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1. IIT–JEE syllabus
Law of mass action; equilibrium constant; exothermic and endothermic reactions; Le – Chatelier’s Principle and its applications
2. Introduction
Chemical equilibrium is the most important characteristic property of a reversible reaction. It is the state of reversible reaction, at given temperature, characterised by constancy of certain observable properties such as pressure, conc., density etc. The equilibrium state is dynamic and not static in nature. A reaction is said to have attained equilibrium when the rate of forward reaction equals that of backward reaction.
A + B C + D
Characteristics of Equilibrium State
 i) It can be attained only if the reversible reaction is carried out in closed vessel.
 ii) It can be attained from either side of the reaction.
iii) A catalyst can hasten the approach of equilibrium but does not alter the state of equilibrium.
 iv) It is dynamic in nature i.e. reaction does not stop but both forward and backward reactions take place at equal rate.
 v) Change of pressure, concentration or temperature favours one of the reaction (forward or backward) resulting in shift of equilibrium point in one direction.
3. Law of Mass Action and Equilibrium Constant
According to the law of mass action, the rate of reaction, at given temperature is directly proportional to the product of the active masses of the reacting substances each raised to power equal to corresponding stoichiometric coefficient appearing in balanced chemical equations.
Active mass of reactant (a) = Conc. × activity coefficient
i.e. a = Molarity × f for dilute solution f = 1
Applying Law of mass action for general reversible reaction
aA + bB mC + nD
Rate of forward reaction µ [A]^{a}[B]^{b}
or R_{f} = K_{f} [A]^{b}[B]^{b}
Similarly for backward reaction
R_{b} = K_{b}[C]^{m}[D]^{n}
At equilibrium K_{f}[A]^{a}[B]^{b} = K_{b}[C]^{m}[D]^{n}
\ = K_{C}
K_{C} = equilibrium constant in terms of concentration
In terms of partial pressures equilibrium constant is denoted by K_{p}
K_{p} = (P = CRT, Pµ C at constant temperature)
Where P terms represent partial pressures at equilibrium
In terms of mole fraction equilibrium constant is denoted by K_{X}
K_{X} =
Where ‘X’ terms represent mole fractions at equilibrium
The equilibrium constants are related as
K_{p}= ^{ }…(1)
K_{P} = …(2)
Where ‘P’ is total pressure at equilibrium and Ttemperature
Dn_{g} = (m + n) – (a + b)
When Dn_{g} = 0, K_{p}= K_{C} = K_{X}
Also K_{p} =
Where n_{A}, n_{B} … etc. are respective number of moles at equilibrium, ån is total number of moles at equilibrium P is total pressure at equilibrium.
Equilibrium Constant of a reaction involving Condensed Phase: The expression for the equilibrium constant of a reaction involving condensed phase (solid or liquid) and gaseous state is obtained by considering the concentrations or partial pressures of only the gaseous species. The concentration of condensed phases being constant are merged with the equilibrium constant.
Some facts about equilibrium constant
 a) A very large value of K_{C} or K_{P} signifies that the forward reaction, as written, goes to completion or very nearly so.
 b) A very small numerical value of K_{C} or K_{P} signifies that the forward reaction, as written, does not go to any significant extent.
 c) A reaction is most likely to reach a state of equilibrium in which both reactants and products are present if numerical value of K_{C} or K_{P} is neither very large nor very small.
 d) The equilibrium constant of a forward reaction and that of its backward reaction are reciprocal of each other.
If A + B C + D K = x
Then C + D A + B K¢ =
 e) If a chemical equation is multiplied by certain factor, its equilibrium constant must be raised to a power equal to that factor in order to obtain the equilibrium constant for the new reaction.
If, for NO + O_{2} NO_{2} K =
Then for 2NO + O_{2} 2NO_{2} K ^{1} = = K^{2}
If K_{1}, K_{2} and K_{3} are stepwise equilibrium constant for
A B (1)
B C (2)
C D (3)
Then for A D K = K_{1} × K_{2} × K_{3}
 The Reaction Quotient ‘Q’
At any point in a reversible reaction, ratio of the concentration terms in the same form as in the equilibrium constant expression can be formulated. This ratio is referred to as reaction quotient or mass action ratio (Q)
For general reaction, aA + bB mC + nD
Q_{C} = (in terms of partial pressure)
 a) A net reaction proceeds from left to right (the forward reaction) if Q_{C} < K_{C}.
 b) The backward reaction takes place if
Q > K_{C}
 c) The reaction is at equilibrium if
Q = K_{C}
Illustration1: At 1000 K, the pressure of iodine gas is found to be 0.112 atm due to partial dissociation of I_{2}(g) into I. Had there been no dissociation, the pressure would have been 0.074 atm. Calculate the value of K_{p} for the reaction:
I_{2}(g) 2I(g).
Solution: analysing in terms of pressure directly
Partial pressure  I_{2}  2I 
Initial  0.074  0 
at equilibrium  0.074 p  2p 
Þ total pressure at equilibrium
= (0.074 – p) + 2p = 0.112 (given)
Þ p = 0.038 atm
K_{p} =
Substituting value of p Þ K_{p} = 0.16 atm
Illustration2: In an evacuated vessel of capacity 110 litres, 4 moles of Argon and 5 moles of PCl_{5} were introduced and equilibriated at a temperature at 250^{o}C. At equilibrium, the total pressure of the mixture was found to be 4.678 atm. Calculate the degree of dissociation, a of PCl_{5} and K_{P} for the reaction
PCl_{5 }PCl_{3} + Cl_{2} _{ }at this temperature.
Solution: PCl_{5}(g) PCl_{3}(g) + Cl_{2}(g)
Initial moles 5 0 0
At equilibrium 5x x x
Total moles = 5 + x + 4 (including moles of Argon) = 9 + x
Since total moles = = 11.99 12
\ x = 3
\ a = = 0.6
K_{P} = = 1.75 atm.
The degree of dissociation of PCl_{5} would have been 0.6 even in the absence of Argon. As one can see, the total pressure of gases constituting equilibrium is equal to 3.11 atm. The observed equilibrium pressure is 4.678 atm which means by the addition of 4 moles of Argon, the total pressure increases. This implies that addition of Argon has been done at constant volume which does not result in any change in degree of dissociation.
5. The Le–Chatelier’s Principle
This principle, which is based on the fundamentals of a stable equilibrium, states that
“When a chemical reaction at equilibrium is subjected to any stress, then the equilibrium shifts in that direction in which the effect of the stress is reduced”.
Confused with “stress”. Well by stress here what I mean is any change of reaction conditions e.g. in temperature, pressure, concentration etc.
This statement will be explained by the following example.
Let us consider the reaction: 2NH_{3} (g) N_{2} (g) + 3H_{2} (g)
Let the moles of N_{2}, H_{2} and NH_{3} at equilibrium be a, b and c moles respectively. Since the reaction is at equilibrium,
= K_{P} =
Where,
X terms denote respective mole fractions and P_{T} is the total pressure of the system.
Þ = K_{P}
Here, = mole fraction of N_{2}
= mole fraction of H_{2}
= mole fraction NH_{3}
Þ = K_{P}
Since P_{T} = ( assuming all gases to be ideal)
\ = K_{P} ———— (1)
Now, let us examine the effect of change in certain parameters such as number of moles, pressure, temperature etc.
If we increase a or b, the left hand side expression becomes Q_{P} ( as it is disturbed from equilibrium) and we can see that Q_{P} > K_{P}
The reaction therefore moves backward to make Q_{P} = K_{P}.
If we increase c, Q_{P} < K_{P} and the reaction has to move forward to revert
back to equilibrium.
If we increase the volume of the container (which amounts to decreasing the pressure), Q_{P} < K_{P} and the reaction moves forward to attain equilibrium.
If we increase the pressure of the reaction then equilibrium shifts towards backward direction since in reactant side we have got 2 moles and on product side we have got 4 moles. So pressure is reduced in backward direction.
If temperature is increased the equilibrium will shift in forward direction since the forward reaction is endothermic and temperature is reduced in this direction.
However from the expression if we increase the temperature of the reaction, the left hand side increases (Q_{P}) and therefore does it mean that the reaction goes backward (since Q_{P} > K_{P})?. Does this also mean that if the number of moles of reactant and product gases are equal, no change in the reaction is observed on the changing temperature (as T would not exist on the left hand side). The answer to these questions is No. This is because K_{P} also changes with temperature. Therefore, we need to know the effect of temperature on both Q_{P} and K_{P} to decide the course of the reaction.
5.1. Effect of Addition of Inert Gases to a Reaction at Equilibrium
 Addition at constant pressure
Let us take a general reaction
aA + bB cC + dD
We know,
Where,
n_{C} n_{D}, n_{A}, n_{B} denotes the no. of moles of respective components and PT is the total pressure and ån = total no. of moles of reactants and products.
Now, rearranging,
Where Dn = (c + d) – (a + b)
Now, Dn can be = 0, < 0 or > 0
Lets take each case separately.
 a) Dn = 0 : No effect
 b) Dn = ‘+ve’ :
Addition of inert gas increases the ån i.e. is decreased and so is . So products have to increase and reactants have to decrease to maintain constancy of Kp. So the equilibrium moves forward.
 c) Dn = ‘–ve’ :
In this case decreases but increases. So products have to decrease and reactants have to increase to maintain constancy of Kp. So the equilibrium moves backward.
 Addition at Constant Volume :
Since at constant volume, the pressure increases with addition of inert gas and at the same time ån also increases, they almost counter balance each other. So can be safely approximated as constant. Thus addition of inert gas has no effect at constant volume.
5.2. Dependence of K_{P} on Temperature
Now we will derive the dependence of K_{P} on temperature.
Starting with Arrhenius equation of rate constant
——————– (i)
Where, k_{f} = rate constant for forward reaction, A_{f} = Arrhenius constant of forward reaction,
= Energy of activation of forward reaction
——————– (ii)
Dividing (2) by (3) we get,
We know that (equilibrium constant )
\ K =
At temperature T_{1}
————– (iii)
At temperature T_{2}
——————– (iv)
Dividing (5) by (4) we get
Þlog
The enthalpy of a reaction is defined in terms of activation energies as = DH
\
\log————— (v)
For an exothermic reaction , DH would be negative. If we increase the temperature of the system ( T_{2}>T_{1}), the right hand side of the equations (V) becomes negative.
\ , that is, the equilibrium at the higher temperature would be less than that at the lower temperature.
Now let us analyse our question. Will the reaction go forward or backward?
Before answering this, we must first encounter another problem. If temperature is increased, the new K_{P} would either increase or decrease or may remain same. Let us assume it increases.
Now, Q_{P} can also increase, decrease or remain unchanged. If K_{P} increases and Q_{P} decreases, than , therefore the reaction moves forward. If K_{P} increase and Q_{P} remains same, then . Again, the reaction moves forward. What, if K_{P} increase and Q_{P} also increases?
Will or or ? This can be answered by simply looking at the dependence of Q_{P} and K_{P} on temperature. You can see from the equation (6) that K_{P} depends on temperature exponentially. While Q’s dependence on T would be either to the power g,l,t…….. Therefore the variation in K_{P} due to T would be more than in Q_{P} due to T.
\ K_{P} would still be greater than Q_{P} and the reaction moves forward again.
Therefore, to see the temperature effect, we need to look at K_{P} only. If it increases the reaction moves forward, if it decreases, reaction moves backward and if it remains fixed, then, no change at all.
 Degree of Dissociation (a)
Let us consider the reaction,
2NH_{3} (g) N_{2} (g) + 3H_{2} (g)
Let the initial moles of NH_{3}(g) be ‘a’. Let x moles of NH_{3} dissociate at equilibrium.
2NH_{3} (g) N_{2} (g) + 3H_{2} (g)
Initial moles a 0 0
At equilibrium a–x
Degree of dissociation (a) of NH_{3} is defined as the number of moles of NH_{3} dissociated per mole of NH_{3}.
\ if x moles dissociate from ‘a’ moles of NH_{3}, then, the degree of dissociation of NH_{3} would be .
We can also look at the reaction in the following manner.
2NH_{3} (g) N_{2} (g) + 3H_{2} (g)
Initial moles a 0 0
At equilibrium a(1–a)
or a–2x¢ x ¢ 3 x ¢
where a=
Here total no. of moles at equilibrium is
a – 2x¢ + x¢ + 3x¢ = a + 2x¢
Mole fraction of NH_{3} =
Mole fraction of N_{2} =
Mole fraction of H_{2}=
The expression of K_{p} is
K_{p} =
=
In this way you should calculate the basic equation. So my advice to you is that, while solving problem follow the method given below:
 Write the balanced chemical reaction (mostly it will be given)
 Under each component write the initial no. of moles.
 Do the same for equilibrium condition.
 Then derive the expression.
Do it and you are the winner.
Exercise1: H_{2} and I_{2} are mixed at 400°C in a 1.0 L container and when equilibrium established, the following concentrations are present: [HI] = 0.49 M, [H_{2}] = 0.08 M and [I_{2}] = 0.06 M. If now an additional 0.3 mol of HI are added, what are the new equilibrium concentrations, when the new equilibrium
H_{2} + I_{2} 2HI is reestablished?
6.1 Dependence of Degree of Dissociation from Density Measurements
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 K_{P} exist, i.e., reactions having at least one gas and having no solution.
Since PV = nRT
PV =
M =
\ VD =
Since P =
\ VD =
For a reaction at eqb., V is a constant and r is a constant. \ vapour Density
\ =
(Q molecular weight = 2 ´ V.D)
Here M = molecular weight initial
m = molecular weight at equilibrium
Let us take a reaction
PCl_{5} PCl_{3} + Cl_{2}
Initial moles C 0 0
At eqb. C(1a) Ca Ca
\ =
Knowing D and d, a can be calculated and so for M and m.
Illustration3: In an experiment 5 moles of HI were enclosed in a 5 litre container. At 717 K equilibrium constant for the gaseous reaction 2HI (g) H_{2} (g) + I_{2 }(g) is 0.025. Calculate the equilibrium concentrations of HI, H_{2} and I_{2}. What is the fraction of HI that decomposes.
Solution: Let 2n be the number of moles of HI which is decomposed, the number of moles of H_{2} and I_{2} produced will be n moles each. Then molar concentrations of various species at equilibrium are –
[HI] = mol/l, [H_{2}] = mol/l, [I_{2}] = mol/l
Also, K_{C} = =
0.025 =
Solving for n, we get n = 0.6
\ [HI] = = = 0.76 mol/l
[H_{2}] = = 0.12 mol /l
[I_{2}] = = 0.12 mol /l
Fraction of HI decomposed = = 0.24 or 24%
Illustration4: 1 mole of N_{2} and 3 moles of PCl_{5} are placed in a 100 litre container heated to 227°C. The equilibrium pressure is 2.05 atm. Assuming ideal behaviour, calculate the degree of dissociation of PCl_{5} and value of K_{p} for its dissociation.
Solution: Dissociation of PCl_{5} is written as
PCl_{5}(g) PCl_{3} + Cl_{2}(g)
Let x be the no. of moles of PCl_{5} decomposed at equilibrium
Moles  PCl_{5}  PCl_{3}  Cl_{2} 
Initial  3  0  0 
At equilibrium  3x  x  X 
Now total gaseous moles in the container = n_{T}
n_{T} = moles of (PCl_{5} + PCl_{3} + Cl_{2}) + moles of N_{2}
n_{T} = 3 – x + x + x + 1 = 4 + x
The mixture behaves ideally, hence PV = n_{T}RT
Let us calculate no. of moles by using gas equation
Þ n_{T} =
Þ n_{T} = 5
now equating the two values of n_{T}, we have
4 + x = 5 Þ x = 1
Þ degree of dissociation = 1/3 = 0.333
Now K_{p} =
P = equilibrium pressure
= = 0.82 atm
atm
Þ K_{p} = = 0.205 atm
K_{p} = 0.205 atm
Note 1: The inert gases like N_{2} or noble gases (He, Ne etc.) though do not take part in the reaction, but still they affect the degree of dissociation and equilibrium concentrations for the reactions in which Dn ¹ 0. They add to the total pressure of the equilibrium mixture (p a n).
Exercise 2: Determine K_{c} for the reaction ½N_{2}(g) + ½O_{2}(g) + ½Br_{2}(g) NOBr(g) from the following information (at 298^{o}K)
K_{c} = 2.4 ´ 10^{30} for 2NO(g) N_{2}(g) + O_{2}(g) ;
= 1.4 for NO(g) + ½Br_{2}(g) NOBr(g)
 Thermodynamics of chemical equilibrium
DG = DG^{0} + 2.303 RT logQ …(1) (Q = Reaction Quotient)
Under equilibrium condition Q = K_{C} or K_{P}
The Gibbs free energy G for a system is defined as G = H – TS
DG = DH – TDS …(2)
DG^{0} = DH^{0} – TDS …(3) (standard condition)
For a process occurring at constant T and P if (1) DG < O, the process is spontaneous.
(2) DG > O, the process is nonspontaneous
(3) DG = 0, the process is at equilibrium
\ At equilibrium
0 =DG^{0} + 2.303 RT logK
\DG^{0} = –2.303 RT logK …(4)
\logK = – Þ lnK =
From equation (3) and (4)
– 2.303 RT logK = DH^{0} – TDS^{0}
Þ logK =
logK_{1} = (At temperature T_{1})
logK_{2} = (At temperature T_{2}, T_{1} < T_{2})
(Assuming DH^{0} to be constant in temperature range T_{1} and T_{2})
\ lok
Þlog
Also
Graph between logK and
K is referred to as thermodynamic equilibrium constant and can be replaced by K_{C} or K_{p} for general reaction.
aA + bB mC + nD
K = (Thermodynamic equilibrium constant)
 i) For pure solids and liquids a = 1
 ii) For gases ideal behaviour is considered and activity is equal to its pressure in (atm)
iii) For components in solution a = molar concentration.
Activation energies for forward and backward reaction: Both forward and backward reactions follow the same reaction path and involve the same activated complex. However, the activation energies of both reactions are different.
DH = DE = E_{a(f)} – E_{a(b) } (at constant volume)
For exothermic reaction Ea_{(f)} < E_{a(b)}
Hence, DE is negative
For endothermic reaction, E_{a(f)} > E_{a(b)}
Hence DE is positive
 Solution to Exercises
Exercise1: First determine the equilibrium constant
K_{C} for H_{2} + I_{2} 2HI
K_{C} =
When 0.3 mol of HI are added, equilibrium is disturbed. At the instant,
[HI] = 0.49 + 0.3 = 0.79 M
Þ Q > K_{C}, since K_{C} =
Þ backward reaction dominate and the equilibrium shifts to the left.
Let 2x = concentration of HI consumed (while going left) then concentration of each of H_{2} and I_{2} formed = x
Þ [HI] = 0.79 – 2x, [H_{2}] = 0.08 + x, [I_{2}] = 0.06 +x and K_{C} = 50
Þ K_{C} =
Þ 46x^{2} + 10.2x – 0.35 = 0
Þ x = 0.033 or – 0.25 (neglecting the –ve value)
Finally, the equilibrium concentrations are:
[HI] = 0.79 – 2x =0.79 – 0.033 ´ 2 = 0.724 M
[H_{2}] = 0.08 + x = 0.08 + 0.033 = 0.113 M
[I_{2}] = 0.06 + x = 0.06 + 0.033 = 0.093
Exercise2: N_{2}(g) + O_{2}(g) 2NO (g)
K_{c}^{/} =
 i) ½N_{2}(g) + ½O_{2}(g ) NO(g)
K_{c}^{//} =
 ii) NO(g) + ½Br_{2}(g) NOBr(g) = 1.4.
(i) + (ii) gives the net reaction:
½N_{2}(g) + ½O_{2}(g) + ½Br_{2}(g) NOBr(g)
K =
= ´ = 0.6455 ´ 10^{15} ´ 1.4 = 9.037 ´ 10^{16}
9. Solved Problems
9.1. Subjective
Problem 1: At 540 K, 0.10 mol of PCl_{5} are heated in a 8 litre flask. The pressure of equilibrium mixture is found to be 1.0 atm. Calculate K_{p} and K_{C} for the reaction.
Solution: PCl_{5} PCl_{3} + Cl_{2}
Initial moles 0.1 0 0
Moles at equilibrium (0.1 – a) a a
(a = degree of dissociation)
Applying (PV = nRT) for the equilibrium mixture at equilibrium
1 × 8 = (0.1 + a) × 0.082 × 540
\ a = 0.08
\ K_{C} = = 4 × 10^{–2} mol/litre
K_{p} = K_{C}(RT)^{D}^{n}
= 4 × 10^{–2} × (0.082 × 540)
= 1.77 atm
Problem 2: For a gaseous phase reaction A + 2B AB_{2} K_{C} = 0.3475 lt^{2} mol^{–2} 200°C. When 2 mol of B are mixed with one mole of A, what total pressure is required to convert 60% of A in AB_{2}.
Solution: A + 2B AB_{2}
Initial moles 1 2 0
Moles at eqm (1 – a) (2 – 2a) a
Total moles at equilibrium (ån) = 1 – a + 2 – 2a + a
K_{p} =
Þ K_{p} =
=
Þ K_{p} = …(1)
Also K_{p} = K_{C}(RT)^{D}^{n}
Þ K_{p} = 0.3475 × (0.0821 × 473)^{–2} …(2)
From equation (1) and (2)
P = 181.5 atm
Problem 3: A sample of air consisting of N_{2} and O_{2} was heated to 2500 K until the equilibrium
N_{2(g)} + O_{2(g)} 2NO_{(g)}
was established with an equilibrium constant K_{C} = 2.1 × 10^{–3}. At equilibrium mole % of NO was 1.8. Estimate the initial composition of air in mole fraction of N_{2} and O_{2}
Solution: N_{2} + O_{2} 2NO Let total no. of mole = 100
Initial mole n (100 – n) 0 Moles of N_{2}
At equilibrium (n – x) (100 – n – x) 2x
ån = (n – x) + (100 – n – x) + 2x = 100
At equilibrium mole % of NO = × 100 = 1.8
\ x = 0.9
As Dm_{g} = 0, \ K_{p}= K_{C}
\K_{p} – K_{C} = = = 2.1 × 10^{–3}
\n = 79%, (10 – n) = 21%
Problem 4: For the reaction [Ag(CN)_{2}] ^{–} Ag^{+} + 2CN^{–} the equilibrium is constant 4.0 × 10^{–19}. Calculate the silver ion concentration in a solution which is originally 0.10 M in KCN and 0.03 M in AgNO_{3}.
Solution: [Ag(CN)_{2}]^{–} Ag^{+} + 2CN^{–} K_{C} = 4 × 10^{–19}
Hence for Ag^{+} + 2CN^{–} [Ag(CN)_{2}]^{–} ¢_{C} = 0.25 × 10^{19}
Very large value of K¢_{C} shows that complex forming equilibrium is spontaneous and almost all the Ag^{+} ion would have reacted leaving xM in solution.
Ag^{+} + 2CN^{– } [Ag(CN)_{2}]^{–}
Initial mole 0.03 0.1 M 0
At equilibrium xM (0.1 – 2 × 0.03)M 0.03 M
K¢_{C} = = 0.25 ×10^{19}
\ 0.25 × 10^{19} =
\ [Ag^{+}] = x = 7.5 × 10^{–18}M
Problem 5: DG^{0} = 77.77 kJ/mol at 100 K for the reaction N_{2(g)} + O_{2(g)} NO_{(g)}. What is the partial pressure of NO under equilibrium at 1000K for air at 1 atm containing 80% N_{2} and 20% O_{2} by volume.
Solution: DG^{0} = – 2.303 RT log_{10}K_{C}
Þ 77.77 × 10^{3} = – 2.303 × 8.314 × 1000 × logK_{C}
\ K_{C}= 8.67 × 10^{–5}
N_{2} + O_{2} NO
Initial pressure 0.8 0.2
At eqm. pressure x
\ 8.67 × 10^{–5}=
\ x = 3.47 × 10^{–5} atm
Problem 6: NH_{3} is heated at 15 atm from 27°C to 347°C assuming volume constant. The new pressure becomes 50 atm at the equilibrium of the reaction
2NH_{3} N_{2} + 3H_{2}. Calculate % of mole of NH_{3} actually decomposed.
Solution: 2NH_{3} N_{2} + 3H_{2}
Initial mol a 0 0
(a – 2a) a 3a
Initial pressure of NH_{3} of a mol = 15 atm, at 27°C
The pressure of a mol of NH_{3} = P atm at 347°C
\ P = 31 atm
At constant volume at 347°C mole µ pressure
a µ 31 atm (before equilibrium)
(a + 2x) µ 50 atm (after equilibrium)
\
\ x = a
\ % of NH_{3} decomposed = = 61.3%
Problem 7: K_{p} for the reaction PCl_{5} PCl_{5} + Cl_{2} at 250°C is 0.82. Calcualte the degree of dissociation at given temperature under a total pressure of 5 atm. What will be the degree of dissociation if the equilibrium pressure is 10 atm at same temperature.
Solution: Let 1 mole of PCl_{5} be taken initially. If ‘x’ moles of PCl_{5} dissociate at equilibrium, its degree of dissociation = x
Moles  PCl_{5}  PCl_{3}  Cl_{2} 
Initially  1  0  0 
at equilibrium  1x  x  X 
Total moles  1 – x + x + x = 1 + x 
P = 5 atm K_{p} = 0.82
=
K_{p} =
K_{p} =
x = 0.375 (0r 37.5%)
Now the new pressure P = 10 atm
Let y be the new degree of dissociation. As the temperature is same (250°C), the value of K_{p} will remain same. i.e. the same manner. Proceeding in the same manner.
K_{p} =
Þ y = or y = 0.275 (or 27.5%)
Note 1: by increasing pressure, degree of dissociation has decreased, i.e., the system shifts to reverse direction. Compare the result by applying Le Chatelier’s principle.
Problem8: The value of K_{C} for the reaction: A_{2}(g) + B_{2}(g) ] 2AB(g) at 100°C is 50. If 1.0 L flask containing one mole of A_{2} is connected with a 2.0 L flask containing two moles of B_{2}, how many moles of AB will be formed at 100°C?
Solutions: A_{2}(g) 2AB(g)
As the two vessels are connected, the final volume for the contents is now 3.0 L. Let x mole each of A_{2} and B_{2} react to form 2x moles of AB_{2} (from stoichiometry of reaction)
Moles  A_{2}  B_{2}  2AB 
Initially  1  2  0 
at equilibrium  1x  2x  2x 
K_{C} = ; concentration of species at equilibrium are:
[A_{2}] = (1 – x)/3, [B_{2}] = (2x)/3, [AB] = 2x/3
K_{C} =
Þ 46x^{2} + 150x + 100 = 0
Þ x = 0.93 or 2.33 (neglecting this value)
Þ moles of AB(g) formed at equilibrium = 2x = 1.86
Problem 9: Variation of equilibrium constant K with temperature T is given by Van’t Hoff equation
log K = log A –
A graph between log K and T^{–1} was a straight line as shown and having OP = 10 and tan q = 0.5. Calculate
(i) equilibrium constant at 298 K, and
(ii) and equilibrium constant at 798 K, assuming DH° to be independent of temperature.

Solution: To calculate equilibrium constant, we need to know A and DH°, which are calculated as –
The given equation represents a straight line of slope =
= –tan q = –0.5
\ DH° = 2.303 ´8.314 ´0.5 = 9.574 J/mol
Intercept = log A = OP = 10
\ log K = log A–
\ K = 9.96 ´10^{9}
Now, to calculate equilibrium constant at some other temperature, we will use the expression
=
log =
\ K_{2} (equilibrium constant at 798) = 9.98 ´ 10^{9}
Problem10: The equilibrium constant K_{P} of the reaction
2SO_{2}(g) + O_{2}(g) 2SO_{3}(g)
is 900 atm^{–1} at 800°C. A mixture containing SO_{3} and O_{2} having initial partial pressures of 1 atm and 2 atm respectively is heated at constant volume to equilibrate. Calculate the partial pressure of each gas at 800°C.
Solution: It can be seen that as SO_{2} is not present initially, so equilibrium cannot be established in the forward direction. Therefore it is established from reverse direction. Let n be the increase in partial pressure of O_{2}. Then at equilibrium the partial pressures of SO_{2}, O_{2} and SO_{3} are (0+2n), (2+n) and (1–2n) in atm respectively.
2SO_{2}(g) + O_{2}(g) 2SO_{3}(g)
(0+2n) (2+n) (1–2n)
Also,
As n is small (because equilibrium constant for the reverse reaction is very small i.e., 1/900), it can be neglected in comparison to 2 and also
1–2n can be taken approximately to 1.
\ 900 =
Solving for n, we get n = 0.0118
Hence, = 2+n = 2.0118 atm
= 2n = 0.0236 atm
= 1–2n = 1–0.0236 = 0.9764 atm
Problem 11: At a given temperature and a total pressure of 1.0 atm for the homogenous gaseous reaction N_{2}O_{4} 2NO_{2}, the partial pressure of NO_{2} is 0.5 atm
 a) Find the value of K_{p}
 b) If the volume of the vessel is decreased to half of its original volume, at constant temperature, what are the partial pressures of the components of the equilibrium mixture.
Solution: For equilibrium system, N_{2}O_{4} 2NO_{2}, the total pressure is 1.0 atm
Þ the total pressure =
Þ
 i) = 0.5 atm
 ii) as volume is decreased to half its original volume, equilibrium is disturbed and the new initial conditions for the reestablishment new equilibrium are
atm & = 1.0 atm
According to Le Chatelier principle, when volume is decreased, the system moves in that direction where there is decrease in number of moles. Hence, the system (here will move in reverse direction, as there is a decrease in mole (Dn = 2 – 1 = 1) i.e. the NO_{2} will be converted to N_{2}O_{4}.
Let the decrease in pressure of NO_{2} be x atm.
Pressure  N_{2}O_{4}  NO_{2} 
Initial  1.0  1.0 
At equilibrium  1 + x/2  1 – x 
Þ K_{p} =
Þ 4x^{2 } – 9x + 2 = 0
Þ x = 2 or 0.25 (x ¹ 2 as initial pressure = 1.0)
Þ x = 0.25
&
= 0.75 atm
Problem12: Ammonium hydrogen sulphide dissociated according to the equation
NH_{4}HS(s) NH_{3}(g) + H_{2}S(g). If the observed pressure of the mixture is 1.12 atm at 106^{o}C, what is the equilibrium constant K_{p} of the reaction ?
Solution: The reaction is
NH_{4}HS(s) NH_{3}(g) + H_{2}S(g)
If a is the degree of dissociation of equilibrium,
Total moles of NH_{3} + H_{2}S = 2a.
Partial pressure =
\ p_{NH3} = ´ P = 0.5P; p_{H2S} = ´ P = 0.5P
K_{p} = p_{NH3} ´ p_{H2S} = 0.5P ´ 0.5P = 0.25P^{2}
Substituting the value of P = 1.12 atm.,
K_{p} = 0.25 ´ 1.12 ´ 1.12 = 0.3136.
Problem 13: In a mixture of N_{2} and H_{2} initially in a mole ratio of 1:3 at 30 atm and 300^{o}C , the percentage of ammonia by volume under the equilibrium is 17.8. Calculate the equilibrium constant (K_{P}) of the mixture, for the reaction
N_{2}(g) + 3H_{2}(g) 2NH_{3}(g)
Solution: Let the initial moles of N_{2} and H_{2} be 1 and 3 respectively (this assumption is valid as K_{P} will not depend on the exact no. of moles of N_{2} and H_{2}. One can even start with x and 3x).
Alternatively
N_{2}(g) + 3H_{2}(g) 2NH_{3}
Initial 1 3 0
At eqb 1x 33x 2x
Since % by volume of a gas is same as % by mole,
\ =0.178
\
\ Mole fraction of H_{2} at equilibrium =
Mole fraction of N_{2} at equilibrium = 1 – 0.6165 – 0.178 = 0.2055
\K_{P} =
= 7.31 ´ 10^{4}
N_{2}(g) + 3H_{2}(g) 2NH_{3}(g)
Initia x 3x 0
At eqb. xa 3x3a 2a
\ =0.178
\ = 0.177
\ = 0.302
Similarly we can calculate the mole fraction of N_{2}(g) and H_{2}(g) at equilibrium.
Problem 14: The density of an equilibrium mixture of N_{2}O_{4} and NO_{2} at 1 atm. and 348 K is 1.84 g dm^{3}. Calculate the equilibrium constant of the reaction
N_{2}O_{4}(g) 2NO_{2}(g).
Solution: Let us assume that we start with C moles of N_{2}O_{4}(g) initially.
N_{2}O_{4}(g) 2NO_{2}(g)
Initial 0 0
At equilibrium C(1a) 2Ca
Where a is the degree of dissociation of N_{2}O_{4}(g)
Since
=
Initial vapour density =
Since vapour density and actual density are related by the equation, V.D. = = = 26.25
\ 1 + a =
\ a = 0.752
 K_{p} =
 = 2 atm.
Problem15: At temperature T, a compound AB_{2}(g) dissociates according to the reaction 2AB_{2}(g) 2AB(g) + B_{2}(g) with a degree of dissociation x, which is small compared with unity. Deduce the expression for x in terms of the equilibrium constant, K_{p} and the total pressure P.
Solution: Let the initial pressure of AB_{2}(g) be P_{1}
2AB_{2}(g) 2AB(g) + B_{2}(g)
P_{1}(1x) P_{1}x
The total pressure is = P_{1 }<< 1)
 P_{1} = P
 So, K_{P} =
9.2 Objective
Problem 1: For the gaseous phase reaction
2A + B C+ D,
The value of K_{p} is . What is the value of DG° at 27°C.
(A) – 600 K cals (B) – 6 Kcals
(C) 10^{3} Kcals (D) 10 Kcals
(E) None
Solution: DG° = – 2.303 RT log K
= –2.303 ´ 2 cal. K^{–1 }´ 300 K log 10
= – 2.303 ´ 2 ´ 300 cals ´
= – 600 Kcals
\(A)
Problem2: For the reaction
A + B 2C + D
The initial concentration of A and B are 1M each. The value of K_{c} is 10^{8}. What is equilibrium concentration of A?
(A) 2 ´ 10^{–4} M (B) 2 ´ 10^{4} M
(C) .005 M (D) .0025 M (e) None
Solution:. A + B 2C + D
t= 0 1M 1M 0 0
–(1–x) (1–x) (2–2x) (1–x)
——————————————————————————
xM x (2–2x) (1–x)
K = » = 10^{8}
\ x = 2 ´ 10^{–4}
\(A)
Problem 3: The solid dissociates as H_{2}NCOONH_{4}(s) 2NH_{3}(g) + CO_{2}(g)
The equilibrium total pressure is 0.3 atm. What is the value of K_{p}
(A) 4 ´ 10^{–3} (B) 4
(C) 10^{–3} (D) 8 ´ 10^{3}
Solution: NH_{2}COONH_{4}(s) 2NH_{3} + CO_{2}
2x x
3x = .3
x = .1
K_{p} = (2x)^{2} ´ x = 4x^{3} = 4 ´ .1^{3} = 4 ´ 10^{–3}
^{ }\(A)
Problem4: For the gas phase reaction 2NO(g) N_{2}(g) + O_{2}(g) , DH = –43.5 kcal, which one of the following is true for N_{2}(g) + O_{2}(g) 2NO(g)
(A) K is independent of T (B) K decreases as T decreases
(c) K increases as T decreases (d) K varies with addition of NO
Solution: The given reaction N_{2}(g) + O_{2}(g) 2NO(g) is endothermic. Therefore, according to Le–Chatlier’s Principle, high temperature favours forward reaction and hence K increases as T increases or K decreases as T decreases.
\ (b)
Problem5: For the reaction PCl_{3} (g) + Cl_{2}(g) PCl_{5}(g), the value of K_{P} at 250°C is 0.61 atm^{–1}. The value of K_{C} at this temperature will be
(A) 15 (mol/l)^{–1} (B) 26 (mol/l)^{–1}
(C) 35 (mol/l)^{–1} (D) 52 (mol /l)^{–1}
Solution: PCl_{3}(g) + Cl_{2}(g) PCl_{5}(g)
Dn = –1, K_{P} = 0.61 atm^{–1}.
K_{C} = K_{P} (RT)^{–}^{D}^{n}
= 0.61 (0.0821 ´523)^{+1} = 26 mol /l.
\(b)
Problem6: In the reaction A_{2}(g) + 4B_{2} (g) 2AB_{4}(g) , DH > 0. The decomposition of AB_{4} (g) will be favoured at
(A) low temperature and high pressure
(B) high temperature and low pressure
(C) low temperature and low pressure
(D) high temperature and high pressure
Solution: 2AB_{4}(g) A_{2}(g) + 4B_{2}(g) DH = –ve
It is an Exothermic reaction and hence favoured at low temperature. Dn for the reaction is +3. Therefore low pressure will favour the forward reaction
\(C)
Problem 7: For the reaction SO_{2}(g) + O_{2}(g) SO_{3}(g) K_{p} = 1.7 ´ 10^{12} at 20°C and 1 atm pressure. Calculate K_{c}.
(A) 1.7 ´ 10^{12} (B) 0.7 ´ 10^{12}
(C) 8.33 ´ 10^{12} (D) 1.2 ´ 10^{12}
Solution: For the reaction SO_{2}(g) + O_{2}(g) SO_{3}(g)
Given K_{P} = 1.7 ´ 10^{12} at 20°C and 1 atm pressure
K_{P} = K_{c} (RT)^{D}^{n}; Dn = 1–
1.7 ´ 10^{12} = K_{c} (0.0821 ´ 293)^{–1/2}
K_{c} =
K_{c} = 7 ´ 10^{12} ´ (0.0821 ´ 293)^{+1/2}
= 8.338 ´ 10^{12}
\(C)
Problem8: Applying the law of mass action to the dissociation of hydrogen Iodide
2HI(g) H_{2} + I_{2}
We get the following expression
Where a is original concentration of H_{2}, b is original concentration of I_{2}, x is the number of molecules of H_{2} and I_{2} reacted with each other. If the pressure is increased in such a reaction then
(A) K = (B) K >
(C) K < (D) none of these
Solution: For the reaction H_{2} + I_{2} 2HI (g) since Dn = 0. Equilibrium constant K_{C} will be uneffected by change in pressure.
\ (A)
Problem9: For a gaseous equilibrium
2A(g) 2B(g) + C(g) , K_{p} has a value 1.8 at 700°K. What is the value of K_{c} for the equilibrium 2B(g) + C(g) 2A at that temperature
(A) » 0.031 (B) » 32
(C) » 44.4 (D) » 1.3 ´ 10^{–3}
Solution: For 2A(g) 2B(g) + C(g)
K_{P} = 1.8 at 700°C
\ For 2B(g) + C(g) 2A(g) at 700°C
K_{P} = Dn = 2 – (2+1) = –1
K_{P} = K_{c}(RT)^{D}^{n}
K_{c} =
K_{c} = K_{P} ´ (0.0821 ´ 973)^{1 }
= ´ 0.0821 ´ 973
K_{c} = 44.3796
\ (C)
Problem10: The following equilibrium when methyl mercaptan (CH_{3}SH – the smell in onions) is dissolved in water, which statement is false?
CH_{3}SH + H_{2}O CH_{3}S^{–} + H_{3}O^{+}
(A) CH_{3}SH is weak acid
(B) H_{2}O is a weak base
(C) CH_{3}S^{–} is the conjugate base
(D) H_{3}O^{+} is the conjugate acid of H_{2}O
Solution: Since CH_{3}SH is acid, CH_{3}S^{–} will be its conjugate base.
\ (C)
Problem11: In a closed container at 1 atm pressure 2 moles of SO_{2}(g) and 1 mole of O_{2}(g) were allowed to react to form SO_{3}(g) under the influence of catalyst . Reaction 2SO_{2}(g) + O_{2}(g) 2SO_{3}(g) occurred
At equilibrium it was found that 50% of SO_{2}(g) was converted to SO_{3}(g). The formal pressure of O_{2} (g) at equilibrium will be
(A) 0.66 atm (B) 0.493 atm
(C) 0.33 atm (D) 0.20 atm
Solution: For H_{2}(g) + O_{2}(g) 2SO_{3}(g)
t = 0 2 moles 1 moles 0
t = equilibrium 2(1 – 0.5) 1(1– 0.5) 1 mole
2–1 + 0.5 +1 = 2.5
P¢ = P ´ mF
= = 1 ´ = 0.20 atm
\ (D)
Problem12: For the reaction N_{2}O_{4}(g) 2NO_{2}(g) the reaction connecting the degree of dissociation (a) of N_{2}O_{4}(g) with its equilibrium constant K_{P} is
(A) a = (B) a =
(C) a = (D) a =
Solution: N_{2}O_{4}(g) 2NO_{2}(g)
t = 0 1 0
t = equilibrium 1–a 2a
ån = 1 + a
K_{p} =
K_{p} =
\(C)
Problem13: 40% of a mixture of 0.2 mol of N_{2} and 0.6 mol of H_{2} react to give NH_{3} according to the equation: N_{2}(g) + 3H_{2}(g) 2NH_{3}(g) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases are
(a) 4:5 (b) 5:4
(c) 7:10 (d) 8:5
Solution: N_{2} + 3H_{2} 2NH_{3}
t = 0 0.2 0.6 0
at = t 0.2–n 0.6–3x 2x
40% of N_{2} = 0.2 ´0.4 = 0.08
40% of H_{2}= 0.6 ´0.4 = 0.24
\ number of moles of N_{2} remaining = 0.2 – 0.08 = 0.12
number of moles of H_{2} remaining = 0.6–0.24 = 0.36
number of moles of NH_{3} formed = 0.16
Total number of moles = 0.12 + 0.36 + 0.16 = 0.64
\ =
\(a)
Problem 14: One mole of N_{2}O_{4}(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when N_{2}O_{4} (g) decomposes to NO_{2}(g). If the resultant pressure is 2.4 atm, the percentage dissociation by mass of N_{2}O_{4} (g) is
(A) 10% (B) 20%
(C) 30% (D) 40%
Solution: N_{2}O_{4}(g) 2NO_{2}(g)
Equilibrium 1–a 2a
Where a is the degree of dissociation
= ( Q V is constant )
\ P_{2} = 2 atm
Ater dissociation of N_{2}O_{4} at 600K,
= 2 (1–a) = 2–2a
= 2 ´ 2a = 4 a
Total pressure = 2–2a+4a = 2+2a
2+2a = 2.4 (Given)
a = 0.2
\ Percentage dissociation = 20%
\(b)
Problem 15: K for the synthesis of HI (g) is 50. The degree of dissociation of HI is
(A) 0.10 (B) 0.14
(C) 0.18 (D) 0.22
Solution: 2HI(g) H_{2}(g) + I_{2}(g)
1–a Where a is the degree of dissociations
a = 2–2a
2+5 a = 2
a = = 0.22
\(d)
 Assignments (Subjective Problems)
Level – I
 The value of K_{p} for the reaction
2H_{2}O(g) + 2Cl_{2}(g) 4HCl(g) + O_{2}(g)
is 0.035 atm at 400^{o}C, when the partial pressures are expressed in atmosphere. Calculate K_{c} for the reaction,
Cl_{2}(g) + H_{2}O(g)
 K_{p} for the reaction N_{2}O_{4} (g) 2NO_{2} (g) is 0.66 at 46°C. Calculate the percent dissociation of N_{2}O_{4} at 46°C and a total pressure of 0.5 atm. Also calculate the partial pressure of N_{2}O_{4} and NO_{2} at equilibrium.
 The density of an equilibrium mixture of N_{2}O_{4} and NO_{2} at 1 atm. and 348 K is 1.84 g dm^{3}. Calculate the equilibrium constant of the reaction,
N_{2}O_{4}(g) 2NO_{2}(g).  Ammonium hydrogen sulphide dissociated according to the equation
NH_{4}HS(s) NH_{3}(g) + H_{2}S(g). If the observed pressure of the mixture is 1.12 atm at 106^{o}C, what is the equilibrium constant K_{p} of the reaction ?
 In a mixture of N_{2} and H_{2} initially in a mole ratio of 1:3 at 30 atm and 300^{o}C, the percentage of ammonia by volume under the equilibrium is 17.8. Calculate the equilibrium constant (K_{P}) of the mixture, for the reaction,
N_{2}(g) + 3H_{2}(g) 2NH_{3}(g)
 25 c.c. of H_{2} and 18 c.c. of I_{2} vapour were heated in a sealed tube at 456°C and at equilibrium 30.8 c.c. of HI were formed. Calculate the degree of dissociation of pure HI at 456°C.
 The pressure of iodine gas at 1273 K is found to be 0.112 atm whereas the expected pressure is 0.074 atm. The increased pressure is due to dissociation
I_{2} 2I. Calculate K_{P}.
 0.0755 gm of Selenium vapour occupying a volume of 114.2 ml. at 700°C exerts a pressure of 185 mm. The Selenium is in a state of equilibrium according to the reaction ; Se_{6}(g) 3Se_{2}(g). Calculate (i) the degree of dissociation of Selenium (ii) K_{P} and (iii) K_{C} . [At. weight of Se = 79]
9 A sample of air, consisting of N_{2} and O_{2} was heated to 2500 K until the equilibrium, N_{2}(g) + O_{2}(g) 2NO (g) was established with an equilibrium constant K_{C} = 2.1 ´10^{–3}. At equilibrium, the mole percentage of NO was 1.8. Estimate the initial composition of air in terms of mole fraction of O_{2} and N_{2} assuming the moles of N_{2} were greater than of O_{2}, initially.
 To 500 ml of 0.150 M AgNO_{3} solution we add 500 ml of 1.09 M Fe^{2+} solution and the reaction is allowed to reach equilibrium at 25°C.
Ag^{+} (aq) + Fe^{2+} (aq) Fe^{3+} (aq) + Ag(s)
For 25 ml. of the solution, 30 ml. of 0.0832 M KMnO_{4} were required for oxidation under acidic conditions. Calculate K_{C} for the reaction.
 N_{2}O_{4} (g) dissociates as follows: N_{2}O_{4}(g) 2NO_{2} (g)
The density of equilibrium mixture is found to be 3.62 gL^{–1} at 298K and at 1 atmospheric pressure. Calculate the equilibrium constant for the reaction.
 When a–D–glucose is dissolved in water, it undergoes a partial conversion to
b–D–glucose to exhibit mutarotation. This conversion stops when 63.6% of glucose is in b–form. Assuming that equilibrium has been attained,
calculate K_{C} for mutarotation.
13. The K_{P} for the reaction, H_{2} + I_{2} 2HI at 460°C is 49. If the initial partial pressure of H_{2} and I_{2} each is 0.5 atm respectively, determine the partial pressure of each gas at equilibrium.
 N_{2}O_{4} is 25% dissociated at 37°C and one atmospheric pressure. Calculate (i) K_{P} and (ii) percent dissociation at 0.1 atmospheric pressure and 37°C.
 When 3.06 g of solid NH_{4}HS is introduced into a two litre evacuated flask at 27°C, 30% of the solid decomposes into gaseous ammonia and hydrogen sulphide. (i) Calculate K_{C} and K_{P} for the reaction at 27°C. (ii) What would happen to the equilibrium when more solid NH_{4}HS is introduced into the flask?
Level– II
 Two solid compounds A and C dissociates into gaseous product at temperature T as follows:
(i) A(s) B(g) + D(g)
(ii) C(s) E(g) + D(g)
At 20°C pressure over excess solid A is 50 atm and that over excess solid C is 68 atm. Find the total pressure of the gases over the solid mixture.
 For the reaction NH_{3} (g) N_{2} (g) + H_{2} (g)
Show that the degree of dissociation of NH_{3} is given as
a =
where P is the equilibrium pressure. If K_{P} of the above reaction is 78.1 atm at 400°C, determine the value of K_{C}.
 Calculate the total pressure at 400°C that must be applied to a mixture of three parts of hydrogen and one part of nitrogen to give a mixture containing 10% ammonia at equilibrium.
Given: N_{2 } (g) + 3H_{2} (g) 2NH_{3 }(g); K_{p }=1.60 ´ 10^{4 } bar^{2}_{ }at 400°C.
 Given the following equilibrium constant at 1476 K
CO (g) + O_{2 }(g) CO_{2} (g); K_{P} = 2.5 ´10^{5}
C (graphite) + CO_{2} (g) 2CO (g); K_{P} = 1.67 ´10^{3}
Find K_{P} for the reaction C (graphite) + O_{2} CO_{2} (g)
 When one mole of benzoic acid (C_{6}H_{5}COOH) and three moles of ethanol (C_{2}H_{5}OH) are mixed and kept at 200°C until equilibrium is reached, it is found that 87% of the acid is consumed by the reaction.
C_{6}H_{5}COOH(l) + C_{2}H_{5}OH (l) C_{6}H_{5}COOC_{2}H_{5}(l) + H_{2}O(l)
Find out the percentage of the acid consumed when one mole of the benzoic acid is mixed with four moles of ethanol and treated in the same way.
 For the reaction, A+3B 2C, the reaction was carried out at a temperature of 400°C and pressure 1 atm in a closed container with the molar ratio of A:B as 1:3. At equilibrium the mole percent of C was found to be 20%. Calculate K_{P} and K_{C}.
 15 g sample of BaO_{2} is heated to 794°C in a closed evacuated vessel of 5 litre capacity. How many gm of peroxide are converted to BaO (s)?
2BaO_{2} (s) 2BaO (s) + O_{2} (g); K_{P} = 0.5 atm
 Consider the vapour phase dissociation of an oxyacid HXO_{3} according to the equation:
4HXO_{3} (g) 4XO_{2} (g) + 2H_{2}O (g) + O_{2}(g)
Derive the expression : K_{P} = , where P is the total pressure and is the partial pressure of O_{2}.
 A mixture of 0.373 atm of NO(g) and 0.310 atm Cl_{2}(g) is prepared at 500K.
The reaction 2NO(g) + Cl_{2} (g) 2NOCl (g) takes place. The total pressure at equilibrium is 0.544 atm. Determine K_{P} for the reaction.
 For the reaction I_{2} (g) 2I (g), K_{C} = 1.0 ´10^{–2} mol lit^{–1}. What volume of the vessel should be taken so that at equilibrium 1 mole of I_{2} and 0.5 mole of I are present at equilibrium?
 Show that K_{P} for the reaction
2H_{2}S(g) 2H_{2} (g) + S_{2} (g) is given by the expression
K_{P} =
Where a is the degree of dissociation and P is the total equilibrium pressure. Calculate K_{C} of the reaction if a at 298 K and 1 atm pressure is 0.055.
 The density of an equilibrium mixture of N_{2}O_{4} and NO_{2} at 101.325k Pa is 3.62 g dm^{–3} at 288K and 1.84 g dm^{–3} at 348K. What is the heat of reaction for
N_{2}O_{4}(g) 2NO_{2}(g)?
13*. For the equilibrium
LiCl.3NH_{3} (s) LiCl. NH_{3} (s) + 2NH_{3},K_{P} = 9 atm^{2} at 40°C. A five litre vessel contains 0.1 mole of LiCl.NH_{3}. How many minimum moles of NH_{3} should be added to the flask at this temperature to derive the backward reaction for completion?
 What is the percent decomposition of H_{2}S if 1.0 mole of H_{2}S is introduced into 1.1L vessel at 1000K? K_{C} for the reaction:
2H_{2}S(g) 2H_{2}(g) + S_{2}(g) is 1.0 ´10^{–6}.
 At a certain temperature, equilibrium constant (K_{C}) is 16 for the reaction:
SO_{2}(g) + NO_{2} (g) SO_{3}(g) + NO(g)
If we take one mole of each of the four gases in one litre container, what would be the equilibrium concentration of NO and NO_{2}?
LEVEL – III
 A mixture of SO_{3}, SO_{2} and O_{2} gases is maintained in a 10 litre flask at a temperature at which the equilibrium constant (K_{C}) for the reaction: 2SO_{2}(g) + O_{2}(g) 2SO_{3}(g) is 100.
 If the number of moles of SO_{2} and SO_{3} in the flask are equal, how many moles of O_{2} are present?
 If the number of moles of SO_{3} in the flask is twice the number of moles of SO_{2}, how many moles of O_{2} are present?
 0.15 mole of CO taken in a 2.5L flask is maintained at 750 K along with a catalyst so that the following reaction can take place.
CO(g)+2H_{2}(g)CH_{3}OH(g)
Hydrogen is introduced until the total pressure of the system is 8.5 atmosphere at equilibrium and 0.08 mole of methanol is formed. Calculate (a) K_{P} and K_{C} and (b) the final pressure if the same amount of CO and H_{2} as before are used, but with no catalyst so that reaction does not take place.
 At temperature T, a compound AB_{2}(g) dissociates according to the reaction, 2AB_{2}(g) 2AB(g) + B_{2}(g) with degree of dissociation, a, which is small compared to unity. Deduce the expression for a in terms of the equilibrium constant K_{P} and the total pressure P.
 The equilibrium constant of the reaction A_{2}(g) + B_{2}(g) 2AB (g) at 100°C is 50. If one litre flask containing one mole of A_{2} is connected to a two litre flask containing two moles of B_{2}, how many moles of AB will be formed at 373K. [IIT–JEE 1985]
 A saturated solution of I_{2} in water contains 0.33 g of I_{2} L^{–1}. More than this can be dissolved in a KI solution because of the following equilibrium :
I_{2} + I^{–}
A 0.1 M KI solution (0.1M I^{–‑}) actually dissolves 12.5 g I_{2}/ litre, most of which is converted to , assuming that the concentration of I_{2} in all saturated solution is the same, calculate the equilibrium constant (K_{C}) for the above reaction. What is the effect of adding water to a clear saturated solution of I_{2} in the KI solution?
 3. 0.1 mole of ethanol and 0.1 mole of butanoic acid are allowed to react. At equilibrium, the mixture is titrated with 0.85 M NaOH solution and the titre value was 100 ml. Assuming that no ester was hydrolysed by the base, calculate K for the reaction.
C_{2}H_{ 5}OH + C_{3}H_{7}COOH C_{3}H_{7}COOC_{2}H_{5} + H_{2}O
 COF_{2(g)} passed over catalyst at 1000°C comes to equilibrium, thus
2COF_{2}(g) CO_{2}(g) + CF_{4}(g)
Analysis of the equilibrium mixture (after quick cooling to ‘freeze’ the equilibrium) shows that 500 c.c of the equilibrium mixture (STP) contains 300 c.c. (STP) of (COF_{2} + CO_{2}) taking the total pressure to be 10 atm. Calculate K_{p}?
 Ammonia under a pressure of 15 atm at 27° C is heated 347°C in a closed vessle in the presence of catalyst. Under these conditions, NH_{3} partially decomposes to H_{2} and N_{2}. The vessel is such that the volume remains effectively constant, whereas the pressure increases to 50 atm. Calculate the % of NH_{3} actually decomposed.
 Solid Ammonium carbamate dissociates as: NH_{2}COONH_{4}(s) 2NH_{3}(g) +CO_{2}(g). In a closed vessel solid ammonium carbamate is in equilibrium with its dissociation products. At equilibrium ammonia is added such that the partial pressure of NH_{3} at new equilibrium now equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to that of original total pressure.
 Two solids X and Y dissociates into gaseous products at a certain temperature as follows
X(s) A(g) + C(g), and Y(s) B(g) + C(g). At a given temperature, pressure over excess solid X is 40 mm and total pressure over solid Y is 60 mm. Calculate.
 a) the values of K_{p} for two reactions
 b) the ratio of mole sof A and B in the vapour state over a mixture of X and Y
 c) the total pressure of gases over a mixture of X and Y
 Calculate the % age dissociation of H_{2}S(g) if 0.1 mole of H_{2}S is kept in a 0.4 L vessel at 1000 K. The value of K_{C} for the reaction 2H_{2}S (g) 2H_{2}(g) + S_{2}(g) is 1.0 ´ 10^{4}
 Consider the equilibrium: P(g) + 2Q(g) R(g). When the reaction between P and Q is carried out at a certain temperature, the equilibrium concentration of P and Q are 3 M and 4 M respectively. When the volume of the vessel is doubled and the equilibrium is allowed to reestablished, the concentration of Q is found to be 3 M. Find the
 a) value of K_{C}
 b) concentration of R at new equilibrium stages
 The degree of dissociation of HI at a particular temperature is 0.8. Find the volume of 1.5 M sodium thiosulphate solution required to react completely with the iodine present at equilibrium in acidic condition, when 0.135 mol each of H_{2} and IO_{2} are heated at 440 K in a closed vessel of capacity 2.0 L.
 At 817°C, K_{p} for chemical reaction between CO_{2} and excess of hot graphite form 2CO_{(g)} is 10.
 i) What is the fraction of the gases at equilibrium at 817°C and a totral pressure of 4 atm? What is the partial pressure of CO_{2} at equilibrium?
 ii) At what total pressure will the gas mixture have 6% CO_{2} by volume?
 At 25°C and 1 atm, N_{2}O_{4} dissociates by the reaction
N_{2}O_{4}(g) 2NO_{2}(g)
If it is 35%, dissociated at a 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 80 gm sample of N_{2}O_{4} confined to a 7 litre vessel.
iii) What volume of above mixture will diffuse if 20 ml pure O_{2} diffuses in 10 minutes at same temperature and pressure.
 Assignments (Objective Problems)
Level – I
Instruction: In each of the following objective questions, 4/5 alternatives are given. In each one, one or more than one may be correct. It is important to do these questions in one attempt and under a time limit. Therefore you are required to do this assignment in 20 minutes.
 In the reaction C(s)+ CO_{2}(g) 2CO(g) the following amounts of substances were formed in 0.2 litre flask at equilibrium: C = 0.1 mole CO = 0.05 mole CO_{2} = 0.06 mole. The equilibrium constant is
(A) 0.208 (B) 4.10
(C) 0.30 (D) 0.416
 The vapour density of PCl_{5} is 104.25 but when heated to 230°C, its vapour density is reduced to 62. The degree of dissociation of PCl_{5} at this temperature will be
(A) 6.8% (B) 68%
(C) 46% (D) 64%
 For the reaction,
PCl_{5}(g)PCl_{3}(g) + Cl_{2}(g), the forward reaction at constant temperature is favoured by
(A) introducing an inert gas at constant volume
(B) introducing chlorine gas at constant volume
(C) introducing an inert gas at constant pressure
(D) increasing the volume of the container
 For a reversible reaction, if the concentrations of the reactants are doubled, then the equilibrium constant will
 A) be halved (B) be doubled
 C) remain the same (D) will become one fourth
 The vapour density of fully dissociated NH_{4}Cl would be
(A) less than half of the vapour density of pure NH_{4}Cl
(B) double of the vapour density of pure NH_{4}Cl
(C) half of the vapour density of pure NH_{4}Cl
(D) one fourth of the vapour density of pure NH_{4}Cl
 vant Hoff’s equation giving the effect of temperature on chemical equilibrium may be represented as
(A) ln K_{P} = (B) ln K_{C} =
(C) ln K_{P} = (D) ln K_{P} =
 For the reaction A + 2B 2C at equilibrium [C] = 1.4 M , [A]_{O} = 1M, [B]_{O} = 2M, [C]_{O} = 3M, The value of K_{C}is
(A) 0.084 (B) 8.4
(C) 84 (D) 840
 The equilibrium constant K_{C} of the reaction : A_{2} (g) + B_{2} (g) 2AB (g) is 50. If 1 mol of A_{2} and 2 mol of B_{2} are mixed, the amount of AB at equilibrium would be
(A) 0.467 mol (B) 0.934 mol
(C) 1.401 mol (D) 1.866 mol
 In an aqueous solution of volume 500 ml, when the reaction of
2Ag^{+} + Cu Cu^{2+} + 2Ag reached equilibrium the [Cu^{2+}] was x M. When 500 ml of water is further added, at the equilibrium [Cu^{2+}] will be
(A) 2 x M (B) x M
(C) between x M and x /2 M (D) less than x/2 M
 A 10 litre box contains O_{3} and O_{2} at equilibrium at 2000K. K_{P} = 4.17 ´ 10^{14} for 2O_{3} 3O_{2}. Assume that >>_{ }and if total pressure is 7.33 atm, then partial pressure of O_{3} will be
(A) 9.71 ´10^{–5} atm (B) 9.71 ´10^{–7} atm.
(C) 9.71 ´10^{–6} atm (D) 9.71 ´10^{–2} atm
 In the system, LaCl_{3}(s) + H_{2}O (g)+heat¾®LaClO(s)+2HCl (g), equilibrium is established. More water vapour is added to reestablish the equilibrium. The pressure of water vapour is doubled. The factor by which pressure of HCl is changed is
(A) 2 (B)
(C) (D)
 Which one of the following statements is correct?
In a reversible reaction, catalyst
(A) increases the rate of the forward reaction
(B) decreases the rate of the reverse reaction
(C) increases the rate of the forward and reverse reactions equally
(D) alters the equilibrium constant of the reaction
 In what manner will increase of pressure affect the following equation?
C(s) + H_{2}O (g) CO(g) + H_{2}(g)
(A) shift in the forward direction (B) shift in the reverse reaction
(C) increase in the yield of H_{2} (D) no effect
 The reaction A + B C + D proceeds to right hand side upto 99.9%. The equilibrium constant K of the reaction will be
(A) 10^{4} (B) 10^{5}
(C) 10^{6} (D) 10^{8}
 A chemical reaction is said to attain equilibrium when
(A) reactants get completely converted into products
(B) equal amounts of reactants and products are formed
(C) the rate of forward reaction becomes equal to the rate of reverse reaction
(D) the concentration of reactants and products are same
16 Chemical equilibrium is dynamic in nature because
(A) the equilibrium is maintained rapidly
(B) the concentration of reactants and products become same at equilibrium
(C) the concentration of reactants and products are constant but different
(D) both forward and reverse reaction occur at all times with same speed
17 For an equilibrium reaction involving gases, the forward reaction is Ist order while the reverse reaction is IInd order. The units of K_{P} for the forward equilibrium is
(A) atm (B) atm^{2}
(C) atm^{–1} (D) atm^{–2}
 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) reduced to one fourth of its original value
(C) doubled
(D) constant
19 In a slow reaction, the rate of reaction generally ——————– with time
(A) decreases
(B) increases
(C) sometimes increases and sometimes decreases
(D) remains constant
20 The yield of product in the reaction A_{2}(g) + 2B (g) C(g) + Q kJ would
be high at
(A) high temperature and high pressure
(B) high temperatue and low pressure
(C) low temperature and high pressure
(D) low temperature and low pressure
Level – II
Instruction: In each of the following objective questions, 4/5 alternatives are given. In each one, one or more than one may be correct. It is important to do these questions in one attempt and under a time limit. Therefore you are required to do this assignment in 20 minutes.
 In which of the following equilibrium, the value of K_{P} is less than K_{C}?
(A) N_{2}O_{4} 2NO_{2} (B) N_{2} + O_{2} 2NO
(C) N_{2} + 3H_{2} 2NH_{3} (D) 2SO_{2} + O_{2} 2SO_{3}
 For the reaction PCl_{3} (g) + Cl_{2}(g) PCl_{5}(g), the value of K_{P} at 250°C is 0.61 atm^{–1}. The value of K_{C} at this temperature will be
(A) 15 (mol/l)^{–1} (B) 26 (mol/l)^{–1}
(C) 35 (mol/l)^{–1} (D) 52 (mol /l)^{–1}
 In the reaction A_{2}(g) + 4B_{2} (g) 2AB_{4}(g) , DH < 0. The decomposition of AB_{4} (g) will be favoured at
(A) low temperature and high pressure
(B) high temperature and low pressure
(C) low temperature and low pressure
(D) high temperature and high pressure
 The equilibrium constant for the reaction
N_{2}(g) + O_{2}(g) 2NO(g) is 4 ´10^{–4} at 200 K. In the presence of a catalyst the equilibrium is attained 10 times faster. Therefore the equilibirum constant in presence of the catalyst at 200 K is
(A) 4 ´10^{–3} (B) 4 ´10^{–4}
(C) 4 ´10^{–5} (D) None
 In a system A (s) 2B (g) + 3C (g)
If the conc. of C at equilibrium is increased by a factor of 2, it will cause the equilibrium concentration of B to change to
(A) two times the original value (B) one half of its original value
(C) times the original value (D) times the original value
 For the decomposition reaction
NH_{2}COONH_{4} (s) 2NH_{3}(g) + CO_{2}(g)
The K_{P} = 2.9 ´10^{–5} atm^{3}. The total pressure of gases at equilibrium when 1 mole of NH_{2}COONH_{4 }(s) was taken to start with would be
(A) 0.0194 atm (B) 0.0388 atm
(C) 0.0582 atm (D) 0.0766 atm
 For the gas phase reaction C_{2}H_{4}(g) + H_{2}(g) C_{2}H_{6}(g); DH = –32.7 kcal carried out in a vessel, the equilibrium concentration of C_{2}H_{4} can be increased by
(A) increasing temperature (B) decreasing pressure
(C) removing some H_{2} (D) adding some C_{2}H_{6}
 K_{P} for a reaction at 25°C is 10 atm. The activation energy for forward and reverse reactions are 12 and 20 kJ / mol respectively. The K_{C} for the reaction at 40°C will be
(A) 4.33 ´10^{–1} M (B ) 3.33 ´10^{–2 }M
(C) 3.33 ´10^{–1} M (D) 4.33 ´10^{–2} M
 For the reaction CaCO_{3}(s)CaO(s) +CO_{2} (g), the pressure of CO_{2} depends on
(A) the mass of CaCO_{3}(s)
(B) the mass of CaO(s)
(C) the masses of both CaCO_{3}(s) and CaO(s)
(D) temperature of the system
 For the reaction H_{2}(g)+I_{2} (g)2HI(g), the equilibrium constant Kp changes with
(A) total pressure (B) catalyst
(C) the amount of H_{2} and I_{2} present (D) temperature
[IIT–JEE 1981]
 An example of reversible reaction is
(A) Pb(NO_{3})_{2}(aq) + 2NaI (aq) Þ PbI_{2}(s) + 2NaNO_{3} (aq)
(B) AgNO_{3}(aq) + NaCl (aq) Þ AgCl (s) + NaNO_{3} (aq)
(C) 2Na(s) + 2H_{2}O(l) Þ 2NaOH (aq) + H_{2}(g)
(D) KNO_{3} (aq) + NaCl (aq) Þ KCl (aq) + NaNO_{3} (aq)
[IIT–JEE 1985]
 Pure ammonia is placed in a vessel at a temperature where its dissociation constant (a) is appreciable. At equilibrium
(A) K_{P} does not change significantly with pressure
(B) a does not change with pressure
(C) concentration of NH_{3} does not change with pressure
(D) concentration of H_{2 }is less than that of N_{2}
[iit–JEE 1984]
 When NaNO_{3} is heated in a closed vessel, O_{2} is liberated and NaNO_{2} is left behind. At equilibrium.
(A) addition of NaNO_{2} favours reverse reaction
(B) addition of NaNO_{3} favours forward reaction
(C) increasing temperature favours forward reaction
(D) increasing pressure favours forward reaction [IIT–JEE 1986]
14.For the reaction, 2NO_{2} (g) 2NO(g) + O_{2} (g), K_{C} = 1.8 ´10^{–6} at 185°C. At 185°C,
the value of K_{C} for the reaction:
NO(g) + O_{2} (g) NO_{2} (g) is
(A) 0.9 ´ 10^{6} (B) 7.5 ´10^{2}
(C) 1.95 ´10^{–3} (D) 1.95 ´10^{3}
 The oxidation of SO_{2} by O_{2} to SO_{3} is an exothermic reaction. The yield of SO_{3} will be maximum, if
(A) temperature is increased and pressure is kept constant
(B) temperature is reduced and pressure is increased
(C) both temperature and pressure are increased
(D) both temperature and pressure are reduced
 Answers to Subjective Assignments
Level – I
 39.7 (mol l^{–1})^{–1/2 }
 (i) Partial pressure of N_{2}O_{4} = 0.167 atm (ii) Partial pressure of NO_{2} = 0.333 atm
 K_{p} = 5.2 atm
 K_{p} = 0.3136 atm^{2}
 K_{p} = 7.31 ´ 10^{–4} atm^{–2}
 Degree of dissociation of HI = 0.245
 0.16
 (i) 2 = 0.59 (ii) K_{P} = 0.1687 (iii) 2.645 ´ 10^{–5}
 Mole fraction of N_{2} = 0.788
Mole of fraction of O_{2} = 0.212
 K_{c} = 3.178
 K_{p} = 6.41 ´ 10^{–3} atm
 K_{c} = 1.747
P_{HI} = 0.776
 (i) K_{p} = 0.266 atm
(ii) 2 = 63.27%
 (i) K_{c} = 8.1 ´ 10^{–5} mol^{2} litre^{–2}
K_{p} = 4.90 ´ 10^{–2} atm^{2}
(ii) Addition of more NH_{4}HS will cause no effect on this equilibrium because concentration of NH_{4}HS is not involved in the expression of K_{C} or K_{P}
Level – II
 84.34 atm
 K_{c} = 1.413 mol / lit
 30 bar
 1.04 ´ 10^{14}
 90%
6.
 9.65 gm
 50.07
 25 L
 3.71 ´ 10^{–6}
 DH = 75.17 ´ 10^{3} J
 0.7837
 1.3%
 [NO] = 1.6 mol / lit, [NO_{2}] = 0.4 mol / lit
Level – III
 (i) 0.10 mole (ii) 0.40 mole
 (a) K_{c} = 178.57 M^{–2}
K_{p} = 0.047 atm^{–2}
(b) 12.546 atm
 a =
 Moles of AB = 1.868
 K_{c} = 707 on adding water, reverse reaction is favoured
 0.0311
 4 atm
 % NH_{3} decomposed = 61.2%
 (a) K_{p} = 900 mm^{2}
(b) =
(c) 72.15
 2% dissociation of H_{2}S
 (a) K_{c} =
(b) Conc. of R = 1.5 M
 V = 144 ml
 (i) X_{CO} = 0.7655, = 0.938 atm
(ii) P = 0.68 atm
 K_{P }= 0.56 (i) 64% (ii) 21.42% (iii) 12.998 ml
 Answers to Objective Assignments
Level – I
 A 2. B
 C,D 4. C
 A,C 6. A
 A 8. D
 D 10. B
 B 12. C
 B 14. C
 C 16. D
 A 18. D
 D 20. C
Level – II
 C,D 2. B
 C 4. B
 D 6. C
 A,B,C,D 8. C
 D D
 D 12. A
 A,B,C 14. B
 B