All questions of Equilibrium - Chemistry for NEET Exam

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

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.
   NH₄HS   ------->  NH₃ + H₂S
Let P be the pressure at eq. of NH₃ and H₂S.
Therefore, K_p = P²
= (20 / 2)²
= 100 mm
= 100
Also, K_p = (15 + P) (P)
100 = 15 P + P2
P² + 15 P – 100 = 0
P = 5
Total pressure = 15 + 2(P)
= 15 + 2(5)
= 25 mm

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.
 

The correct answer is option C
CaCO₃ ​→ CaO + CO_2​
K_p​ = k₁ ​= Pco₂​​
total pressure of container P
k_1​ = p
NH₄​HS → NH₃ ​+ H2​S
PNH_3​​ = PH_2​S ​= P0​
P_0​ + P₀​ = p (total pressure)
P0 ​= p/₂
k_2​ = kp ​= [P_NH3​​][P_H2​s​] p2₄​
NH₂​CoNH₂ ​→ 2NH₃ ​+ CO_2​
P_NH3​​ = 2P₀​        P_CO2​ ​= P0​
2P_0​ + P₀ ​= P

K= (partial pressure of co²/(partial pressure of co²)
since k =1.9
So 1.9 = (partial pressure of co)²/2.105
(partial pressure of co)² =2.105×1.9
= 3.99 = 4
(partial pressure of co) =2

=> 10 bar approximately

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.

Correct answer is A.

 NH4HS (s)  ⇋ NH3 (g) + H2S (g)
Initial    1                   -               -
At eqm     1-x                 x+0.02     x
Kc = [NH₃][H₂S]   (Since NH4HS is solid, we ignore it.)
1.8×10-4    = (x+0.02)(x)
x2+0.02x-1.8×10-4 = 0
Applying quadratic formula; x = -0.02+√{(0.02)2-4×1.8×10-4}
= 0.033-0.020/2 = 0.0065
Therefore, concn of NH₃ at equilibrium = x+0.020 = 0.0265
concn of H2S at equilibrium = x = 0.0065
So, option b is correct

The state of HCl is given wrong. It will be in gaseous state.
So, the reaction be like;-
NH4Cl(s)  ⇌  NH3(g) + HCl(g)        kp = 1.00×10-2 at 275° C
Kp = kc(RT)2
1.00×10-2 = kc(0.0821×548)2
Or kc = 4.94×10-6
                          NH4Cl(s)  ⇌  NH3(g) + HCl(g)
Initial  1                     -             -
At eqm 1-x                  x            x 
Kc = x2
x = √(4.94×10-6)
=  2.22×10-3
Therefore, NH4Cl dissociated at eqm = 2.22×10-3 × 53.5 = 0.118
%age decomposition = 0.118/0.980×100 = 12.13%

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

C(s) + CO₂(g) <=========> 2CO(g)
K_p = pCO²/pCO²
GIven K_p = 63 and p_CO = 10p_CO2
Putting the value of pCO in above equation,
63 = 100(_pCO²)²/p_CO2
Or p_CO2 = 0.63
p_CO = 6.3
Therefore, total pressure = 6.3+0.63 = 6.93 atm

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

 

We know that with increase in pressure on one side, reaction shifts to that side which has less no. of moles(of gaseous species). However the no. of moles are same on both sides. So the REACTION WILL REMAIN IN EQUILIBRIUM.

The correct answer is Option A.

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

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%

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 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.

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.

Initial moles of H₂ = 0.2
Initial moles of S = 1  
Kp = 6.8 * 10⁻²
Given equation:
H2(g) + S(s) ⇋ H2S(g)
Initial moles:          0.2        1
At equilibrium: (0.2-α)   (1-α)       α
Here, in the above equation we can see that hydrogen is the limiting reagent.  
∴ Kp = α/(0.2 – α)
⇒ 6.8 * 10⁻²  = α/(0.2 – α)
⇒ 1.36*10⁻² – (6.8*10⁻²)α = α
⇒ α + 0.068α = 1.36*10⁻²
⇒ α = 1.36*10⁻² / 1.068 = 1.273 * 10⁻² ← moles of H₂S
So, at equilibrium moles of H₂ = 0.2 – α = 0.2 – 1.273 * 10⁻² = 0.1873
Now, using the Ideal Gas equation,
PV = nRT ….. (i)
Where P = total pressure of the vessel
n = total no. of moles = (0.2-α) + (1-α) + α = 1.2 – α = 1.2 – 1.273*10⁻² = 1.1873
V = volume of vessel = 1 litre
R = Ideal gas constant = 0.082 L atm K⁻¹mol⁻¹
T = total temperature = 90℃ = 90+273 = 363 K
Substituting all the values in eq. (i), we get
P * 1 =  1.1873 * 0.082 * 363  
⇒ P = 35.34 atm
Thus,  
The partial pressure of H₂S at equilibrium
= (mole fraction of H₂S) * (total pressure)
= [1.273*10⁻² /  1.1873] * 35.34
= 0.3789 atm
≈ 0.38 atm

The correct answer is Option A - A, B and D only

In a dynamic chemical equilibrium the rate of the forward reaction equals the rate of the reverse reaction, so there is no net change in amounts of reactants and products over time.

Because the forward and reverse rates are equal, the concentration of each species remains constant with time even though individual molecules continue to react in both directions.

The idea that no molecules are converting in either direction is incorrect because microscopic reactions continue; the equilibrium is dynamic, not static.

At equilibrium the macroscopic variables such as temperature and, in a closed steady system, pressure remain constant; additionally, the equilibrium constant is defined for a given temperature (so maintaining temperature is essential for a fixed equilibrium position).

There is no general requirement that concentrations of reactant and product be equal; they are only constant, not necessarily equal, unless dictated by stoichiometry or the value of the equilibrium constant.

The reaction is as follow:-
NH₄Cl(s)  ⇋  NH₃(g) + HCl(g)
Kp = (pNH₃)(pHCl)
1×10^-2 = p₂   (SInce reactant dissociates into same ratio, so the partial pressure will be same for both)
100×10^-4 = p₂
Or p = 10×10^-2 = 0.10
So,  the partial pressure of NH₃ and HCl are 0.10 atm.

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

N₂​O₄​(g) ⇌ 2NO₂​(g)

We are given:

We need to find the total pressure in the flask.

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

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

We can use the ideal gas law equation:

Where:

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.

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

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

For the given reaction,

None of the provided options exactly match these concentrations.

The equilibrium reaction is:

At equilibrium, we know:

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.

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:

Thus, the equilibrium pressures can be written as:

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.

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.

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