Equilibrium Test 7

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Equilibrium Test 7

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Weekly Quiz Competition

Question 1
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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

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
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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
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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
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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
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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
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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
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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 = 25K_x

C
K_x = 25K_c

D
K_c = 5K_x

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
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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
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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
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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
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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
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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