# Chemical Equilibrium_ Hints & Solutions

###### Chemical Equilibrium

** **

**Hints to Subjective Problems**

** **

**LEVEL – I**

- Apply K
_{p}= K_{c}(Rt)^{Dn}

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

- % by volume of a gas is same as % by mole

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

- To calculate K
_{c}calculate the conc. Of Fe^{2+}at equilibrium from volumetric data.

**LEVEL – II**

- To calculate K
_{p}, first find out partial pressure of all the component of equilibrium. Then apply K_{p}= K_{c}(RT)^{Dn}.

- As Dn
_{g}is negative, equilibrium will shift towards right

- Find number of moles of O
_{2}at equilibrium from PV = nRT

**LEVEL – III**

- To calculate initial number of moles of H
_{2}introduced, equate total number of moles at equilibrium to the total number of moles from gaseous data.

- 2AB
_{2}(g) 2Ag(g) + B_{2}(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–a^{2}) and (1+a/2) in expression for equilibrium constant

- Apply . Also,

Where a_{1} = initial no. of moles P_{1} = initial pressure

a_{2} = moles at equilibrium P_{2} = final pressure

- Calculate reaction quotient from given data, and relate it to K
_{c}find the direction of change.

** Solutions to Subjective Problems **

* *

#### Level – I

- 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 O_{2}(g) + 2HCl(g) Cl_{2}(g) + H_{2}O(g)

is **39.7 (mol.l ^{-1})^{-½}**

- This problem can be solved by two methods.

** Method (1) **Let the number of moles of N

_{2}O

_{4}initially be 1 and a is the degree of dissociation of N

_{2}O

_{4}.

N_{2}O_{4 } 2NO_{2}

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 N_{2}O_{4} = **0.167 atm.**

and partial pressure of NO_{2} = **0.333 atm.**

** Method (2) **Let the partial pressure of NO

_{2}at equilibrium be p atm, than the partial pressure of N

_{2}O

_{4}at equilibrium will be (0.5–p)atm.

\ Kp =

p^{2} + 0.66 p–0.33 = 0

On solving p = 0.333 atm.

** ****\ = 0.333 atm and = 0.167 atm. **

- Let us assume that we start with C moles of N
_{2}O_{4}(g) initially.

N_{2}O_{4}(g) 2NO_{2}(g)

Initially C 0

At equilibrium C(1-a) 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} =

** K _{P}= 5.2 atm.**

- 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 atm ^{2}.**

** Alternatively, **

At equilibrium = 1.12 atm** **

** **As

\ = 0.56 atm

\K_{P }= 0.56 ´ 0.56 = **0.3136 atm ^{2}.**

- 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).

N_{2}(g) + 3H_{2}(g) 2NH_{3}

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 H_{2} at equilibrium

=

Mole fraction of N_{2} at equilibrium

= 1 – 0.6165 – 0.178 = 0.2055

\K_{P} =

** K _{P} = 7.31 **

**´ 10**

^{-4}atm^{-2}.** Alternatively,**

N_{2}(g) + 3H_{2}(g) 2NH_{3}(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 N_{2}(g) and H_{2}(g) at equilibrium and finally K_{P} comes out to be **7.31 ****´10 ^{–4} atm^{–2}**

- H
_{2}(g) + I_{2}(g) 2HI (g)

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

Equilibrium (25–a) (18–a) 2a

Where a is the volume of H_{2} reacted at equilibrium. This is same as the volume of I_{2} 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) H_{2}(g) + I_{2} (g)

1–a

Equilibrium constant of the above reaction, K¢ =

Also, =

= = 0.1622

a = 0.245

\ Degree of dissociation of HI = **0.245**

- Initial pressure of I
_{2}(P_{O}) = 0.074 atm

If P¢ is the decrease in pressure of I_{2} pressure of I_{2} 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

I_{2} 2I

0.074 – 0.038 2´ 0.038

K_{P} = = **0.16**

- Let the combined moles of N
_{2}and O_{2}, be 1, out of which N_{2}was x moles and O_{2}was (1–x) moles.

Let a is the number of moles of N_{2} reacted at equilibrium.

N_{2}(g) + O_{2}(g) ¾® 2NO(g)

(x–a) (1–x–a) 2a

2a = = 1.8 ´ 10^{–2}

a = 0.9 ´ 10^{12} = 9 ´ 10^{–3}

K_{C} = =

(2.1 ´ 10^{–3}) =

On solving we get x = 0.788

\ Mole fraction of N_{2} in the air initially present = **0.788 **

mole fraction of O_{2} in the air initially present = **0.212**

- Initial conc. of AgNO
_{3}= = 0.075

Initial conc. of Fe^{2+} solution = = 0.545

Final conc. of Fe^{2+} solution = = 0.499

Ag^{+} + Fe^{2+} Fe^{3+} + Ag

0.029 0.499 0.046

K_{C} = = **= 3.178**

- a–D– glucose b–D–glucose

t = 0 1 0

equilibrium 1–0.636 0.636

K_{C} = = **1.747**

- H
_{2}+ I_{2}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**

- N
_{2}O_{4}2NO_{2}

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 % N_{2}O_{4} dissociates at 0.1 atm pressure

N_{2}O_{4 }_{ } 2NO_{2}

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**

- 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 atm

^{2}

\ = 34 ´34 = 1156 atm^{2}

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.**

- Let n and a be the initial number of moles and degree of dissociation of NH
_{3}(g) respectively.

NH_{3} (g) N_{2} (g) + H_{2} (g)

Initially n 0 0

At equilibrium n(1–a)

Total number of moles = n(1+a)

Partial pressure of NH_{3}, =

Partial pressure of N_{2}, =

Partial pressure of H_{2}, = ´ P

\ K_{P} =

=

= or, **a**** = **

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

K_{P} = K_{C}(RT)^{D}^{n}

or, K_{C}= K_{P} (RT)^{–}^{D}^{n}

K_{C }= 78.1 [0.0821´673]^{–1} = **1.413 mol/l**

- Let p be the initial partial pressure of N
_{2}and p¢ be the decrease in its partial pressure at equilibrium. Therefore, the partial pressures of various gases at equilibrium are

N_{2 } (g) + 3H_{2} (g) 2NH_{3 }(g)

Initially p 3p 0

At equilibrium p-p¢ 3(p–p¢) 2p¢

Total pressure at equilibrium = 4p–2p¢

% of NH_{3} = = 10

On solving, we get p¢ =

K_{P} =

On solving we get p = 8.26 bar

Total pressure = 4p–2p¢ = = **30 bar**

- For the given reaction CO(g) + O
_{2}(g) CO_{2}(g) …(1)

=

For the given reaction C(graphite) + CO_{2} (g) 2CO (g) …(2)

=

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

C(graphite) + O_{2}(g) CO_{2}(g)

Equilibrium constant K_{P} of the above reaction is given by

K_{P} = = (2.5´10^{5})^{2} (1.67 ´10^{3}) = **1.04 ****´****10 ^{14}**

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

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*)

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.

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*)

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%**

- 2BaO
_{2}(s) 2BaO(s) + O_{2}(g)

K_{P} = 0.5 atm,

weight of BaO_{2} = 15 gm

Molecular weight of BaO_{2} = 169.3

K_{P} = = 0.5 atm, V = 5*l*, T = 1067K

Number of moles of O_{2} produced = = = 0.0285

\ Number of moles of BaO_{2} used up = 2 ´ 0.0285 = 0.057

Weight of BaO_{2} used up = 0.057 ´ 169.3 = **9.65 gm **

- 2H
_{2}S(g) 2H_{2}(g) + S_{2}(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}**

- LiCl.3NH
_{3}(s) LiCl.NH_{3}(s) + 2NH_{3}(g)

K_{P} = 9 atm, Vol = 5*l* , Temperature = 40 °C

Number of moles of LiCl.NH_{3} = 0.1

2NH_{3}(g) + LiCl.NH_{3}(s) LiCl.3NH_{3}(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 NH _{3} required = 0.7837**

- 2H
_{2}S(g) 2H_{2}(g) + S_{2}(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%**

- SO
_{2}(g) + NO_{2}(g) SO_{3}(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**

[NO_{2}] = 1– = = **0.4 M**

** **

**LEVEL – III**

- 2SO
_{2}(g) + O_{2}(g) 2SO_{3}(g) K_{c}= 100

[O_{2}]

K_{c} =

[O_{2}] =

No. of moles of O_{2} = x

C =

=

n = 0.1

100 =

[O_{2}] =

= 0.4

- CO(g) + 2H
_{2}(g) CH_{3}OH(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] = , [H_{2}] = , [CH_{3}OH] =

\ 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**

- 2AB
_{2}(g) 2AB (g) + B_{2}(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 =**

- A
_{2}(g) + B_{2}(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

23x^{2} – 75a + 50 = 0

x = 0.934 or 2.326

Only 0.934 value is permissible

So, moles of AB = **1.868**

- I
_{2}+ I^{–}I_{3}^{–}

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**

- C
_{2}H_{5}OH + C_{3}H_{7}COOH C_{3}H_{7}COOC_{2}H_{5}+ H_{2}O

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

- = 200CC, = 200CC, = 100

Vol. fraction = mole fraction

Hence K_{p} can calculated

- Ammonia decomposes to N
_{2}and H_{2}as follows

2NH_{3} N_{2} + 3H_{2}

NH_{3} ¾® NH_{3}

At 27°C at 347°C

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

First, let us find initial pressure of NH_{3 }at 347°C

Þ = 31 atm

Partial pressure | 2NH_{3} |
3H_{2} |
N_{2} |

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 % NH_{3} decomposed = = **61.2%**

** Alternative method:**

2NH_{3} 3H_{2} + N_{2}

Let x be the degree of dissociation

Moles | 2NH_{3} |
3H_{2} |
N_{2} |

At equilibrium | 1-x | 3/2x | x/2 |

Þ total mole = 1 + x

Þ x =

Þ % dissociation = ´ 100 = **61.2%**

- NH
_{2}COONH_{4}(s) 2NH_{3}(g) + CO_{2}(g)

Let P = original equilibrium pressure, from the mole ratio of NH_{3} and CO_{2} at equilibrium, we have

&

Þ K_{p} =

Now NH_{3} is added such that, = P

Find the pressure of CO_{2}

K_{p} =

Þ

Þ

Total new pressure = P_{new} =

Þ

Þ ratio =

- 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} = 20^{2} = 400 mm^{2}

Similarly Y(s) B(g) + C(g)

P_{B} = P_{C} = 30 mm Þ K_{p} = P_{B} . P_{C} = 30^{2} = **900 mm ^{2}**

- 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.

- 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**

- 2H
_{2}S 2H_{2}+ S_{2}volume of vessel = V = 0.4 L

Let x be the degree of dissociation

Moles | 2H_{2}S |
2H_{2} |
S_{2} |

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 H _{2}S**

- 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} =

- First find the value of K
_{C}for dissociation of HI from its degree of dissociation

2HI H_{2} + I_{2} (degree of dissociation is 0.8)

Concentrations | 2HI | H_{2} |
I_{2} |

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 H_{2} and I_{2} and the following equilibrium will be established.

H_{2} + I_{2} HI with K_{C} = 1/4

Concentrations | H_{2} |
I_{2} |
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 I_{2}left unreacted at equilibrium

n(I_{2}) = 0.135 – 0.027 = 0.108

I_{2} reacts with sodium thiosulphate (Na_{2}S_{2}O_{3}) as follows:

2Na_{2}S_{2}O_{3} + I_{2} Na_{2}S_{4}O6 + 2NaI

applying the mole concept, we have,

2 moles of Na_{2}S_{2}O_{3} º 1 mole of I_{2}

Þ 0.018 moles of I_{2} º 2 ´ 0.108 = 0.216 moles of Na_{2}S_{2}O_{3} 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**

- CO
_{2(g) }+ C_{(graphite)}= 2CO_{(g)}

K_{p} = 10

By K_{p}, fraction can be calculated, knowing fraction, partial pressure can be calculated.

- 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**

- K =

K = 0.208

- PCl
_{5}(g) PCl_{3}(g) + Cl_{2}(g)

1 + a =

1 + a =

1 + a = 1.68

\ a = 0.68 or 68%

- 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

- 2Ag
^{+}+ Cu Cu^{2+}+ 2Ag

aM xM if V is 500 ml

If volume is doubled, then for K_{c} = to have same value [Cu^{2+}] will be less than x/2

**Level – II**

- 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)**

- PCl
_{3}(g) + Cl_{2}(g) PCl_{5}(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)**

- 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)**

- 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)**

- A(s) 2B(g) + 3C(g)

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

\K_{C} = *x*^{2}y^{3} ——————– (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}(8y^{3}) = *x*^{2}y^{3}

\ *x**¢* =

\Equilibrium concentration of B changes to times the original value

** ****\ (d)**

- NH
_{2}COONH_{4}(s) 2NH_{3}(g) + CO_{2}(g)

1 2 1

K_{P} = 2.9 ´ 10^{–5} atm^{3}

If the P is the total pressure at equilibrium

K_{P} =

\P^{3} = = 1.9575

P = = 0.058

** ****\(C)**

- C
_{2}H_{4}(g) + H_{2}(g) C_{2}H_{6}(g) DH = –32.7 K cal.

Concentration of C_{2}H_{4} will increase by all the factors favouring backward reaction. High temperature, low pressure, decrease in concentration of H_{2 }and increase in concentration of C_{2}H_{6} will favour backward reaction.

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

- 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)**

** **