Chemical Equilibrium

 

Hints to Subjective Problems

 

LEVEL – I

1. Apply K_p = K_c(Rt)^Dn

 

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

 

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

 

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

 

10. To calculate K_c calculate the conc. Of Fe²⁺ at equilibrium from volumetric data.

 

LEVEL – II

 

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

 

3. As Dn_g is negative, equilibrium will shift towards right

 

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

 

LEVEL – III

 

2. 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.

 

3. 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

 

8. Apply . Also,

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

a₂ = moles at equilibrium        P₂ = final pressure

 

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

    Solutions to Subjective Problems

 

Level – I

1. K_P = K_C (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 = K_C (0.082 ´ 673)

K_C = 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 (K_C)ⁿ or (K_P)ⁿ of the original reaction.

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

is 39.7 (mol.l^-1)^-½

 

2. 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.

3. 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

\  K_p =

            K_P=   5.2 atm.

4. 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 =

\ p_NH3 = ´ P = 0.5P;    p_H2S = ´ P = 0.5P

K_p = p_NH3 ´ p_H2S = 0.5P ´ 0.5P = 0.25P²

Substituting the value of P = 1.12 atm.,

K_p = 0.25 ´ 1.12 ´ 1.12 = 0.3136 atm².

            Alternatively,

At equilibrium  = 1.12 atm     

            As

\  = 0.56 atm

\K_P = 0.56 ´ 0.56 = 0.3136 atm².

5. Let the initial moles of N₂ and H₂  be 1 and 3 respectively (this assumption is valid as K_P 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

\K_P =

 

            K_P = 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 K_P comes out to be 7.31 ´10^–4 atm^–2

6. 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

7. Initial pressure of I₂ (P_O) = 0.074 atm

If P¢ is the decrease in pressure of I₂ pressure of I₂ at equilibrium be (P_O–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

K_P =  =  0.16

9. 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

K_C =  =

(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

10. 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

K_C =  =  = 3.178

12. a–D– glucose b–D–glucose

t = 0           1                                  0

equilibrium 1–0.636                       0.636

K_C =  = 1.747

13. H₂ +        I₂         2HI      K_P = 49

Initial (p.p.)            0.5             0.5             0

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

K_P =  = 49

On solving p = 0.388 atm

So, =  = 0.112  atm,  = 0.776 atm

14. N₂O₄   2NO₂

Initial (mole)    100                  0

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

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

\ K_P =  = 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.)

\K_P =

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.

2. 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

\ K_P =

=

 

=   or, a =

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

K_P = K_C(RT)^Dn

or, K_C= K_P (RT)^–Dn

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

3. 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)  2NH₃  (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¢ =

K_P =

 

On solving we get p = 8.26 bar

Total pressure = 4p–2p¢ =  = 30 bar

4. 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 K_P of the above reaction is given by

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

5. 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

\ K_C =  =   = 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

K_C =  = 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%

7. 2BaO₂ (s)  2BaO(s) + O₂ (g)

K_P = 0.5 atm,

weight of BaO₂ = 15 gm

Molecular weight of BaO₂ = 169.3

K_P =  = 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

11. 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

= ,  = , =

K_P =

If a = 0.055 and P = 1 atm

K_P =  = 9.114 ´ 10^–5

K_P = K_C (RT)^Dn, [Here Dn =1]

K_C =  = 3.71 ´ 10^–6

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

K_P = 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

K_P^¢ = =

= 5.1395 (a–0.2)

K_P^¢ =

(a–0.2)² =         Þ a = 0.7837

            \ Minimum number of moles of NH₃ required = 0.7837

14. 2H₂S(g)      2H₂(g)       +  S₂(g)     K_C­ = 1´ 10^–6

            Initial (mole)    1                            0                0

            At eq. (mole)   1–a                        a                a/2

            At eq. (conc.)

\K_C =  = 1 ´ 10^–6

10^–6 =

When a << 1

10^–6 =

a = 0.013

So, % dissociation = 1.3%

15. SO₂(g) + NO₂(g)           SO₃(g) + NO(g)          K_C =16

Initial (mole)    1                1                            1                1

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

K_C =  = 16

= 4

a =

\[NO] = 1+  =  = 1.6 M

[NO₂] = 1– =  = 0.4 M

 

LEVEL – III

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

[O₂]

K_c =

[O₂] =

No. of moles of O₂ = x

C =

=

n = 0.1

100 =

[O₂] =

= 0.4

2. 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] =

\ K_C =

K_C =  = 178.57 M^–2

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

 

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

\ P_final  =    = 12.546 atm

3. 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)

K_P =

K_P =

But 1>> a

\ K_P =

            a =

4. A₂(g) +    B₂(g)         2AB(g)       K_C = 50

Initial mole       1                2                            0

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

At eq. conc.

K_C = = 50

23x² –  75a + 50 = 0

x = 0.934 or 2.326

Only 0.934 value is permissible

So, moles of AB = 1.868

5. 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

\ K_C =

K_C = 707.2

6. C₂H₅OH + C₃H₇COOH C₃H₇COOC₂H₅ + H₂O

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

7. = 200CC,  = 200CC,  = 100

Vol. fraction  = mole fraction

Hence K_p can calculated

8. 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%

9. NH₂COONH₄(s) 2NH₃(g) + CO₂(g)

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

&

Þ  K_p =

Now NH₃ is added such that,  = P

Find the pressure of CO₂

K_p =

Þ

Þ

Total new pressure = P_new =

Þ

Þ  ratio =

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

Also     P­_A + P_C = 40 Þ P_A = P_C = 20 mm

Þ  K_p = P_A.P_C = 20² = 400 mm²

Similarly Y(s)  B(g) + C(g)

P_B = P_C = 30 mm Þ K_p = P_B . P_C = 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 P_A = a mm, P_B = 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 P_C = (a + b) mm

K_p (for X) = P_A . P_C = a ( a + b) = 400                  (I)

K_p (for Y) = P_B.P_C = 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 = P_T = P_A + P_B + P_C

= 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

11. 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

K_C =  = 10^–6

Þ  x = 0.02

            2% dissociation of H₂S

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

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

K_C =                       (1)

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

Q =

Þ Q> K_c 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 K_C with (I)

Þ  Þ x = 4 M

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

and K_C =

13. First find the value of K_C 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

K_C =

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 K_C = 1/4

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

Þ  K_C =

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 = MV_n (M = molarity, V_a = volume in litres)

Þ  0.216 = 1.5 V

Þ V = 0.144 lt = 144 ml

 

14. CO_2(g) + C_(graphite) = 2CO_(g)

K_p = 10

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

 

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

 

 

 

    Solutions to Objective Problems

 

Level – I

1. K =

K = 0.208

 

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

1 + a =

1 + a =

1 + a = 1.68

\ a = 0.68 or 68%

 

7. 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

K_c = , putting values K_c = 0.084

 

9. 2Ag⁺ +    Cu      Cu²⁺    +    2Ag

aM                               xM                         if V is 500 ml

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

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Level – II

1. K_P = K_C(RT)^Dn

            K_P <K_C if Dn<0

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

            \ (c) & (d)

2. PCl₃(g) +    Cl₂(g)        PCl₅(g)

Dn = –1, K_P = 0.61 atm^–1.

K_C = K_P (RT)^–Dn

= 0.61 (0.0821 ´523)⁺¹ = 26 mol /l.

            \(b)

3. 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)

4. 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)

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

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

\K_C = 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

\K_C = (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)

6. NH₂COONH₄(s) 2NH₃(g) + CO₂(g)

1                            2                1

K_P = 2.9 ´ 10^–5 atm³

If the P is the total pressure at equilibrium

K_P =

\P³ =  = 1.9575

P =  = 0.058

            \(C)

7. 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)

8. 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

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

QK_P = K_C (RT)^Dn,    K_C =  = 0.4 M

K_C at 40°C is given by

log

=  = –0.06719

(K_C)₄₀/(K_C)₂₅ = 0.85

(K_C)₄₀ = 0.85 ´0.4 = 0.34 M

            \ (c)