# Hints to Subjective Problems

LEVEL – I

1. Apply Kp = Kc(Rt)Dn

1. Activity of component in condensed phase (solid or liquid) is unity

1. % by volume of a gas is same as % by mole

1. As volume is proportional to number of moles under given condition of temperature and pressure hence calculation can be done in terms of volume.

1. To calculate Kc calculate the conc. Of Fe2+ at equilibrium from volumetric data.

LEVEL – II

1. To calculate Kp, first find out partial pressure of all the component of equilibrium. Then apply Kp = Kc(RT)Dn.

1. As Dng is negative, equilibrium will shift towards right

1. Find number of moles of O2 at equilibrium from PV = nRT

LEVEL – III

1. To calculate initial number of moles of H2 introduced, equate total number of moles at equilibrium to the total number of moles from gaseous data.

1. 2AB2(g)      2Ag(g)             +          B2(g)

1                                  0                                  0

(1–a)                           a                                  a/2

Total number of moles at equilibrium (1+a/2)

As a <<1,  neglect the term a in

(1–a2)  and (1+a/2) in expression for equilibrium constant

1. Apply . Also,

Where a1 = initial no. of moles            P1 = initial pressure

a2 = moles at equilibrium        P2 = final pressure

1. Calculate reaction quotient from given data, and relate it to Kc find the direction of change.

# Solutions to Subjective Problems

#### Level – I

1. KP = KC (RT)Dn

Dn = 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

\  for the reverse reaction would be

\   =  = 1576.8 (mol l-1)-1

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

\ KC for O2(g) + 2HCl(g)  Cl2(g) + H2O(g)

is 39.7 (mol.l-1)

1. This problem can be solved by two methods.

Method (1)      Let the number of moles of N2O4 initially be 1 and a is the degree of dissociation of N2O4.

N2O4       2NO2

Initial moles                             1                      0

Moles at equilibrium                1–a                  2a

Total moles at equilibrium =    1–a + 2a = 1+a

\ Kp =

a = 0.5 i.e.,  50% dissociation

Hence, partial pressure of N2O4 = 0.167 atm.

and partial pressure of NO2 = 0.333 atm.

Method (2)      Let the partial pressure of NO2 at equilibrium be p atm, than the partial pressure of N2O4 at equilibrium will be (0.5–p)atm.

\ Kp =

p2 + 0.66 p–0.33 = 0

On solving p = 0.333 atm.

\  = 0.333 atm and  = 0.167 atm.

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

N2O4(g)    2NO2(g)

Initially                   C                     0

At equilibrium        C(1-a)                  2Ca

where a is the degree of dissociation of N2O4(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

\  Kp =

KP=   5.2 atm.

1. The reaction is

NH4HS(s)  NH3(g) + H2S(g)

If a is the degree of dissociation of equilibrium,

Total moles of NH3 + H2S = 2a.

Partial pressure =

\ pNH3 = ´ P = 0.5P;    pH2S = ´ P = 0.5P

Kp = pNH3 ´ pH2S = 0.5P ´ 0.5P = 0.25P2

Substituting the value of P = 1.12 atm.,

Kp = 0.25 ´ 1.12 ´ 1.12 = 0.3136 atm2.

Alternatively,

At equilibrium  = 1.12 atm

As

\  = 0.56 atm

\KP = 0.56 ´ 0.56 = 0.3136 atm2.

1. 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).

N2(g)  +    3H2(g)    2NH3

Initially    1                   3            0

At equib  1-x         3-3x            2x

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

\  =0.178

\

\  Mole fraction of H2 at equilibrium

=

Mole fraction of N2 at equilibrium

= 1 – 0.6165 – 0.178 = 0.2055

\KP =

KP = 7.31 ´ 10-4 atm-2.

Alternatively,

N2(g)   +  3H2(g)    2NH3(g)

Initially       x                3x                0

At equib.   x-a                  3x-3a          2a

\  =0.178

\  = 0.178

\   = 0.302

Similarly we can calculate the mole fraction of N2(g) and H2(g) at equilibrium and finally KP comes out to be 7.31 ´10–4 atm–2

1. H2(g) +    I2(g)          2HI (g)

t = 0     25 c.c.       18 c.c.             0

Equilibrium      (25–a)                    (18–a)              2a

Where a is the volume of H2 reacted at equilibrium. This is same as the volume of I2 reacted or half the volume of HI  produced at equilibrium.

2a = 30.8 c.c.

Þ a = 15.4 c.c.

\ Equilibrium constant, K =  = 38.0064

If a is the degree of dissociation of HI at 456°C

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

1–a

Equilibrium constant of the above reaction, K¢ =

Also,  =

=   = 0.1622

a = 0.245

\ Degree of dissociation of HI = 0.245

1. Initial pressure of I2 (PO) = 0.074 atm

If P¢ is the decrease in pressure of I2 pressure of I2 at equilibrium be (PO–P¢)
= 0.074 – P¢

\ Pressure of I – atoms = 2P¢

Total pressure at equilibrium = 0.074 –P¢+2P¢ = 0.112

\ P¢ = 0.112 – 0.074 = 0.038

I2                 2I

0.074 – 0.038        2´ 0.038

KP =  =  0.16

1. Let the combined moles of N2 and O2, be 1, out of which N2 was x moles and O2 was (1–x) moles.

Let a is the number of moles of N2 reacted at equilibrium.

N2(g)      +       O2(g)         ¾®     2NO(g)

(x–a)          (1–x–a)            2a

2a =  = 1.8 ´ 10–2

a = 0.9 ´ 1012 = 9 ´ 10–3

KC =  =

(2.1 ´ 10–3) =

On solving we get x = 0.788

\ Mole fraction of N2 in the air initially present  = 0.788

mole fraction of O2 in the air initially present = 0.212

1. Initial conc. of AgNO3 = = 0.075

Initial conc. of  Fe2+ solution =  = 0.545

Final conc. of Fe2+ solution =  = 0.499

Ag+      +    Fe2+          Fe3+     +          Ag

0.029         0.499               0.046

KC =  =  = 3.178

1. a–D– glucose b–D–glucose

t = 0           1                                  0

equilibrium 1–0.636                       0.636

KC =  = 1.747

1. H2 +        I2         2HI      KP = 49

Initial (p.p.)            0.5             0.5             0

At eq. (p.p.)           (0.5–p)       (0.5–P)      2p

KP =  = 49

On solving p = 0.388 atm

So, =  = 0.112  atm,  = 0.776 atm

1. N2O4   2NO2

Initial (mole)    100                  0

At eq. (mole)   (100–25)          50        = 75

At eq. (p.p.)     1´            1´   =  =

\ KP =  = 0.267 atm.

(ii) Suppose x % N2O4 dissociates at 0.1 atm pressure

N2O4                2NO2

Initial (mole )   100                        0

At eq. (mole)   100–x                    2x   total moles at equilibrium = 100 + x

At eq. (p.p.)

\KP =

On solving, we get  x = 63.25 %

Level – ii

1. When A(s) dissociates alone in a vessel, the pressure over excess solid A is 50 atm. The product gases B and D are present in the molar ratio of 1:1, this means that the pressure of each gas will be half of total pressure.

\  = 25 ´ 25 = 625 atm2

\ = 34 ´34 = 1156 atm2

But when A(s) and C(s) are placed together in the container, the degree of dissociation of each will be suppressed by the presence of other.

A(s)  B(g) + D(g)

x         (x +y)

C(s)  E(g) + D(g)

y          (x +y)

625 = x (x +y)                    …(1)

1156 = y(x +y)                  …(2)

From eq. (1) and eq. (2) we get

x = 14.8 atm, y = 27.37 atm.

\ Total pressure of the gases over the solid mixture=2(x +y)=84.34 atm.

1. Let n and a be the initial number of moles and degree of dissociation of NH3 (g) respectively.

NH3 (g)   N2 (g)   +      H2 (g)

Initially                   n                      0                     0

At equilibrium       n(1–a)

Total number of moles = n(1+a)

Partial pressure of NH3,  =

Partial pressure of N2,   =

Partial pressure of H2,  = ´ P

\ KP =

=

=   or, a =

Dn, change in number of the moles of the given reaction = +1

KP = KC(RT)Dn

or, KC= KP (RT)Dn

KC = 78.1 [0.0821´673]–1 = 1.413 mol/l

1. Let p be the initial partial pressure of N2 and p¢ be the decrease in its partial pressure at equilibrium. Therefore, the partial pressures of various gases at equilibrium are

N2  (g)     +      3H2 (g)  2NH(g)

Initially                 p                      3p                    0

At equilibrium        p-p¢                  3(p–p¢)             2p¢

Total pressure at equilibrium = 4p–2p¢

% of NH3 =  = 10

On solving,  we get  p¢ =

KP =

On solving we get p = 8.26 bar

Total pressure = 4p–2p¢ =  = 30 bar

1. For the given reaction CO(g) + O2(g) CO2 (g)                      …(1)

=

For the given reaction C(graphite) + CO2 (g)   2CO (g)             …(2)

=

Multiplying equation (1) by 2 and adding (2) we get

C(graphite) + O2(g)  CO2(g)

Equilibrium constant KP of the above reaction is given by

KP =  = (2.5´105)2 (1.67 ´103) = 1.04 ´1014

1. In the given reaction 87% of the acid is  consumed at equilibrium

C6H5COOH(l) + C2H5OH (l)  C6H5COOC2H5(l) + H2O(l)

Initially                   1                            3                            0                      0

At equilibrium        1–0.87                   3–0.87                   0.87                 0.87

\ KC =  =   = 2.73

Let n be the number of moles of acid consumed when 1 mole of the acid is mixed with 4 moles of ethanol.

C6H5COOH (l) + C2H5OH (l)  C6H5COOC2H5(l) + H2O(l)

At equilibrium 1–n            4–n            n              n

KC =  = 2.73

= 2.73

On solving, we get n = 0.90 or 6.99

6.99 is not possible as we started with 1 mole of the acid

\% of acid consumed  = 90%

1. 2BaO2 (s)  2BaO(s) + O2 (g)

KP = 0.5 atm,

weight of BaO2 = 15 gm

Molecular weight of BaO2 = 169.3

KP =  = 0.5 atm,         V = 5l, T = 1067K

Number of moles of O2 produced =   =   =  0.0285

\ Number of  moles of BaO2 used up = 2 ´ 0.0285 = 0.057

Weight of BaO2 used up = 0.057 ´ 169.3  = 9.65 gm

1. 2H2S(g)       2H2(g)             +  S2(g)

2(1–a)                   2a            a

Where a is the degree of dissociation

Total number of moles at equilibrium = 2–2a + 2a+a = 2+a

= ,  = , =

KP =

If a = 0.055 and P = 1 atm

KP =  = 9.114 ´ 10–5

KP = KC (RT)Dn, [Here Dn =1]

KC =  = 3.71 ´ 10–6

1. LiCl.3NH3 (s) LiCl.NH3(s) + 2NH3(g)

KP = 9 atm, Vol = 5l , Temperature = 40 °C

Number of moles of LiCl.NH3 = 0.1

2NH3(g)     +    LiCl.NH3(s)                 LiCl.3NH3(s)

a–2x                0.1–x                                 x

0.1–x = 0         Þ x = 0.1

a–2x = a–0.2

KP¢ = =

= 5.1395 (a–0.2)

KP¢ =

(a–0.2)2 =         Þ a = 0.7837

\ Minimum number of moles of NH3 required = 0.7837

1. 2H2S(g)      2H2(g)       +  S2(g)     KC­ = 1´ 10–6

# At eq. (conc.)

\KC =  = 1 ´ 10–6

10–6 =

When a << 1

10–6 =

a = 0.013

So, % dissociation = 1.3%

1. SO2(g) + NO2(g)           SO3(g) + NO(g)          KC =16

Initial (mole)    1                1                            1                1

At eq. (mole)   1–a            1–a                        1+a           1+a

KC =  = 16

= 4

a =

\[NO] = 1+  =  = 1.6 M

[NO2] = 1– =  = 0.4 M

LEVEL – III

1. 2SO2(g) +    O2(g)        2SO3(g)           Kc = 100

[O2]

Kc =

[O2] =

No. of moles of O2 = x

C =

=

n = 0.1

100 =

[O2] =

= 0.4

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

Initial (mole)    0.15                 a                            0

At eq. (mole)   (0.15–x)           (a–2x)                    x

(0.15–0.08)      (a–0.16)                 0.08           But, x = 0.08 mole

= 0.07

Total number of moles at equilibrium  =  a – 0.01

number of moles at equilibrium =   =  = 0.35

\   a– 0.01 = 0.35

a= 0.36

At equilibrium [CO] =  ,  [H2] = , [CH3OH] =

\ KC =

KC =  = 178.57 M–2

\KP = 178.57  (0.082 ´ 750)–2 = 0.047 atm–2

(b) When ro reaction takes place, so total moles = 0.51

\ Pfinal  =    = 12.546 atm

1. 2AB2(g)  2AB (g) + B2(g)

Initial (mole)    1                      0                0

At eq. (mole)   1–a                  a                a/2   Total moles at equilibrium = (1+a/2)

At eq. (p.p)

KP =

KP =

But 1>> a

\ KP =

a =

1. A2(g) +    B2(g)         2AB(g)       KC = 50

Initial mole       1                2                            0

At eq. mole      1–x                  2–x                              2x

At eq. conc.

KC = = 50

23x2 –  75a + 50 = 0

x = 0.934 or 2.326

Only 0.934 value is permissible

So, moles of AB = 1.868

1. I2 +          I         I3

Initial (conc.)            0.1             0

At eq. (conc.)   (0.1–a)     a

But –a =

0.0492 – a = 0.0013

a = 0.0479

At eq. conc.     (0.0013)     0.0521       0.0479

\ KC =

KC = 707.2

1. C2H5OH + C3H7COOH C3H7COOC2H5 + H2O

At equilibrium remaining acid reacts with NaOH. By which amount of acid and K can be calculated

1. = 200CC,  = 200CC,  = 100

Vol. fraction  = mole fraction

Hence Kp can calculated

1. Ammonia decomposes to N2 and H2 as follows

2NH3  N2 + 3H2

NH3     ¾®  NH3

At 27°C             at 347°C

(15 atm)    (p = ?) V remains constant

First, let us find initial pressure of NH3 at 347°C

Þ   = 31 atm

 Partial pressure 2NH3 3H2 N2 Initial 31 0 0 At equilibrium 31-x 3/2x X/2

Now final equilibrium pressure = 50 atm

Þ 50 = 31 – x +

Þ  x = 19 atm

Þ  % NH­3 decomposed =  = 61.2%

Alternative method:

2NH3  3H2 + N2

Let x be the degree of dissociation

 Moles 2NH3 3H2 N2 At equilibrium 1-x 3/2x x/2

Þ  total mole = 1 + x

Þ  x =

Þ  % dissociation =  ´ 100 = 61.2%

1. NH2COONH4(s) 2NH3(g) + CO2(g)

Let P = original equilibrium pressure, from the mole ratio of NH3 and CO2 at equilibrium, we have

&

Þ  Kp =

Now NH3 is added such that,  = P

Find the pressure of CO2

Kp =

Þ

Þ

Total new pressure = Pnew =

Þ

Þ  ratio =

1. X(s) A(g) + C(g) at equilibrium A & C are in equal proportions, so  their pressures will be same.

Also     P­A + PC = 40 Þ PA = PC = 20 mm

Þ  Kp = PA.PC = 202 = 400 mm2

Similarly Y(s)  B(g) + C(g)

PB = PC = 30 mm Þ Kp = PB . PC = 302 = 900 mm2

1. b) Now for a mixture of X and Y, we will have to consider both the equilibrium simultaneously.

X(s)  A(g) + C(g) and

Y (s)  B(g) + C(g)

Let PA = a mm, PB = b mm

Note that the pressure of C due to dissociation of X will also be a mm and similarly the pressure of C due to dissociation of Y will also be b mm.

Þ PC = (a + b) mm

Kp (for X) = PA . PC = a ( a + b) = 400                  (I)

Kp (for Y) = PB.PC = b (a + b) = 900                     (ii)

From (I) and (ii), we get

as volume and temperature are constant, the mole ratio will be same as the pressure ratio.

1. c) The total pressure = PT = PA + PB + PC

= a + b + 9a + b) = 2(a + b)

to find a and b, solve the equations

& a (a + b) = 400

Þ a = 11.1 mm, b = 24.97 mm

Þ Total pressure = 2 (a + b) = 72.15

1. 2H2S 2H2 + S2 volume of vessel = V = 0.4 L

Let x be the degree of dissociation

 Moles 2H­2S 2H2 S2 Initially 0.1 0 0 At equilibrium 0.1 – 0.1x 0.1x 0.1x/2

KC =  = 10–6

Þ  x = 0.02

2% dissociation of H2S

1. P(g) + 2Q(g) R(g)

At equilibrium, [P] = 3 M, [Q] = 4 M & let [R] = x M,

KC =                       (1)

Now the volume is doubled, hence the concentrations are halved and a new equilibrium will be re-established with same value of KC., Calculate Q and determine the direction of equilibrium.

Q =

Þ Q> Kc Hence the system will predominantely move in backward direction so as to achieve new equilibrium state. Let y M be the decrease in concentration of R.

 Concentrations P 2Q R Initially 1.5 2 x/2 At new equilibrium 1.5 + y 2 + 2y x/2 – y

Given: [Q] = 3 M at new equilibrium Þ 2 + 2y = 3 Þ y = 0.5 M

Þ at new equilibrium, [P] = 1.5 + 0.5 = 2 M;

[Q] = 3M (given); [R] = x/2 – 0.5 M

Þ Q =

equating this value of KC with (I)

Þ  Þ x = 4 M

Hence [R] = 4M and at new equilibrium [R] = x/2 – 0.5 = 1.5 M

and KC =

1. First find the value of KC for dissociation of HI from its degree of dissociation

2HI  H2 + I2 (degree of dissociation is 0.8)

 Concentrations 2HI H2 I2 Initially 1.0 0 0 At new equilibrium 1.0 – 0.8 0.4 0.4

KC =

Now we have to start with 0.135 mol each of H2 and I2 and the following equilibrium will be established.

H2 + I2  HI with KC = 1/4

 Concentrations H2 I2 2HI Initially 0.135 0.135 0 At new equilibrium 0.135 – x 0.135 – x 2x

Þ  KC =

Þ x = 0.135/5 = 0.027

now find the moles of I2left unreacted at equilibrium

n(I2) = 0.135 – 0.027 = 0.108

I2 reacts with sodium thiosulphate (Na2S2O3) as follows:

2Na2S2O3 + I2  Na2S4O6 + 2NaI

applying the mole concept, we have,

2 moles of Na2S2O3 º 1 mole of I2

Þ 0.018 moles of I2 º 2 ´ 0.108 = 0.216 moles of Na2S2O3 are used up

Þ moles = MVn (M = molarity, Va = volume in litres)

Þ  0.216 = 1.5 V

Þ V = 0.144 lt = 144 ml

1. CO2(g) + C(graphite) = 2CO(g)

Kp = 10

By Kp, fraction can be calculated, knowing fraction, partial pressure can be calculated.

1. From the data given Kp can be calculated. Keeping Kp constant, % dissociation can be calculated at 0.2 atm

Solutions to Objective Problems

Level – I

1. K =

K = 0.208

1. PCl5 (g)   PCl3(g)            +    Cl2(g)

1 + a =

1 + a =

1 + a = 1.68

\ a = 0.68 or 68%

1. Since [C]o = 3M, at eq. it is 1.4M. so reaction must go in backward direction

A    +          2B       2C

Initial conc.      1                2                0

Eq. conc.         1 + x/2       2 + x          3 – x

Given 3 –x = 1.4M Þ x = 1.6M

Kc = , putting values Kc = 0.084

1. 2Ag+ +    Cu      Cu2+    +    2Ag

aM                               xM                         if V is 500 ml

If volume is doubled, then for Kc =  to have same value [Cu2+] will be less than x/2

Level – II

1. KP = KC(RT)Dn

KP <KC if Dn<0

For reaction (C) the value of Dn is –2 and for reaction (D) the value of
Dn is–1

\ (c) & (d)

1. PCl3(g) +    Cl2(g)        PCl5(g)

Dn = –1, KP = 0.61 atm–1.

KC = KP (RT)Dn

= 0.61 (0.0821 ´523)+1 = 26 mol /l.

\(b)

1. 2AB4(g) A2(g) + 4B2(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)

1. Equilibrium constant of a reaction remains unchanged when the same reaction remains unchanged when the same reaction is carried out in presence of a catalyst because  the catalyst will catalyse both the forward and backward reactions equally.

\(b)

1. A(s) 2B(g) + 3C(g)

Let x and y be the concentrations of B and C at equilibrium respectively

\KC = x2y3 ——————– (1)

Now the concentration of C is changed from y to y¢ such that y¢ = 2y. If x¢       is the new concentration of B

\KC = (x¢)2 (y¢)3 = (x¢)2(2y)3 ————————-(2)

From (1) and (2)

(x¢)2(8y3) = x2y3

\ x¢ =

\Equilibrium concentration of B changes to times  the original value

\ (d)

1. NH2COONH4(s) 2NH3(g) + CO2(g)

1                            2                1

KP = 2.9 ´ 10–5 atm3

If the P is the total pressure at equilibrium

KP =

\P3 =  = 1.9575

P =  = 0.058

\(C)

1. C2H4(g) + H2(g) C2H6(g)      DH = –32.7 K cal.

Concentration of C2H4 will increase by all the factors favouring backward reaction. High temperature, low pressure, decrease in concentration of H2 and increase in concentration of C2H6 will favour backward reaction.

\(a), (b), (C) & (d)

1. Enthalpy changed of a reaction is given by

DH = (Ea)f – (Ea)b

Where (Ea)f and (Ea)b are energies of activation for the forward and backward reactions.

DH = 12–20 = –8 kJ / mol

KP for the reaction at 25°C = 10 atm. Since KP is expressed in
atmosphere, Dn = +1

QKP = KC (RT)Dn,    KC =  = 0.4 M

KC at 40°C is given by

log

=  = –0.06719

(KC)40/(KC)25 = 0.85

(KC)40 = 0.85 ´0.4 = 0.34 M

\ (c)