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QUESTION: 1

**Direction (Q. Nos. 1-12) This section contains 12 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.**

**Q.**Conversion factor for converting partial pressures (in K_{p}) to active masses (in K_{c}) is

Solution:

Kp = kc(RT)∆ng

So to convert kp to kc we have, kc = kp(1/RT)∆ng.

So, the converting factor is 1/RT

QUESTION: 2

For which of the following reactions does the equilibrium constant depends on the unit of concentration?

Solution:

For the reaction COCl_{2} (g) ⇌ CO(g) + Cl_{2 }(g), the equilibrium constant depends on the units of concentration.

For this reaction, the number of moles of reactants is not equal to the number of moles of products. So in the equilibrium constant expression, the units of concentration do not cancel out.

K = [CO] [Cl_{2}] ÷ [COCl_{2}]

K = mol/L × mol/L ÷ mol/L = mol/L

QUESTION: 3

The concentration of the oxides of nitrogen are monitored in air-pollution reports. At 25°C, the equilibrium constant for the reaction,

is 1.3 x 106 and that for

is 6.5 x 10^{-16} (when each species is expressed in terms of partial pressure).

For the reaction,

equilibrium constant is

Solution:

Given equations are

NO (g) + ½ O_{2} (g) ⇌ NO_{2} (g) ----------(i) k_{1} = 1.3×106

And ½ N_{2} (g) + ½ O_{2} (g) ⇌ NO(g) ----------(ii) k_{2} = 6.5×10^{-16}

To get the reaction, N_{2}(g) + 2O_{2}(g) ⇌ 2NO_{2}(g) ----------(iii) k_{3}

We multiply eqn (i) and eqn (ii) by 2 and adding both reaction, we get eqn (iii)

k_{3} = k_{1}^{2}×k_{2}^{2}

= (1.3×10^{6})^{2}×(6.5×10^{-16})^{2}

= 1.69×10^{12}×42.25×10^{-32}

= 7.14×10^{-19}

QUESTION: 4

For the following gaseous phase equilibrium,

K_{p} is found to be equal to K_{x} (K_{x} is equilibrium constant when concentration are taken in terms of mole fraction. This is attained when pressure is

Solution:

The correct answer is Option A.

K_{p} = Equilibrium constant in terms of partial pressure

K_{c} = Equilibrium constant in terms of concentration

K_{x} = Equilibrium constant in terms of mole fraction

K_{p} = K_{c}RT^{Δn} ---(1)

K_{p} = K * (P_{t})^{Δn} ---(2)

a) 1 atm

Given PCl_{5} (g) ---> PCl_{3} (g) + Cl_{2} (g)

Δn = 2 – 1

Given Kp = Kx

From (2)

K_{p} = K_{x} when P_{T} = 1

QUESTION: 5

For the reaction in equilibrium, A B

Thus, K is

Solution:

From the reaction

-d[A]/dt = d[B] /dt

⇒2.3 × 10^{6} [A] = k [B]

as given in question [B] /[A] = 4 × 10^{8}

so [A] / [B] = 1/ 4 ×10^{8}

⇒ 2.3 × 10^{6} . [A] /[B] = k

⇒ 2.3 × 10^{6} / 4 × 10^{8} = k

Or k = 5.8 × 10^{-3} /sec¹

QUESTION: 6

For the following equilibrium, N_{2}O_{4} (g) 2NO_{2}(g)

K_{p} = K_{C}. This is attained when

Solution:

The correct answer is option B

K_{P}=K_{C}(RT)^{Δn}

Δ_{n}=1

K_{P}=K_{C}(RT)

RT=1

T=1/R=1/0.0821=12.18

QUESTION: 7

is a gaseous phase equilibrium reaction taking place at 400 K in a 5 L flask. For this

Solution:

The correct answer is option B

K, is the equilibrium constant w.r.t the mole fractions of each substance at equilibrium.

K_{p} = K_{c}(RT)Δn

K_{p} = K_{x} pΔn

by solving it using partial pressure and above formula you get k_{c}=25k_{x}

QUESTION: 8

A sample of pure PCI_{5}(g)was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCI_{5} was found to be 0.05 mol L^{-1}.

Thus, [PCI_{3}] and [Cl_{2}](in mol L^{-1})at equilibrium are

Solution:

The correct answer is Option B

Let x M be the equilibrium concentration of PCl_{3}.

The equilibrium reaction is shown below.

PCl_{5}(g) ⇌ PCl_{3}(g) + Cl_{2}(g)

The equilibrium concentrations of PCl_{5}, PCl_{3} and Cl_{2} are 0.5×10^{−2}M, x and x respectively.

The expression of the equilibrium constant is

K_{c}= ( [PCl3][Cl2] ) / [PCl3].

Substitute values in the above expression.

8.3 × 10^{−3} = x^{2} / (0.5 × 10^{−1})

x^{2} = 4.15m × 10^{−4}

x ≃ 0.02M.

QUESTION: 9

Ag^{+}(aq)+NH_{3}(aq) [Ag(NH_{3})(aq)]^{+} ; K, = 3.5x 10^{-3}

[Ag(NH_{3})]^{+} (aq)+NH_{3}(aq) [Ag(NH_{3})_{2}]^{2+}(aq); K_{2} = 1.7x 10^{-3}

Formation constant of [Ag(NH_{3})_{2}]^{+}(aq) is

Solution:

The correct answer is Option C.

To get the formation constant add both reactions-

So the resultant K = K_{1} × K_{2}

= 3.5×1.7×10^{−3}

= 5.95×10^{−6}

QUESTION: 10

For the reversible reaction,

at 500° C, the value of K_{p} is 1.44x 10^{-5}, when partial pressure is measured in atmosphere. The corresponding value of K_{c} with concentration in mol L^{-1} is

**[IITJEE 2000]**

Solution:

QUESTION: 11

For the reaction,

if K_{p} = K_{c} (RT)^{X}, when the symbols have usual meaning, the value of x is (assuming ideality)

**[jee Main 2014]**

Solution:

The correct answer is Option B.

SO_{2(g)} + 1/2O_{2(g)} ⇌ SO_{3(g)}

KP = KC(RT)Δn

Δn= no. of gaseous moles of product minus no. of gaseous moles of reactant

Δn = 1−1−1/2

∴Δn = −1/2

QUESTION: 12

Given that for the equilibrium constants of two reactions,

areK_{1} and K_{2}. Equilibrium constant k_{3} of the following reaction in terms of k_{1}, and K_{2}.

Solution:

The correct answer is Option C

XeF_{6}(g) + H_{2}O(g) ⇌ XeOF_{4}(g)^{+}2HF(g)

K_{1} = ([XeOF_{4}][HF]^{2} ) / ( [XeF_{6}][H_{2}O] ) ...(i)

XeO_{4}(g) + XeF_{6}(g) ⇌ XeOF_{4}(g) + XeO_{3}F_{2}(g)

K_{2} = ([XeOF_{4}][XeO_{3}F_{2}] ) / ( [XeO_{4}][XeF_{6}] ) ...(ii)

For the reaction,

XeO_{4}(g) + 2HF(g) ⇌ XeO_{3}F_{2}(g) + H_{2}O(g)

K=[XeO_{3}F_{2}][H_{2}O]) / ([XeO_{4}][HF]^{2} ) ...(iii)

∴ From Eqs. (i), (ii) and (iii)

K= K_{2} / K_{1}

*Multiple options can be correct

QUESTION: 13

**Direction (Q. Nos. 13-14) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THANT ONE is correct.**

**F**or the following type of reversible reaction,

Solution:

*Multiple options can be correct

QUESTION: 14

Select the correct statement (s) about the following equilibrium.

Solution:

The correct answers are Options A, B and C.

During constant vapor pressure the no of molecules of H2O leaving the surface = the no of H2O molecules coming to the surface from the atmosphere. This process is dynamic as there is continuous movement of molecules.

QUESTION: 15

**Direction (Q. Nos. 15-18) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d)**

**Passage I**

**At 573 K, PCI _{5} dissociates as,
PCI_{5} (g) ⇔ PCI_{3}(g)+ Cl_{2} (g), Kp= 11.5 atm
Given, [PCI_{3}]_{eq}, = [Cl_{2}]eq = 0.01 mol L^{-1} **

**Q**.

K_{c} of this equilibrium is

Solution:

The correct answer is 0.27 mol.L-1

The given reversible gaseous reaction & ice table:

PCl_{5(g)}⇌PCl_{3(g)} + Cl_{2(g)}

I- 2mol 0 mol 0mol

C- 2α mol 2α mol 2α mol

E- 2(1-α ) mol 2α mol 2α mol

Where degree of dissociation, α=40% =0.4

Volume of the equilibrium mixture, V=2L

At equilibrium the molar concentrations of the components of the mixture are

**Equilibrium Constant:
**

QUESTION: 16

**Q. What is [PCI _{5}] eq?**

Solution:

The correct answer is Option A.

P_{PCl3} = [PCl_{3}]RT

PCl_{2} = [Cl_{2}]RT

QUESTION: 17

Passage II

The complex ion of Fe^{2+} with the chelating agent dipyridyl (abbreviated dipy) has been studied kinetically in both the forward and backward directions. For the complexion reaction,

the rate of the formation of the complex at 298 K is given by rate = (1. 45 x 10^{13} L^{3}mol ^{-3}s^{-}^{1}) [Fe ^{2+}] [dipy]^{3} and for the reverse reaction , the rate of disapperance of the complex is

**Q. What is equilibrium constant for the equilibrium ? **

Solution:

The correct answer is Option C.

Fe^{2+} + 3 dipy -----> Fe(dipy)_{3}^{2+}

R_{f} = K_{f} [Fe^{2+}] [dipy]_{3}

R_{b} = K_{b} [Fe(dipy)_{3}^{2+}]

K_{eq} = K_{b} [Fe(dipy)_{3}^{2+}] / K_{f} [Fe^{2+}] [dipy]_{3}

R_{f} = R_{b}

K_{eq} = K_{f} / K_{b}

= 1. 45 x 10^{13} / 1.24 x 10^{-4}

= 1.885 x 10^{17}

QUESTION: 18

Passage II

The complex ion of Fe^{2+} with the chelating agent dipyridyl (abbreviated dipy) has been studied kinetically in both the forward and backward directions. For the complexion reaction,

the rate of the formation of the complex at 298 K is given by rate = (1. 45 x 10^{13} L^{3}mol ^{-3}s^{-}^{1}) [Fe ^{2+}] [dipy]^{3} and for the reverse reaction , the rate of disapperance of the complex is

**Q.** If half-life Period of the backward reaction is (k = rate constant), then half-life period is about

Solution:

*Answer can only contain numeric values

QUESTION: 19

Direction (Q. Nos. 19 and 20) This section contains 2 questions. when worked out will result in an integer from 0 to 9 (both inclusive)

Q. At elevated temperature, PCl_{5} dissociates as,

At 300°C,K_{p} = 11.8 and

[PCI_{3}]= [Cl_{2}]= 0.01 moi L^{-1} at equilibrium.

[PCI_{5}] = x x 10^{-4} mol L^{-1} what is the value of x?

Solution:

Applying Kp = kcRT∆ng

11.8 = kc(0.0821)(273+300)(1)

Kc = 0.25

Applying Kc = [PCl_{3}][Cl_{2}]/[PCl_{5}]

0.25 = 10^{-4} / [PCl_{5}]

[PCl_{5}] = 10^{-4}/0.25

= 4×10^{-4}

Therefore x = 4

*Answer can only contain numeric values

QUESTION: 20

What is Kp for the equation,

When the system contains equal number of Cl (g)atom and Cl_{2}(g) molecules at 1 bar and 300 K?

Solution:

The correct answer is 2

2Cl(g) ⇌ Cl_{2}(g)

Initially take moles of Cl on right hand side to be 1 then

2Cl Cl2

1 -x

=1 - x.x

kP will be equal to K_{c} because no. of moles are equal so

1 - x = x

2x = 1

X = ½

K_{p} = x/1 - x. 1 - x

= ½ ÷ (½ * ½ )

=2

QUESTION: 21

Direction (Q. Nos. 21) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.

The progress of the reaction with time t is shown below.

Match the parameters in Column l with their respective values in Column II.

Codes

Solution:

The correct answer is Option A.

Loss in concentration of A in I hour = = 0.1

Gain in concentration of B in I hour =0.2

(i) ∵0.1 mole of A changes to 0.2 mole of B in a given time and thus, n=2

(ii) Equilibrium constant,

= 1.2mollitre^{−1}

(iii) Initial rate of conversion of A = changes in conc. of A during I hour =

= 0.1 mol litre−1hour^{−1}

(iv) ∵ Equilibrium is attained after 5 hr, where [B]=0.6 and [A]=0.3

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