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 reaction is:
N₂​(g) + 3H₂​(g) ⇌ 2NH₃​(g)

Given:
K_c​ = 4

Step 1: Calculate the reaction quotient (Q)
The expression for Q is:

Given concentrations:

• [N₂​] = 1mol/L
• [H₂​] = 3mol/L
• [NH₃​] = 2mol/L

So,

Step 2: Compare Q with K_c​

• K_c ​= 4
• Q = 4/27

Since Q < K_c​, the system is not at equilibrium and will shift forward to make more NH₃​.

Step 3: Analyze the statements

• Statement I: "If K_c ​= 4, then the mixture is in equilibrium."
  This is incorrect because Q ≠ K_c

• Statement II: "The reaction quotient is less than the equilibrium constant."
  This is correct because Q = 274​ is less than K_c ​= 4.

Step 1: Analyze Statement I
Statement I says:
1 mole of A(g) and 1 mole of B(g) produce 0.5 mole of C(g) and 0.5 mole of D(g) at equilibrium.
This implies the reaction:
A(g) + B(g) ⇌ C(g) + D(g)
Initially:

• Moles of A(g) = 1
• Moles of B(g) = 1

At equilibrium, we have:

• 0.5 moles of C(g)
• 0.5 moles of D(g)

So, 0.5 moles ofA and 0.5 moles of B have reacted, meaning the reaction has gone halfway to completion.
Step 2: Determine the effect of doubling the initial moles
Now, if we take 2 moles each of A(g) and B(g), the reaction should still proceed in the same manner as before, and the dissociation (reaction extent) will also double.
So, instead of 0.5 moles of A and B reacting, 1 mole of AA and 1 mole of B will react. This means the percentage dissociation ofA and B doubles, which is in line with Statement I.
Step 3: Analyze Statement II
Statement II says that the equilibrium constant K_c​ = 1.
The equilibrium constant is given by the expression:

At equilibrium, if we have:

• 0.5 moles of C and D
• 0.5 moles of A and B remaining

Then, assuming the volume is constant and we can use molar concentrations directly, the equilibrium constant would be:

Thus, Statement II is correct: K_c​ = 1.
Conclusion

• Statement I is correct because the percentage dissociation of A and B doubles when their initial amounts are doubled.
• Statement II is also correct because the equilibrium constant K_c​ = 1.

However, Statement II does not explain Statement I. The relationship between initial moles and dissociation is a separate concept from the equilibrium constant value.
Therefore, the correct answer is: Answer: D
Statement II is correct, but Statement II does not explain Statement I.

The correct answer is Option A.

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

Step 1: Understanding the equilibrium reaction

The equilibrium reaction is:

At equilibrium, we know:

• The equilibrium constant K_p​ = 0.148 at 298 K.
• The total equilibrium pressure is reduced to 1.00 atm due to an increase in the volume of the container.

Step 2: Use the equilibrium expression for K_p​

The equilibrium expression for this reaction is:

Where P_NO2​​ is the partial pressure of NO_2​ and P_N2​O4​​ is the partial pressure of N₂​O₄​ at equilibrium.

Step 3: Let’s assume an initial amount of N₂​O_4​

Let’s assume the initial pressure of N_2​O₄​ before dissociation is P₀​, and the partial pressure of NO₂​ at equilibrium is P_NO2​​. Let’s denote the fraction dissociated as x.

At equilibrium, the changes in pressures would be:

• The partial pressure of N₂​O_4​ will decrease by xP₀​.
• The partial pressure of NO_2​ will increase by 2xP_0​ (since two moles of NO₂​ are produced for each mole of N₂​O₄​).

Thus, the equilibrium pressures can be written as:

• P_N2​O4​​ = P₀ ​− xP₀​
• P_NO2​​ = 2xP₀​

Step 4: Total pressure at equilibrium

The total pressure at equilibrium is the sum of the partial pressures:

We are told that the total pressure is 1.00 atm, so:

This equation allows us to solve for P₀​ in terms of x.

Step 5: Apply the equilibrium constant expression

Substitute the equilibrium pressures into the K_p​ expression:

We know that K_p​ = 0.148, so:

Simplify the equation:

Now, substitute ​ into this equation and solve for x.

Step 6: Solve for the fraction dissociated x

After solving the equation, you will find that the fraction dissociated xxx is approximately 0.189.

K_p = K_c(RT)^∆n

K_p = 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,
⇒ K_c = 6.67×0.0821×300
K_c = 164.28.

Step 1: Write the equilibrium reaction

The reaction between N₂​O₄​ and NO₂​ is:

N₂​O₄​(g) ⇌ 2NO₂​(g)

Step 2: Use the given data

We are given:

• Volume of the flask = 15 L
• Temperature = 300 K
• Mass of the mixture = 64.4 g (which is the mass of a mixture of N₂​O₄​ and NO₂​)

We need to find the total pressure in the flask.

Step 3: Calculate moles of N₂​O₄​ and NO_2​

Let's assume the mixture contains xxx grams of N₂​O_4​ and (64.4 − x) grams of NO₂​.

• Molar mass of N_2​O₄​ = 92 g/mol
• Molar mass of NO_2​ = 46 g/mol

The moles of N₂​O_4​ and NO₂​ can be calculated as:

Step 4: Use the ideal gas law to calculate pressure

We can use the ideal gas law equation:

Where:

• P is the pressure,
• n is the total number of moles,
• R is the gas constant (0.0821 L·atm/mol·K),
• T is the temperature (300 K),
• V is the volume (15 L).

The total number of moles is the sum of the moles of N₂​O₄​ and NO_2​:

Now, we can use the ideal gas law equation:

Substitute the values into the equation and solve for P.

Step 5: Calculate the total pressure

After performing the calculations, we find that the total pressure in the flask is approximately 1.3400 atm.

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