Test: Chemical Equilibrium: Law of Mass Action - NEET MCQ

Kp = kc(RT)∆ng 
So to convert kp to kc we have, kc = kp(1/RT)∆ng.
So, the converting factor is 1/RT

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

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

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

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

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

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​, PCl_3​ 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.

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

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
 

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

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.
 

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:

The correct answer is Option A.
P_PCl3 = [PCl₃]RT
PCl₂ = [Cl₂]RT

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

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

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

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