Test: Chemical Equilibrium: Homogeneous Equilibrium - NEET MCQ

None of the provided options exactly match these concentrations.

Molar mass of CH₃COOH=60gmol^-1
Molar mass of C₂H₅OH=46gmol^-1
Molar mass of CH₃COOC₂H₅=88gmol^-1
∴[CH₃COOH)Initial=6060×5=0.2molL^-1
 [C₂H₅OH]Initial=4646×5=0.2molL^-1
[CH₃COOC₂H₅]eqm=4488×15=0.1molL-1
CH₃COOH+C₂H₅OH⇔CH₃COOC₂H₅+H₂O
Initial
0.2M          0.2M
0.2M           0.2M
At eqm.
(0.2−0.1)M (0.2−0.1)M 0.1M 0.1M
(0.2-0.1)M (0.2-0.1)M 0.1M 0.1M
∴K=[CH₃COOC₂H₅][H₂O]/[CH₃COOH][C₂H₅OH]
=0.1×0.1/0.1×0.1=1
In second case,
[CH₃COOH]Initial=0.4M
[C₂H₅OH]Initial=0.2M
If x is the amount of acid and alcohol reacted
[CH₃COOH]eqm.=(0.40-x)M
[C2H₅OH]eqm.=[0.2-x]M
[CH₃COOC₂H₅]eqm.=[H₂O]eqm.=xM
∴K=x²/(0.4-x)(0.2-x)=1
x=860M
∴Moles of ethyl acetate produced =860×5=23
Mass of ethyl acetate produced =23×88=58=58.66g.

On substituting the values of conc. of NO, O₂ and NO₂ in given rate equation, we get a +ve (positive) value indicating that the reaction takes place in forward direction.

Squaring on both sides 



⇒
⇒
⇒
=> 10 bar approximately

Let the initial volume of N₂O₄ be x and initial volume of NO₂ is 0
If the degree of dissociation is a, then the final volume of N₂O₄ is x(1−a) and NO2 is 2ax.
Initial
It equilibrium
N2O4            ⟶              2NO2
x                                       0
x(1−a)                               2ax
Total initial volume =x+0=x
Final volume =x(1−a)+2ax=x+ax=x(1+a)
It is given that the initial volume is 25% less than the final volume
x=0.75×(1+a)
1+a=1.33
a=0.33
So %age dissociation = 33.33%

The correct answer is Option A.
    
                 2H₂S(g) ⇌ 2H₂(g) + S_2(g)
​
Pressure
at t=0           Pi                −           −
at eqm       Pi−P            2P          P

as P=0.02    thus Pi−P=10−0.02
     Pi=10                     2P=0.04

Kp = 3.23×10⁻⁷ atm.

Correct answer is A.

The correct answer is Option A.    
                N2O4  ⇌  2NO2
Initial            1                 0           
Equilibrium  1−x             2x

Total moles = 1 - x + 2x 

NO2 is 50% of the total volume when equilibrium is set up.
Thus, the volume fraction (at equilibrium) of NO2 = 50/100 = 0.5 = ½
So,    2x / (1+x) = ½
     => x = ⅓

For 1 litre;
Kc = [NO2] / [N2O4]
    = [4*(1/9)] / [⅔]
    = 0.66; 

For 5 litres; 
Kc = 0.66 / 5
= 0.133
Thus, option A is correct.
 

For the equilibrium reaction:
A+2B ⇌ 2C+D
volume of flask = 1L
Initial moles of A = 2 mol
initial concentration of A=[A]i = 2 M
initial mole of B = 1 mol 
[B]i = 1 M
[A]_eq = 2-x, [B]_eq = 1-2x, [C]_eq = x, [D]_eq = 3x
Given [D]_eq = 1 * 1L
= 1 M
Thus x = 1M
[A]_eq = 1, [B]_eq = -1, [C]_eq = 1, [D] = 3
Kc = {([D]eq)³ * ([C]eq)}/{[A]eq * ([B]eq)²} 
= Kc = {(3)^3*1}/{1*(-1)²}
= 27/1
= 27

The correct answer is option A
2.03x 10^-4
The given equation is :-
 N2​(g)+3H2​(g) ⇌ 2NH_3​(g)
Initial moles : 1             3         0
At eqm ;       (1−x)    (3−3x)   (2x)               
(let)
Total moles of equation
 =1 − x + 3 − 3x + 2x = (4−2x)
Now, X(NH_3​) = 
⇒ 2x = 2 − x
⇒ 3x = 2 ⇒ x = 0.66 = 
32​
Now, at equation, moles of N₂​= 1/3, moles of NH_3​ = 4/3
             moles of H2 ​ =3 − 2 = 1

 

The correct answer is Option A.

NH₄OH + H+ ⇌ NH₄+ + H_2​O
∴
Again,
NH₃ + H₂O → NH₄OH ⇌ NH
4++ OH-

 = [OH
-] [H+] (∵H2​O is in excess)
       =K_w
          =1×10^-14
∴K = K×1×10^-14
        =1.8×10⁹×10^-14
        =1.8×10^-5

0.4m of KMnO₄ = 1 mole of SO₂
​= 1 mole of SO₃ 
2SO₂ + O₂ ⇌ 2SO₃
2      1         0
1      0.5       1
K = [SO₃]²/[SO₂]² [O₂]
= 12/(1²*0.5)
= 2

The correct answer is Option A.

Fraction of N₂O₄ dissociated = x = 0.155 (x = mole fraction)

Kp = Kc(RT)^∆n

Kp = 6.67 ,
∆n = moles of products - moles of reactants = 1-2 = -1
R = 0.0821 L atm mol-¹K-¹
T = 300K

Substitute these values in the formula,
=> Kc = 6.67×0.0821×300
Kc = 164.28.
 

The correct answer is 4
2SO₂(g) + O₂(g) ⇋ 2SO3
Initial moles      2            1
At equilibrium 2 - 2x     1 - x    2x
Net moles at equilibrium  =  2 - 2x + 1 - x + 2x
=(3 - x)moles
Initial:
         moles = 3, 
    Pressure = 3 atm,
      Volume = 2L,
             PV = nRT
          3 x 2 = 3RT  -------- 1
At equilibrium
     Moles = 3 - x,
Pressure = 2.5 atm
  Volume = 2L
        P‘V = n’RT ---------- 2
Divide eqn  2 by 1

⇒2.5 = 3 - x
⇒x = 0.5