = QP / PΔn

Here with increase in external pressure, will shift the equilibrium towards forward direction to maintain Kp. Similarly, decrease in pressure will shift the equilibrium in the reverse direction.

Case C: When Δn = +ve

Kp = QP × P–Δn and, effect will be opposite to that of Case ‘B’

EFFECT OF CATALYST ON EQUILIBRIUM

Since the catalyst is associated with forward as well as reverse direction reaction. So, at equilibrium, rate of forward reaction will be equal to rate of reverse reaction and hence catalyst effect will be same on both forward as well as reverse. Hence catalyst never effect the point of equilibrium but it reduces the time to attain the equilibrium.

Effect on equilibrium due to the addition of inert gases

Following are the cases, where this effect in different manner.

Case A: When Δn = 0 and total volume change at equilibrium remain constant.

So, addition of inert gases will increase the number of moles in the mixture but partial pressure of each component remains constant. Hence equilibrium remains unaffected.

Case B: When Δn = +ve and ΔVT ¹ 0

The total volume will increase with addition of inert gases. This will shift the equilibrium in forward direction.

Case C: When Δn = –ve and ΔVT ¹ 0

In this case, the addition of inert gases will increase the total volume and the reaction will shift in reverse direction.

Dependence of KP or Kc on Temperature

With the increase of temperature, equilibrium favours forward reaction in case of an endothermic (ΔH > 0) reaction while it favours backward reaction in the case of an exothermic (ΔH < 0) reaction.

Sample problem . For the equilibrium

NH4I(g) NH3(g) + HI(g) ΔH = +ve

What will be the effect on the equilibrium constant on increasing the temperature.

Solution: Since the forward reaction is endothermic, so increasing the temperature, the forward reaction is favoured. Thereby the equilibrium constant will increase.

Sample problem Kp for the reaction 2BaO2(s) 2BaO(s) + O2(g) is 1.6 × 10–4 atm, at 400°C. Heat of reaction is – 25.14 kcal. What will be the no. of moles of O2 gas produced at 500°C temperature, if it is carried in 2 litre reaction vessel?

Solution: We know that

log LP2/KP1 = ΔH / 2.303R [1/T1 – 1/T2]

log KP2/1.6 × 10–4 = –25.14 / 2.330 × 2 × 10–3 [773 – 673 / 773 × 673]

=> KP2/1.46 × 10–5 atm

KP2 = p02 = 1.46 × 10–5 atm

Since, PV = nRT

1.46 × 10–5 × 2 = n × 0.0821 × 773

n = 1.46 × 10–5 / 0.0821 × 773 = = 4.60 × 10-7

Let ΔG0 be the difference in free energy of the reaction when all the reactants and products are in the standard state and Kc or, Kp be the thermodynamic equilibrium constant of the reaction. Both related to each other at temperature T by the following relation.

ΔG0 = – 2.303 RT logKc and ΔG0 = – 2.303 RT log Kp(in case of ideal gases)

Now, we know that thermodynamically,

ΔG0 = ΔH0 – TDS0

here ΔH0 is standard enthalpy of reaction, and ΔS0 is standard entropy change

(i) When ΔG0 = 0, then, Kc = 1

(ii) When, ΔG0 > 0, i.e. +ve, then Kc < 1, in this case reverse reaction is feasible showing thereby a less concentration of products at equilibrium rate.

(iii) When ΔG0 < 0, i.e. -ve, then, Kc > 1; In this case, forward reaction is feasible showing thereby a large concentrations of product at equilibrium state.

Sample problem. A system is in equilibrium as

PCl5 + Heat —> PCl3 + Cl2

Why does the temperature of the system decreases, when PCl3 are being removed from the equilibrium mixture at constant volume?

Solution: When PCl3 are being removed from the system, the reaction moved to the right. This consumed heat and therefore, temperature is decreased.

Sample problem. NO and Br2 at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at 300K. At equilibrium the total pressure was 110.5 torr.

Calculate the value of the equilibrium constant and the standard free energy change at 300K for the reaction 2NO(g) + Br2(g) 2NOBr(g).

Solution: 2NO(g) + Br2(g) 2NOBr(g)

Initial pressure 98.4 41.3 0

At eqlbm. 98.4-x 41.3 - x/2 x

The total pressure at equilibrium is 110.5 torr.

98.4-x + 41.3- x/2 + x = 110.5

x = 58.4 torr

Now, 1 atm = 760.4 torr,

x = 7.68 × 10-2 atm.

pNOBr = 7.68 × 10-2 atm ; pNO = 98.4-x = 40 torr = 5.26 × 10-2 atm.

pBr2 = 41.3 – x/2 = 12.1 torr = 1.59 × 10-2 atm.

ΔGo = -2.303 RT log K = -2.303 (1.99) (300) (log 134)

= -2.92 k cal = -12.2 kJ.

[If R is used as 1.99 cal/mol K, then ΔGo will be in cal. If R is used as 8.314 J/mol K, then ΔGo will be in Joules. But KP must always be in (atm)Δn.]

Dependence of Degree of Dissociation on Density Measurements

The degree of dissociation of a substance is defined as the fraction of its molecules dissociating at a given time. The following is the method of calculating the degree of dissociation of a gas using vapour densities. This method is valid only for reactions whose KP exist, i.e., reactions having at least one gas and having no solution.

Since PV = nRT

PV = w/M nRT 0

M = wRT / PV = ρRT / P

VD = ρRT / 2P

Since P = NRT / V

VD = ρRT / 2nRT × V = ρV / 2n

For a reaction at equilibrium V is a constant and ρ is a constant. vapour Density α 1/n

∴ = Total moles at equulibrium / Initial total moles = vapour density intial / vapour ensity at equilibrium = D/d = M/m

(molecular weight = 2 × V.D)

Here M = molecular weight initial

m = molecular weight at equilibrium

Let us take a reaction

PCl5 PCl3 + Cl2

Initial moles C 0 0

At eqb. C(1-α) Cα Cα

∴ = C(1+α) / C = D/d; 1 + α = D/d = M/m

Knowing D and d, α can be calculated and so for M and m.

Sample problem. When PCl5 is heated it dissociates into PCl3 and Cl2. The density of the gas mixture at 200oC and at 250oC is 70.2 and 57.9 respectively. Find the degree of dissociation at 200oC and 250oC.

Solution: PCl5(g) PCl3(g) + Cl2(g)

We are given the vapour densities at equilibrium at 200oC and 250oC.

The initial vapour density will be the same at both the temperatures as

it would be MPCl5 / 2.

∴ Initial vapour density = (31 + 5 × 35.5) / 2 = 104.25

Vapour density at equilibrium at 200oC = 70.2

∴ Total moles at equilibrium / Total moles initial = 1 + α = Vapour density initial / Vapour density at equilibium

= 104.25 / 70.2 = 1.485

∴ a = 0.485

At 250oC, 1 + α = 104.25 / 57.9 = 1.8

∴ a = 0.8

Sample problem. The degree of dissociation of PCl5 is 60%, then find out the observed molar mass of the mixture.

Solution: PCl5(g) PCl3(g) + Cl2(g)

Initially moles 1 0 0

At eqm. 1 – α α α

Where α = degree of dissociation = 0.6

Total moles at equilibrium = 1 + a = observed mole

observed moles / Theoretica moles = molecular weight theoertical / molecular weight observed

=> 1 + a = 206.5 / molecular weight observed

=> molecular weight observed = 206.5 / 1 + a = 206.5 / 1.6 = 129.06

Exercise .

Vapour density of N2O4 which dissociates according to the equilibrium N2O4(g) 2NO2(g) is 25.67 at 100°C and a pressure of 1 atm. Calculate the degree of dissociation and Kp for the reaction.

Exercise .

How much PCl5 must be added to a one litre vessel at 250oC in order to obtain a concentration of 0.1 mole of Cl2? Kc for PCl5 –> PCl3 + Cl2 is 0.0414 mol/litre.

Exercise 1.

Calculate the equilibrium pressure of each reactant of given reaction with the help of following data.

A + B C + D

Initial pressure of A and B are 6 atm and 5 atm. The equilibrium pressure of C and D are 2 atm in each case.

Exercise 2 .

The equilibrium constant for the reaction

PCl5(g) PCl3(g) + Cl2(g) is 2.4 × 10–3. Determine the equilibrium constant for the reverse reaction at same temperature.

Exercise 3

At 21.5°C and a total pressure of 0.0787 atm, N2O4 is 48.3% dissociated into NO2. Calculate Kp and Kc for the reaction.

N2O4(g) 2NO2(g)

Exercise 4 .

(i) Write the expression for Kp and Kc for the following reactions.

N2 + 3H2 2NH3

(ii) Find out the heterogeneous equilibrium from the following reaction

(A) CaCO3 (s) CaO(s) + CO2 (g)

(B) H2(g) + I2(g) 2HI (g)

Exercis 5

What is the value of equilibrium constant for the following reaction

If A D

A B K1 = 2

B C K2 = 5

C D K3 = 3

Exercise 6.

In an equilibrium A + B —> C + D; A and B are mixed in a vessel at temperature T. The initial concentration of A was twice the initial concentration of B. After the equilibrium has established, concentration of C was thrice the equilibrium concentration of B. Calculate Kc.

Exercise .7

Ammonium hydrogen sulphide dissociates as follows

NH4HS(s) NH3(g) + H2S(g)

If solid NH4HS is placed in an evacuated flask at certain temperature it will dissociate until the total pressure is 600 torr.

(a) Calculate the value of equilibrium constant for the dissociation reaction

(b) Additional NH3 is introduced into the equilibrium mixture without changing the temperature until partial pressure of NH3 is 750 torr, what is the partial pressure of H2S under these conditions? What is the total pressure in the flask?

Exercise .8

At 700 K hydrogen and bromine react to form hydrogen bromide. The value of equilibrium constant for this reaction is 5 × 108. Calculate the amount of H2, Br2 and HBr at equilibrium if a mixture of 0.6 mole of H2 and 0.2 mole of Br2 is heated to 700 K.

Exercise 9

The equilibrium constant is 0.403 at 1000K for the process

FeO(s) + CO(g) Fe(s)+ CO2)g)

A steam of pure CO is passed over powdered FeO at 1000K until equilibrium is reached. What is the mole fraction of CO in the gas stream leaving the reaction zone?

Exercise 10 .

N2 + 3H2 2NH3; Δ H = negative

What are the conditions of temperature and pressure favorable for this reaction?

Exercise 11: Equilibrium constant Kp for the reaction 3/2 H2(g) + 1/2 N2(g) NH3(g) are 0.0266 and 0.0129 at 310°C and 400°C respectively. Calculate the heat of formation of gaseous ammonia. Standard Free Energy Change of a Reaction and its Equilibrium Constant

Exercise 12.

What is the value of R if value of ΔGO is in?

(i) Joules (ii) Calories

Exercise 13.

N2O4 is 25% dissociated at 27°C and 1 atm pressure.

Calculate (i) Kp and (ii) the percentage dissociation at 0.1 atm and 27°C.

ANSWERS TO EXERCISES

Exercise 1:

A ——> 4 atm

B ——> 3 atm

Exercise 2:

4.16 × 102

Exercise 3:

Kp = 0.0956 atm

Kc = 3.96 × 10–3 mole/lit

Exercise 4: (i) Kc = [NH3]2 / [N2] [H2]3

Kp =

(ii) (A) is a heterogeneous equilibrium.

Exercise 5:

K = 30

Exercise 6:

1.8

Exercise 7:

(a) 0.155 atm2

(b) 119.32 torr, 869.32 torr

Exercise 8:

0.4, 0, 0.4

Exercise 9:

0.713

Exercise 10:

Low temperature, high pressure and high concentration of N2 and H2.

Exercise 11.

–12140 cals

Exercise 12:

(i) 8.314 J mole-1K-1

(ii) 2 cal mole-1K-1

Exercise 13:

(i) 0.266 atm

(ii) 93.27%

MISCELLANEOUS EXERCISES

Exercise 1: For an equilibrium, the rate constant for the forward and the backward reaction are 2.38 × 10-4 and 8.15 × 10-5 respectively. Calculate the equilibrium constant for the reaction.

Exercise 2: In an experiment starting with 1 mole C2H5OH, 1 mole CH3COOH and 1 mole of water, the equilibrium mixture on analysis shows that 54.3% of the acid is esterified. Calculated Kc.

Exercise 3: From the values of the equilibrium constants, indicate in which case, does the reaction go farthest to completion: K1 = 10-10, K2 = 1010, K3 = 105.

Exercise 4: The value of the equilibrium constant is less than zero. What does it indicate?

Exercise 5: Would you expect equilibrium constant for the reaction l2(g) 2l(g) to increase or decrease as temperature increases? Assign reason.

Exercise 6: Acetic acid is highly soluble in water but still a weak electrolyte. Why?

Exercise 7: Which of the following reactions involve homogenous and which involve heterogenous equilibria?

(a) 2N2O(g) 2N2 (g) + O2 (g)

(b) 2NO3(g) N2 (g) + 3H2 (g)

(c) 2Cu(NO3)2(s) 2CuO(s) + 4O2 (g) + O2(g)

(d) CH3COOC2H5(l) + H2O(l) CH3COOH (l) + C2H5OH(l)

(e) Fe3+ (aq) + 3OH– (aq) Fe (OH)3(s)

Exercise 8: For the reaction, CH3(g) + 2H2S(g) CS2(g) + 4H2(g) at 1173 K, the magnitude of Kc is 3.6. For each of the following composition, decide whether the reaction mixture is at equilibrium. If not, decide to which direction the reaction should go?

Exercise 9: Kp for the reaction H2(g) + l2(g) 2Hl(g) at 460°C is 49. If the partial pressure of H2 and I2 is 0.5 atm respectively, determine the partial pressure of each gas in the equilibrium mixture.

Exercise 10: 0.16 g N2H4 (hydrazine) are dissolved in water and the total volume is made upto 500 mL. Calculate the percentage of N2H4 that has reacted with water in solution. The Kb for N2H4is 4.0 × 10-6 M.

ANSWERS TO MISCELLANEOUS EXERCISES

Exercise 1: 2.92

Exercise 2: Kc = 4.012

Exercise 3: High value of equilibrium constant (K) means that the molar concentration of the products is quite large. In other words, the reaction has proceeded to a large extent in the forward direction before attaining the equilibrium. Thus for K2 = 1010, the forward reaction has proceeded maximum before attaining the equilibrium.

Exercise 4: This indicates that the forward reaction has proceeded only to a small extent before the equilibrium is attained.

Exercise 5: In the forward reaction, energy is needed to bring about the dissociation of I2(g) molecules. This means that the forward reaction is endothermic in nature. The increase in temperature will favour the forward reaction. Therefore, the equilibrium constant will increase with rise in temperature.

Exercise 6: The strength of the electrolyte does not depend upon its amount present in solution but on its ionization in solution. Since acetic acid is a weak acid (organic acid), it is ionized only to small extent. Therefore, it is a weak electrolyte.

Exercise 7: Homogenous equilibria : (a), (b) (d)

Heterogenous equilibria : (c) and (e)

Exercise 8: (a) Backward direction

(b) Backward direction

Exercise 9: Partial pressure of H2 (g) = 0.112 atm

Partial pressure of I2(g) = 0.112 atm

Partial pressure of HI (g) = 0.776 atm

Exercise 10: 2.0%

Objective:

Prob 1. Reaction between iron and steam is reversible if it is carried out

3Fe + 4H2O —> Fe3O4 + 4H2

(A) at constant T (B) at constant P

(C) in an open vessel (D) in a closed vessel

Sol. In open vessel H2 gas will escape.

(D)

Prob 2. For the reaction

PCl4(g) + Cl2(g) ——> PCl5(g)

The value of Kc at 250°C is 26. The value of Kp at this temperature will be

(A) 0.61 (B) 0.57

(C) 0.83 (D) 0.46

Sol. Kc = Kc (RT)Δn = 26 × (0.082 × 523)–1 = 0.61

(A)

Prob 3. In a reversible reaction, two substances are in equilibrium. If the concentration of each one is doubled, the equilibrium constant will be

(A) Reduced to half, its original value

(B) becomes (original)/4

(C) doubled

(D) constant

Sol. Kcor Kp do not depend on concentration, but only on temperature.

(D)

Prob 4. The equilibrium constant for the reaction,

N2(g) + O2(g) ——> 2NO is 4 × 10–4 at 2000 K

In presence of a catalyst, equilibrium is attained ten times faster. Therefore, the equilibrium constant, in presence of the catalyst at 2000K is

(A) 40 × 10–4 (B) 4 × 10–4

(C) 4 × 10–5 (D) difficulty to compute

Sol. Equilibrium is constant at constant temperature for a reaction

(B)

Prob 5. 64g of HI are present in a 2 litre vessel. The active mass of HI is:

(A) 0.5 (B) 0.25

(C) 1 (D) none

Sol. Molecular mass of HI = 128

HI] = [Hl] = 64/128 × 1/2 = 0.25

(B)

Prob 6. For which of the following Kp may be equal to 0.5 atm

(A) 2HI H2 + I2 (B) PCl5 PCl3 + Cl2

(C) N2 + 3H2 2NH3 (D) 2NO2 N2O4

Sol. For Kp = 0.5 atm

Δn = 1 (since the unit is atm)

and PCl5 PCl3 + Cl2

Δn = 1

(B)

Prob 7. The vapour density of undecomposed N2O4 is 46. When heated, vapour density decreases to 24.5 due to its dissociation to NO2. The % dissociation of N2O4 at the final temperature is

(A) 80 (B) 60

(C) 40 (D) 70

Sol. N2O4 2NO2

1 0 at initial

1 – α 2α at equilibrium

46 / 25.4 = 1 + α/1

1.8 = 1 + α => α = 0.8 or 80%

(A)

Prob 8. If pressure is applied to the following equilibrium, liquid vapours the boiling point of liquid

(A) will increase (B) will decrease

(C) may increase or decrease (D) will not change

Sol. Boiling point of a liquid is the temperature at which vapour pressure became equal to atm pressure. If the pressure is applied to the above equilibrium the reaction will go to the backward direction, i.e. vapour pressure decrease hence the boiling point increase.

(A)

Prob 9. For the reaction

A(g) + B(g) 3C(g) at 250°C, a 3 litre vessel contains 1, 2, 4 mole of A, B and C respectively. If KC for the reaction is 10, the reaction will proceed in

(A) Forward direction (B) Backward direction

(C) In either direction (D) In equilibrium

Sol. Q = [C]3 / [A][B] = 43 × 3 × 3 / 33 × 1 × 2 = 10.66

[C] = 4/3

[A] = 1/3 => [B] = (2/3) KC = 10, and Q > KC

reaction will proceed in backward direction

(B)

Prob 10. If CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(l) Kp = 1.086 × 0 10–4 atm2 at 25°C. The efflorescent nature of CuSO4.5H2O can be noticed when vapour pressure of H2O in atmosphere is

(A) > 7.29 mm (B) < 7.92 mm

(C) > 7.92 mm (D) None

Sol. An efflorescent salt is one that loses H2O to atmosphere.

For the reaction

CuSO4.5H2O(s) CuSO4.3H2O(s) + 2H2O(l)

Kp = ( PH2O )2 = 1.086 × 10–4

PH2O = 1.042 × 10–2 atm = 7.92 mm

If PH2O at 25°C < 7.92 mm only then, reaction will proceed in forward direction.

\ (B)

SOLVED PROBLEMS

Subjective:

Board Type Questions

Prob 1. At 444°C, 15g mole of hydrogen are mixed with 5.2g mole of I2 vapour. When equilibrium was established, 10g mole of HI was formed. Calculate the equilibrium constant for the reaction.

Sol. H2(g) + l2(g) 2Hl

a b 0

a – x b – x 2x

So,

= 4x2 / (a–x) (b–x) = 4 × 52 / 10 × 0.2 = 50

Prob 2. The degree of dissociation is 0.4 at 400K and 1 atm for the gaseous reaction

PCl5 PCl3 + Cl2

Assuming ideal behaviour of all the gases, calculate the density of equilibrium mixture at 400K and 1 atm pressure.

(Atomic mass of P = 31 and Cl = 35.5)

Sol. PCl5 PCl3 + Cl2

at t = 0 1 0 0

at eqm. 1 – 0.4 0.4 0.4

Total no. of moles after dissociation = 0.6 + 0.4 + 0.4

Now, V = nRT / P = 1.4 × 0.082 × 400 / 1 = 45.92 litres

So, density = Molecular weight / Volume = 208.5 / 45.92 = 4.54 / litre

Prob 3. When 3.06g of solid NH4HS is introduced into a 2 litre evacuated flask at 27°C, 30% of the solid decomposed into gaseous ammonia and hydrogen sulphide.

(i) Calculate Kc for reaction at 27°C.

(ii) What would happen to the equilibrium when more solid NH4HS is introduced into the flask?

Sol. Moles of NH4HS introduced into flask = 3.06 / 51 = 0.08

NH4HS(s) NH3(g) + H2S (g)

0.06 (1 – x) 0.06x 0.06x

as x = 30%, so, x = 0.3

So, 0.06 × 0.7 0.06´0.3 0 0.6 × 0.3

So, Kc = [NH3][H2S]

0.018 × 0.018 = 3.24 × 10–2

Addition of NH4HS(s) does not change position of equilibrium.

Prob 4. Determine Kc for the reaction ½N2(g) + ½O2(g) + ½Br2(g) —> NOBr(g) from the following information (at 298oK)

Kc = 2.4 × 1030 for 2NO(g) —> N2(g) + O2(g) ;

K'c = 1.4 for NO(g) + ½Br2(g) —> NOBr(g)

Sol. N2(g) + O2(g) —> 2NO (g)

Kc/ = 1/Kc = 1/2.4 × 1030

(i) ½N2(g) + ½O2(g ) NO(g)

Kc// =

(ii) NO(g) + ½Br2(g) NOBr(g) = 1.4.

(i) + (ii) gives the net reaction:

Prob 5. The heat of reaction at constant volume for an endothermic reaction in equilibrium is 1200 cal more than at constant pressure at 300K. Calculate the ratio of equilibrium constant Kpand Kc.

Sol. Given that ΔE – ΔH = 1200 cal

Also we have

nRT = –1200

or Δn = –1200 / 2 × 300 = –2 (R = 2calK–1 mol–1)

Now, Kp = Kc (RT)Δn

Kp = Kc (0.0281 × 300) = 1.648 × 10–3

IIT Level Question

Prob 6. Anhydrous CaCl2 is often used as dessicant. In presence of excess of CaCl2, the amount of water taken up is governed by Kp = 1.28 × 1085 for the following reaction at room temperature, CaCl2(s) + 6H2O(g) CaCl2.6H2O(s)

What is the equilibrium pressure of water in a closed vessel that contains CaCl2(s)?

Sol. CaCl2(s) + 6H2O(g) CaCl2.6H2O(s)

Prob 7. NO2(g) NO(g) + 1/2O2(g)

Consider the above equilibrium where one mole of NO2 is placed in a vessel and allowed to come to equilibrium at a total pressure of 1 atm. Experimental analysis shows that

Temperature 700K 800K

PNO / PNO2 0.872 2.50

(a) Calculate Kp at 700K and 800K

(b) Calculate ΔG0

Sol. (a) NO2(g) NO(g) + 1/2 O2(g)

at t = 0 1 mole 0 0

t eq. 1 – 2x 2x x

partial (1–2x/1+x)PT (2x / 1+x) PT (x/1+x)PT

at equilibrium pressure.

Given that

PNO / PNO2 = 0.872 at 700K

=> (2x/1+x)PT / (–2x/1+x) PT = 0.872

=> 2x / 1 – 2x = 0.872

=> x = 0.2329

PO2 = x / 1 + x PT = 0.1889

KPt = PNO / PNO2 × (PO2)x/2

= = 0.372 atm1/2

= 0.872 × (0.1889)1/2 = 0.372 atm1/2

Similarly KB2 can be calculated

(b) ΔG0 = – RTlogKp

= –2.303 × 8.314 × 700log 0.379

= 5.65kJ/mol

Prob 8. At 25°C and 1 atm, N2O4 dissociates by the reaction

N2O4(g) 2NO2(g)

If it is 35% dissociated at given condition, find

(i) The percent dissociation at same temperature if total pressure is 0.2 atm.

(ii) The percent dissociation (at same temperature) in a 80g of N2O4 confined to a 7 litre vessel.

(iii) What volume of above mixture will diffuse if 20 ml of pure O2 diffuses in 10 minutes at same temperature and pressure?

Sol. N2O4(g) 2NO2(g)

t equilibrium 1 – a 2a

Partial pressure at equilibrium [1–α / 1 + α] PT [2α/1+α]PT

= 4×(0.35)2 / 1 – (0.35)2 × 1 = 0.56 atm

(i) KP = 4α2 / 1 – α2 × P

0.56 = 4α2 / 1 – α2 × 0.2

Solving, α = 0.64

% dissociation = 64%

(ii) Since

KP = Kc (RT)Δn

Here Δn = 1

Kc = Kc / RT = 0.56 / 0.0821 × 298 = 0.0228 mole/litre

Kc = 4 × α2 / (1–α)V where x = no. of moles of N2O4(g) = 80/92

Neglecting α with respect to 1

i.e. 1 – a ~ 1

Kc = 4 × α2/ V

0.0228 = 4 × 80 / 92 α2/ 7 = α = 0.2142

% dissociation = 21.42%

Let V(ml) volume of mixture diffused in Now, from Graham’s law of diffusion.

Prob 9. Solid NH4I on rapid heating in a closed vessel at 357°C develops a constant pressure of 275 mm of Hg owing to the partial decomposition of NH4I into NH3 and HI but the pressure gradually increases further (when excess solid residually remains in the vessel) owing to the dissociation of HI. Calculate the final pressure developed under equilibrium.

NH4I(s) NH3(g) + HI(g)

2HI(g) H2(g) + I2(g) Kc = 0.015 at 357°C

Sol. P = 275/760 = 0.3618 atm

= 0.181 atm

Kp = (0.181)2 = 0.033 atm2

NH4I(s) NH3(g) + HI(g)

(0.181 + p) (0.181 + p – p') Kp = 3.3 × 10–2 atm2

2HI(g) H2(g) + I2(g)

(0.181 + p – p') p'/2 p'/2 Kp = 0.015

0.033 = (0.018 – p) (0.181 + p – p')

0.015 = pr2/4(0.181 + p – p)2 => p' = 0.04 atm

p = 0.026 atm

∴ total pressure = 2(0.181 + p) = 0.414 = 314.64 mm Hg

Prob 1. The enthalpies of two reaction are ΔH1 and ΔH2 (both positive) with ΔH2 >ΔH1. If the temperature of reacting system is increased from T1 to T2, predict which of the following alternatives is correct?

Sol. As the temperature of reacting system is increased the equilibrium constant of reaction is also increased for endothermic reactions so for two reactions on increasing the temperature by equal amounts

K'1 / K1 < K'2 / K2

Hence, (C) is correct.

Prob 2. At a certain temperature 2 moles of carbonmonoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction CO + Cl2 —> COCl2 in a 5 lit vessel. At equilibrium if one mole of CO is present then equilibrium constant for the reaction is:

(A) 2 (B) 2.5

(C) 3.0 (D) 4

Sol. CO + Cl2 —> COCl2

At t = 0 2 3 0

At equilibrium (2–1) (3–1) 1

Concentrations (V = 5 lit) 1/5 2/5 1/5

Hence, (B) is correct.

Prob 3. The reaction: 3O2 2O3, ΔH = + 69,000 calories is favoured by:

(A) high temperature and low pressure (B) high temperature and high pressure

(C) low temperature and high pressure (D) low temperature and low pressure

Sol. According to Le Chatalier principle formation of ozone is favoured by high temperature (endothermic reaction) and high pressure.


Prob4. The equilibrium constant at 323°C is 1000. What would be its value in the presence of a catalyst in the forward reaction?

A + B C + D + 38 Kcal

(A) 1000 × concentration of catalyst (B) 1000

(C) 1000 / concentration of catalyst (D) impossible to tell

Sol. The equilibrium constant varies only with temperature. At constant temperature it will not vary.


Prob 5. Which of the following reactions represent a heterogenous equilibrium?

(A) H2(g) + l1 (g) 2Hl(g)

(B) CaCO3(s) CaO(s) + CO2(g)

(C) N2O4(g) 2NO2(g)

(D) O3(g) 3O2(g)

Sol. In heterogenous equilibrium, physical state of all the reactants and products are not same.


True/False

Prob 6. For a reversible system at constant temperature the value of Kc increase if the concentration are changed at equilibrium.

Sol. False

Prob 7. The apparent molecular weight if PCl5 on dissociation shows lower value.

Sol. True

Prob 8. The value of Kc for the reaction 2SO2 + O2 ——> 2SO3 will increase if the volume of reaction vessel is increased.

Sol. False

Fill in the Blanks

Prob9. Free energy change at equilibrium is …………..

Sol. Zero

Prob 10. The equilibrium constant is ………. of velocity constant of forward and backward reactions.

Sol. Ratio

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Chemical Equillibrium