NEET Previous Year Questions (2014-2025): Equilibrium

Ans: (a)
The relationship between the rate constants for the forward and backward reactions is:

Given that the backward rate constant is 2500 times the forward rate constant, we have:

Next, we use the equation to find Kp :
K_p = K_C (RT)^Δn_g 
Given that Δn_g = 1 (because there is a change in the number of moles from 1 mole of reactants to 2 moles of products), we substitute the known values:

Solving the equation:

Therefore, the correct value of K_p is 0.033.

Ans: (b)

• Statement A: log K = log K_a1 + log K_a2 + log K_a3 is correct because the overall ionization constant is the product of the individual ionization constants, and the logarithm of a product is the sum of the logarithms of the individual factors.
• Statement B: H₃PO₄ is stronger than H₂PO₄^- and HPO₄^2- is correct because the acid strength decreases with successive ionizations. The first ionization constant (K_a1) is the largest, making H₃PO₄ the strongest acid among its ionized forms.
• Statement C: K_a1 > K_a2 > K_a3 is correct because the ionization constants decrease sequentially as the molecule loses protons, reflecting decreasing acid strength with each step.
• Statement D: K = (K_a1 + K_a2)/2 is incorrect because the overall ionization constant is the product of K_a1, K_a2, and K_a3, not an average of the first two constants.

Therefore, the correct answer is A, B, and C only.

Ans: (b)

• Higher temperature: Since the reaction is endothermic, increasing the temperature supplies more energy to drive the reaction forward.
• Higher concentration of N₂: Adding more N₂ shifts the equilibrium toward the products.
• Higher concentration of O₂: Adding more O₂ also shifts the equilibrium toward the products.

Therefore, the correct answer is A, C, D only.

Ans: (a)
To determine in which of the given equilibria K_p and K_c are not equal, it's important to understand the relationship between these two equilibrium constants. This relationship is expressed by the equation:
K_p = K_c(RT)^Δn
where R is the gas constant,
T is the temperature in Kelvin, and
Δn is the change in the number of moles of gas (number of moles of gaseous products minus number of moles of gaseous reactants).
If Δn = 0, then
K_p and K_c are equal because (RT)⁰ = 1. However, if Δn ≠ 0, the constants will not be the same, and the degree to which they differ will depend on the temperature and the value of Δn.
Now, let's analyze each option:
Option A: 
Reactant side moles = 1, Product side moles = 2; Δn$=2-1=1$.
Option B:  
Reactant side moles = 2, Product side moles = 2; Δn$=2-2=0$.
Option C: 
Reactant side moles = 2, Product side moles = 2; Δn$=2-2=0$. 

Option D:  
Reactant side moles = 2, Product side moles = 2; Δn$=2-2=0$.
From this analysis, it is evident that K_p and K_c are not equal in Option A where Δn = 1. In all other options, since Δn = 0, K_p is equal to K_c. Thus, the correct answer is Option A.

Ans: (c)
To determine which option is correct regarding the reaction state and its direction, we need to calculate the reaction quotient Q_c and compare it to the equilibrium constant K_c. The reaction given is:
2A ⇌ B + C
The equilibrium constant expression K_c for this reaction is:

Given that K_c = 4 × 10^-3 and the concentrations of A, B, and C at this time are each 2 × 10^-3 M, we can substitute these values into the expression for K_c to calculate the reaction quotient Q_c:

Simplifying, we find:

Since Q_c > K_c (1 > 0.004), the reaction quotient is greater than the equilibrium constant. This indicates that the concentration of products (B and C) is too high relative to the concentration of reactants (A) for the system to be at equilibrium under these conditions.
This means that the reaction has a tendency to move in the backward direction to reach equilibrium, reducing the concentration of the products (B and C) and increasing the concentration of the reactant (A). Therefore, the correct answer to the given question is:
Option C: Reaction has a tendency to go in backward direction.

Ans: (d)

Ans: (c)
The equilibrium expression for the given reaction is:
K_c = [PCl₃][Cl₂] / [PCl₅]
Substitute the given concentrations into the expression:
K_c = (1.60)(1.60) / (1.40)
K_c = 2.56 / 1.40
K_c = 1.83
So, the value of K_c is 1.83.

Ans: (b)
To convert from K_c to K_p, we use the relationship:
K_p = K_c × (RT)^(Δn)
Where:

• K_p is the equilibrium constant in terms of partial pressure.
• K_c is the equilibrium constant in terms of concentration.
• R is the universal gas constant (8.314 J/mol·K).
• T is the temperature in Kelvin (1000 K in this case).
• Δn is the change in the number of moles of gas between the products and reactants.

For the reaction:
2 NOCl_(g) ⇌ 2 NO_(g) + Cl₂_(g)

• The number of moles of gas on the left side is 2 (NOCl).
• The number of moles of gas on the right side is 2 (NO) + 1 (Cl₂) = 3.

So, Δn = 3 - 2 = 1.
Now, substitute the values into the equation:
K_p = K_c × (RT)^(Δn)
K_p = 3.0 × 10⁻⁶ × (8.314 × 1000)¹
K_p = 3.0 × 10⁻⁶ × 8314
K_p = 2.494 × 10⁻²

Thus, K_p = 2.494 × 10⁻².

Ans: (d)
To find the relation between the equilibrium constants K₁ and K₂, let's first analyze the given equilibria.

1. The first equilibrium is:
   3 A₂ + B₂ ⇌ 2 A₃B, with equilibrium constant K₁.
2. The second equilibrium is:
   A₃B ⇌ 3/2 A₂ + 1/2 B₂, with equilibrium constant K₂.

Now, notice that the second equilibrium is essentially the reverse of the first one, but with the coefficients halved.
To relate K₁ and K₂, we need to consider the effects of reversing the reaction and adjusting the coefficients:

• When a reaction is reversed, the equilibrium constant is inverted. So, for the reaction A₃B ⇌ 3/2 A₂ + 1/2 B₂, the equilibrium constant will be 1/K₁ (since it's the reverse of the first reaction).
• Also, if the coefficients of the reaction are divided by 2, the equilibrium constant changes by a factor of √(K₁). This is because the equilibrium constant is related to the powers of the concentrations of the species involved, and halving the coefficients takes the square root of K₁.

Therefore, the relationship between K₁ and K₂ is: K₂ = 1 / √K₁.
Thus, the correct answer is (d) K₂ = 1 / √K₁.

Ans: (d)

• The reaction is exothermic (ΔH = -Q), which means it releases heat. According to Le Chatelier's Principle, the system will shift in the direction that opposes the change. To favor the forward reaction (formation of ammonia), conditions should favor the products.
• High pressure: In this reaction, there are 4 moles of reactants (1 N₂ + 3 H₂) and 2 moles of product (2 NH₃). An increase in pressure will shift the equilibrium towards the side with fewer moles of gas (towards NH₃), thus favoring the forward reaction.
• Low temperature: Since the reaction is exothermic (ΔH = -Q), lowering the temperature will favor the forward reaction, as it would release heat and oppose the decrease in temperature.
• Higher concentration of H₂: Increasing the concentration of one of the reactants (H₂) will shift the equilibrium towards the formation of more ammonia, as per Le Chatelier's Principle.

Ans: (a)
To find the ratio of solubility of AgCl in 0.1 M KCl solution to its solubility in water, we will use the solubility product (Ksp) and the common ion effect.
Step 1: Solubility of AgCl in Water
The solubility product expression for AgCl is:AgCl ⇌ Ag⁺ + Cl⁻
The solubility product (Ksp) of AgCl is given as K_sp = 10⁻¹⁰. If the solubility of AgCl in water is S, the concentration of Ag⁺ and Cl⁻ will both be S in pure water. Hence, the expression for the solubility product becomes: K_sp = [Ag⁺][Cl⁻] = S × S = S²
Therefore : S² = 10⁻¹⁰So, S = 10⁻⁵ M.

Step 2: Solubility of AgCl in 0.1 M KCl Solution
Now, in the presence of 0.1 M KCl, the concentration of Cl⁻ ions will already be 0.1 M, and the solubility of AgCl will decrease due to the common ion effect. Let the new solubility be S'.
For the dissociation of AgCl:AgCl ⇌ Ag⁺ + Cl⁻
The solubility product expression becomes: K_sp = [Ag⁺][Cl⁻]
Since the concentration of Cl⁻ is now 0.1 M (from KCl), we have: K_sp = S' × 0.1
Substitute the value of K_sp = 10⁻¹⁰:10⁻¹⁰ = S' × 0.1
Solve for S':S' = 10⁻¹⁰ / 0.1 = 10⁻⁹ M

Step 3: Find the Ratio of Solubilities
The ratio of the solubility of AgCl in 0.1 M KCl solution to its solubility in water is:(S' / S) = (10⁻⁹ / 10⁻⁵) = 10⁻⁴
Thus, the correct answer is (a) 10⁻⁴.

Ans: (d)

• Lewis acid is a substance that can accept an electron pair.
• OH⁻ (Hydroxide ion) is a Lewis base because it can donate an electron pair, not accept one.
• NH₃ (Ammonia) is also a Lewis base because the nitrogen atom in ammonia has a lone pair of electrons that it can donate.
• H₂O (Water) can act as both a Lewis acid and base, but it is more often considered a Lewis base because it can donate a lone pair of electrons.
• BF₃ (Boron trifluoride) is a Lewis acid. The boron atom in BF₃ has an incomplete octet and can accept an electron pair from a Lewis base.

Hence, BF₃ is the Lewis acid among the given options.

Ans: (d)
To calculate the standard Gibbs free energy change (ΔG⁰) for the reaction, we can use the following equation:
ΔG⁰ = ΔG⁰° + RT ln(Q)
Where:

• ΔG° is the standard Gibbs free energy change (which we are calculating),
• R is the gas constant (2 cal/mol·K),
• T is the temperature in Kelvin (300 K),
• Q is the reaction quotient.

The reaction is:
A + B ⇌ C + D
Step 1: Write the reaction quotient (Q).
The reaction quotient Q is given by:
Q = [C][D] / [A][B]
From the given equilibrium concentrations:

• [A] = 2 mol/L,
• [B] = 3 mol/L,
• [C] = 10 mol/L,
• [D] = 6 mol/L.

Substitute these values into the equation for Q:
Q = (10 × 6) / (2 × 3) = 60 / 6 = 10.

Step 2: Use the equation for ΔG⁰.
The formula for the standard Gibbs free energy change at equilibrium is:
ΔG⁰ = ΔG⁰ + RT ln(Q)
At equilibrium, ΔG⁰ = 0, and therefore:
0 = ΔG⁰ + RT ln(Q)
Rearranging:
ΔG⁰ = -RT ln(Q)
Substitute the values:

• R = 2 cal/mol·K,
• T = 300 K,
• Q = 10.

ΔG° = - (2 cal/mol·K × 300 K) × ln(10)
We know ln(10) ≈ 2.3026.
ΔG⁰ = - (600) × 2.3026
ΔG⁰ ≈ -1381.56 cal

Step 3: Round to the nearest value.
The calculated value of ΔG° is approximately -1381.80 cal.
Thus, the correct answer is (d) -1381.80 cal.

Ans: (a)

• A buffer solution resists changes in pH when small amounts of acid or base are added. A weak acid and its salt with a strong base form an acidic buffer because the salt provides the conjugate base, and the weak acid provides the acidic component.
• Option (a): A weak acid and its salt (formed from the reaction of the weak acid with a strong base) create an acidic buffer solution. For example, acetic acid (CH₃COOH) reacts with sodium hydroxide (NaOH), forming sodium acetate (NaOAc), which is the salt. The weak acid and the conjugate base (from the salt) balance each other and create an acidic buffer.
• Option (b): Mixing equal volumes of equimolar solutions of a weak acid and weak base may not form a buffer unless specific conditions are met. For it to be a buffer, the pKa of the acid and pKb of the base need to be similar. This combination does not always result in a buffer solution.
• Option (c): A strong acid and its salt with a strong base do not form a buffer solution because both the acid and base completely dissociate, and there is no weak component to resist changes in pH.
• Option (d): A strong acid and its salt with a weak base will not form a buffer either, as the strong acid dissociates completely, and no weak acid/base pair is present to maintain the pH.

Thus, the combination that will form an acidic buffer is weak acid and its salt with a strong base.

Ans: (c)
The percentage of dissociation for a weak acid is given as approximately 1%. This means that 1% of the initial acid concentration dissociates into ions at equilibrium.
Let's calculate the K_a (acid dissociation constant) using the following steps:

Given concentration of the acid (HA) = 0.1 mol/L
Percentage dissociation = 1% = 0.01 (as a decimal)
The amount dissociated at equilibrium is:
[HA dissociated] = 0.1 mol/L × 0.01 = 0.001 mol/L
At equilibrium, the concentration of dissociated ions [H⁺] and [A⁻] will both be 0.001 mol/L.
The concentration of undissociated HA will be:
[HA] = 0.1 mol/L - 0.001 mol/L = 0.099 mol/L
The expression for the acid dissociation constant Ka is:
K_a  = [H⁺][A⁻] / [HA]
Substituting the values:
K_a  = (0.001) × (0.001) / 0.099 = 1 × 10⁻⁶ / 0.099 = 1 × 10⁻⁵
Thus, the K_a  of the acid is approximately 1 × 10⁻⁵.

Ans: c
Given solution is acidic buffer solution.

Ans: c

Ans: (d)

Ans: (c)

Ans: (a)

Calculate the pH of dimethylammonium acetate is as follows:

pH = 7.75
Hence, option 1 is the correct answer.

Ans: (c)

Ans: (a)

pH = 9 Hence pOH = 14 - 9 = 5
[OH^-] = 10^-5 M

Thus K_sp = [Ca²⁺][OH^-]²

Ans: (c)

HF on loss of H⁺ ion becomes F^- is the conjugate base of HF
Example :

Ans: c

This is basic solution due to NaOH.
This is not basic buffer.

Hydrolysis of salt takes place.
This is not basic buffer.

This is basic buffer

⇒ Neutral solution

Ans: (a)

Ans: (d)
(A) 60 mL M/$\frac{10}{}$ HCl + 40 mL M/$\frac{10}{}$ NaOH
Mili. moles of HCl = 60 $\times$ 1/$\frac{10}{}$ = 6
Mili. moles of NaOH = 40 $\times$ 1/10 = 4
Mili. moles of HCl remaining = 6 - 4 = 2
Total volume will be 60 + 40 = 100 mL
Concentration of [H⁺] = $\frac{2}{}$2/100 = 2 $\times$ 10^-2
$\therefore$ pH = 2 - log2 = 1.7
(B) 55 mL M/$\frac{10}{}$ HCl + 45 mL M/$\frac{10}{}$ NaOH
Mili. moles of HCl = 55 $\times$ 1/10 = 5.5
Mili. moles of NaOH = 45 $\times$ $\frac{1}{}$/10 = 4.5
Mili. moles of HCl remaining = 5.5 - 4.5 = 1
Concentration of [H⁺] = $\frac{1}{100}$ = 10^-2
$\therefore$ pH = 2
(C) 75 mL M/$\frac{5}{}$ HCl + 25 mL M/$\frac{5}{}$ NaOH
Mili. moles of HCl = 75 $\times$ $\frac{1}{}$/5 = 15
Mili. moles of NaOH = 25 $\times$1/5 = 5
Mili. moles of HCl remaining = 15 - 5 = 10
Total volume will be 75 + 25 = 100 mL
Concentration of [H⁺] = 10/100 = 10^-1
$\therefore$ pH = 1
(D) 100 mL M/$\frac{10}{}$ HCl + 100 mL M/ 1$\frac{0}{}$ NaOH
Mili. moles of HCl = 100 $\times$ $\frac{1}{}$/10 = 10
Mili. moles of NaOH = 100 $\times$ $\frac{1}{}$/10 = 10
Mili. moles of HCl remaining = 10 - 10 = 0
So, it is neutral solution.
$\therefore$ pH = 7

Ans: (a)
For reaction ΔH = - ve and Δn_g = - ve
∴ High P, Low T, favour product formation.

Ans: c

Ans: a

Ans:(d)

Ans: (c)
For MY,

Ans: (b)
This is Clausius-Clapeyron equation.

Ans: (b)
C₅H₅N + H₂O → C₅H₅N⁺H + OH^-
So, the amount of pyridine that forms pyridinium ion is α. 

So, percentage of pyridine that forms pyridinium ion
= 1.30 × 10^-4 × 100
= 1.30 × 10^-2 = 0.013%

Ans: (b)
PF₃ is lewis base because on P atom there is lone pair.

Ans: (b)

Concentration of Cl^- is (s + 0.1) mol L^-1 because s mol L^-1 from ionization of AgCl and 0.1 mol L^-1 from ionization of 0.1 M NaCl.
Now, K_sp = [Ag⁺][Cl^- ]
$\Rightarrow$ 1.6 × 10^-10 = s (s + 0.1)
$\Rightarrow$ 1.6 × 10^-10 = s (0.1) {$\because$ s << 0.1}
$\Rightarrow$ s = 1.6 × 10^-9 M

Ans: (a)

Ans: (d)

Ans: (d)

As equal volumes of HCl and NaOH are added so the volume of resulting solution becomes double and the concentration of the solution becomes half.
$\therefore$ [OH^-] = 0.09/2 = 0.045 M
$\therefore$ pOH = - log[OH^- ] = -log [0.045] = 1.35
$\therefore$ pH = 14 - pOH = 14 - 1.35 = 12.65 

Ans: (d)
An acidic buffer is a mixture of a weak acid and its salt with a strong base. Among CH₃COOH, H₂CO₃, H₃PO₄ and HClO₄, the HClO₄ is a strong acid while all other are weak acid thus, HClO₄ and NaClO₄ does not constitute to form an acidic buffer. 

Ans: (b)
As the forward reaction is exothermic and leads to lowering of pressure (produces lesser number of gaseous moles). Hence, According to Le Chatelier's principle, at high pressure and low temperature, the given reversible reaction will shift in forward direction to form more product.

Ans: (d)

Ans: (c)
For exothermic reactions, on increasing the temperature the value of equilibrium constant decreases.

Ans: (a)
The highest pH refers to the basic solution containing OH^- ions. Therefore, the basic salt releasing OH^- ions on hydrolysis will give highest pH in water.
Only the salt of a strong base and weak acid would release OH^- ion on hydrolysis. Among the given salts, Na₂CO₃ corresponds to the basic salt as it is formed by the neutralisation of NaOH [strong base] and H₂CO₃ [weak acid]