Equilibrium Test

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Equilibrium Test - 6

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Question 1
4 / -1

Conversion factor for converting partial pressures (in K_p) to active masses (in K_c) is

A
nRT

B
1/RT

C
(RT)²

D
1/RT²

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
4 / -1

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

A
CO(g) + H₂O(g) CO₂(g) + H₂(g)

B

C
COCI₂(g) CO (g)+ Cl₂(g)

D
None as K_C is unitless

Solution

For the reaction COCl₂ (g) ⇌ CO(g) + Cl₂(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] [Cl₂] ÷ [COCl₂]
K = mol/L × mol/L ÷ mol/L = mol/L
Question 3
4 / -1

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

A

B

C

D

Solution

Given equations are
NO (g) + ½ O₂ (g) ⇌ NO₂ (g) ----------(i) k₁ = 1.3×106
And ½ N₂ (g) + ½ O₂ (g) ⇌ NO(g) ----------(ii) k₂ = 6.5×10^-16
To get the reaction, N₂(g) + 2O₂(g) ⇌ 2NO₂(g) ----------(iii) k₃
We multiply eqn (i) and eqn (ii) by 2 and adding both reaction, we get eqn (iii)
k₃ = k₁²×k₂²
= (1.3×10⁶)²×(6.5×10^-16)²
= 1.69×10¹²×42.25×10^-32
= 7.14×10^-19
Question 4
4 / -1

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

A
1 atm

B
0.5 atm

C
2 atm

D
4 atm

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_cRT^Δn ---(1)
K_p = K * (P_t)^Δn ---(2)
a) 1 atm
Given PCl₅ (g) ---> PCl₃ (g) + Cl₂ (g)
Δn = 2 – 1
Given Kp = Kx
From (2)
K_p = K_x when P_T = 1
Question 5
4 / -1

For the reaction in equilibrium, A B

Thus, K is

A

B

C

D

Solution

From the reaction
-d[A]/dt = d[B] /dt
⇒2.3 × 10⁶ [A] = k [B]
as given in question [B] /[A] = 4 × 10⁸
so [A] / [B] = 1/ 4 ×10⁸
⇒ 2.3 × 10⁶ . [A] /[B] = k
⇒ 2.3 × 10⁶ / 4 × 10⁸ = k
Or k = 5.8 × 10^-3 /sec¹
Question 6
4 / -1

For the following equilibrium, N₂O₄ (g) 2NO₂(g)
K_p = K_C. This is attained when

A
T = 1.0K

B
T = 12.18 K

C
T = 27.3 K

D
T = 273 K

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
4 / -1

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

A
K_c = K_x

B
K_c = 5K_x

C
K_x = 25K_c

D
K_c = 25K_x

Solution
Question 8
4 / -1

A sample of pure PCI₅(g)was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCI₅ was found to be 0.05 mol L^-1.
Thus, [PCI₃] and [Cl₂](in mol L^-1)at equilibrium are

A
a

B
b

C
c

D
d

Solution

The correct answer is Option B
Let x M be the equilibrium concentration of PCl₃.
The equilibrium reaction is shown below.
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)
The equilibrium concentrations of PCl₅, PCl₃ and Cl₂ are 0.5×10⁻²M, 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 = x² / (0.5 × 10⁻¹)
x² = 4.15m × 10⁻⁴
x ≃ 0.02M.
Question 9
4 / -1

Ag⁺(aq)+NH₃(aq) [Ag(NH₃)(aq)]⁺ ; K, = 3.5x 10^-3
[Ag(NH₃)]⁺ (aq)+NH₃(aq) [Ag(NH₃)₂]²⁺(aq); K₂ = 1.7x 10^-3
Formation constant of [Ag(NH₃)₂]⁺(aq) is

A
2.06

B
5.2 x 10^-3

C
5.95 x 10^-6

D
None of these

Solution

The correct answer is Option C.
To get the formation constant add both reactions-
So the resultant K = K₁ × K₂
= 3.5×1.7×10⁻³
= 5.95×10⁻⁶
Question 10
4 / -1

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]

A

B

C

D

Solution
Question 11
4 / -1

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]

A
-1

B
-1/2

C
+1/2

D
+1

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
4 / -1

Given that for the equilibrium constants of two reactions,

areK₁ and K₂. Equilibrium constant k₃ of the following reaction in terms of k₁, and K₂.

A
K₁ K₂

B
K₁ / K₂

C
K₂/ K₁

D

Solution

The correct answer is Option C
XeF₆(g) + H₂O(g) ⇌ XeOF₄(g)⁺2HF(g)
K₁ = ([XeOF₄][HF]² ) / ( [XeF₆][H₂O] ) ...(i)
XeO₄(g) + XeF₆(g) ⇌ XeOF₄(g) + XeO₃F₂(g)
K₂ = ([XeOF₄][XeO₃F₂] ) / ( [XeO₄][XeF₆] ) ...(ii)
For the reaction,
XeO₄(g) + 2HF(g) ⇌ XeO₃F₂(g) + H₂O(g)
K=[XeO₃F₂][H₂O]) / ([XeO₄][HF]² ) ...(iii)
∴ From Eqs. (i), (ii) and (iii)
K= K₂ / K₁
Question 13
4 / -1

Once the equilibrium is reached under given condition:

A
Cone, remains the same in spite of the change in temperature

B
Cone, of all the substances presents do not change

C
Cone, of reactants remairfs same

D
Cone, of products remains same

Solution

In a chemical reaction, chemical equilibrium is the state in which the forward reaction rate and the reverse reaction rate are equal. The result of this equilibrium is that the concentrations of the reactants and the products do not change. However, just because concentrations aren’t changing does not mean that all chemical reaction has ceased. Just the opposite is true; chemical equilibrium is a dynamic state in which reactants are being converted into products at all times, but at the exact rate that products are being converted back into reactants. The result of such a situation is analogous to a bridge between two cities, where the rate of cars going over the bridge in each direction is exactly equal. The result is that the net number of cars on either side of the bridge does not change.
Question 14
4 / -1

Ca(HCO₃)₂ is strongly heated and after equilibrium is attained, temperature changed to 25° C.

Ca(HCO₃)₂(s)⇌CaO(s) + 2CO₂ (g) + H₂O(g)
Kp = 36 (pressure taken in atm)
Thus, pressure set up due to CO₂ is

A
36 atm

B
18 atm

C
12 atm

D
6 atm

Solution

The reaction is as follow:-
Ca(HCO₃)₂(s)⇌CaO(s) + 2CO₂ (g) + H₂O(g)
At 25° C H₂O goes in liquid state
Kp = (PCaO)1×(PCO₂)₂
(PCa(HCO₃)₂)
Since, Ca(HCO₃)₂, CaO and H₂O are not in gaseous state, so their partial pressure is taken 1.
Putting all values, we have
36 = (PCO₂)₂
Or PCO₂ = 6 atm
Question 15
4 / -1

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, PCI₅ dissociates as,
PCI₅ (g) ⇔ PCI₃(g)+ Cl₂ (g), Kp= 11.5 atm
Given, [PCI₃]_eq, = [Cl₂]eq = 0.01 mol L^-1

Q.
K_c of this equilibrium is

A
2.444 x 10^-3

B
4.09

C
4.09 x 10²

D
0.27 mol L^-1

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
4 / -1

Q. What is [PCI₅] eq?

A
4.09 x 10^-4mol L^-1

B
2.444 x 10³ mol L^-1

C
4.09 x 10^-6 mol L^-1

D
2.444 x 10⁵mol L^-1

Solution

The correct answer is Option A.
P_PCl3 = [PCl₃]RT
PCl₂ = [Cl₂]RT
Question 17
4 / -1

Passage II
The complex ion of Fe²⁺ 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 10¹³ L³mol ^-3s^-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 ?

A
8.4138x 10^-18

B
2.3770x 10¹⁷

C
1.1885x 10¹⁷

D
5.9426x 10⁶

Solution

The correct answer is Option C.
Fe²⁺ + 3 dipy -----> Fe(dipy)₃²⁺
R_f = K_f [Fe²⁺] [dipy]₃
R_b = K_b [Fe(dipy)₃²⁺]
K_eq = K_b [Fe(dipy)₃²⁺] / K_f [Fe²⁺] [dipy]₃
R_f = R_b
K_eq = K_f / K_b
= 1. 45 x 10¹³ / 1.24 x 10^-4
= 1.885 x 10¹⁷
Question 18
4 / -1

Passage II
The complex ion of Fe²⁺ 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 10¹³ L³mol ^-3s^-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

A
95 min

B
5680 min

C
0.85 min

D
None of the above
Question 19
4 / -1

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, PCl₅ dissociates as,

At 300°C,K_p = 11.8 and
[PCI₃]= [Cl₂]= 0.01 moi L^-1 at equilibrium.
[PCI₅] = x x 10^-4 mol L^-1 what is the value of x?

A
4

B
5

C
7

D
9

Solution

Applying Kp = kcRT∆ng
11.8 = kc(0.0821)(273+300)(1)
Kc = 0.25
Applying Kc = [PCl₃][Cl₂]/[PCl₅]
0.25 = 10^-4 / [PCl₅]
[PCl₅] = 10^-4/0.25
= 4×10^-4
Therefore x = 4
Question 20
4 / -1

What is Kp for the equation,

When the system contains equal number of Cl (g)atom and Cl₂(g) molecules at 1 bar and 300 K?

A
5

B
4

C
3

D
2

Solution

The correct answer is 2
2Cl(g) ⇌ Cl₂(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
4 / -1

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

A
a

B
b

C
c

D
d

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