XI Chemistry Numericals on Equilibrium Constant (Kc & Kp, Ksp) & Equilibrium Concentration | Step-by-Step Important Solved Questions with Chemistry Tricks 2025

Learn Class XI Chemistry Equilibrium Constant (Kc, Kp) and Equilibrium Concentration numericals with easy formulas, solved examples, and smart tips for MDCAT & Board Exams 2025.

The concept of Chemical Equilibrium is one of the most important and scoring topics in XI Chemistry. Students often get confused while solving numericals related to Kc, Kp, and equilibrium concentrations.

Welcome to Learn Chemistry by Inam Jazbi, where complex concepts turn simple and every numerical becomes easy! 💡

Today we’ll dive into one of the most important topics of Class XI Chemistry — Equilibrium Constant (Kc, Kp) and Equilibrium Concentration.

These numericals often confuse students, but don’t worry — by the end of this post, you’ll solve them like a pro.

We’ve covered step-by-step solved examples, smart exam tips, and MDCAT-style numericals that are most expected in 2025 Board Exams.

Let’s break down the mystery of equilibrium and make Chemistry learning fun, easy, and unforgettable — only at Learn Chemistry by Inam Jazbi! ⚗️✨

⚡ Tips & Tricks for Solving Equilibrium Numericals

Write the balanced chemical equation first.

Coefficients in the equation are crucial for setting up concentration terms.

Use the ICE Table method (Initial, Change, Equilibrium):

Write initial moles/concentrations

Show the change (–x or +x)

Write equilibrium expressions

For equilibrium constant (Kc):

Kc​ = [Reactants]/[Products]​

Raise each concentration to the power of its coefficient.

For equilibrium pressure (Kp):

Kp=Kc(RT)Δn

where Δn = moles of gaseous products – moles of gaseous reactants

If Kc is small (Kc < 10⁻³), reaction barely proceeds → reactants dominate.

If Kc is large (Kc > 10³), products dominate.

Temperature affects equilibrium — not Kc formula itself, but the position of equilibrium.

Always keep units consistent:

Concentration in mol dm⁻³, Pressure in atm, R = 0.0821 or 8.314 (choose carefully).

Formulas Used in Problems of Chemical Equilibrium

 

Types of equilibrium Constants

 Relationships between Three equilibrium Constants

Where                                                                                    

R   = General gas constant                                               

T   = Absolute temperature                                             

P   = Pressure of the system                                            

N   = Total number of moles of reactants and products

∆n = difference of number of moles reactants and products (i.e. ∆n = n _product – n _reactant)

When               

∆n = 0

K_p = K_c = K_x = K_n

 

Reaction Quotient      

Formulas on K_sp     

     

Formulas Molar Solubility

🧩 Example Numerical (Solved)

Q1: For the reaction:N₂(g)+3H₂(g)⇌2NH₃(g)

At equilibrium, concentration of H₂ = 0.2 mol dm⁻³, I₂ = 0.2 mol dm⁻³, and HI = 0.6 mol dm⁻³.

Find the value of Kc.

Solution:

$$K_c=\frac{[HI{]}^2}{[H_2][I_2]}$$$$K_c=\frac{(0.6{)}^2}{(0.2)(0.2)}=\frac{0.36}{0.04}=9$$

✅ Kc = 9

Tip: Always check powers (squares, cubes) based on balanced coefficients.

---

Q2: The equilibrium constant (Kc) for

$$N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$$

is 50 at 400 K. Find Kp.

Solution:

$$K_p=K_c(RT{)}^{\mathrm{\Delta }n}$$

Here, Δn = (2 – (1+3)) = –2

$$K_p=50(0.0821\times 400{)}^{-2}=50\mathrm{/}(32.84{)}^2\approx 50\mathrm{/}1078=0.0464$$

✅ Kp = 0.0464 atm⁻²

Writing Equilibrium Constant Expression for Reaction

General Consideration

To write an equilibrium constant (K_c) expression from given balanced chemical equation, all products given in the balanced equation are placed on numerator while reactants on denominator enclosed in square bracket separately. Each concentration term of given substance is raised to the power of its stoichiometric coefficient in the balanced chemical equation.

To write K_p expression from given balanced chemical equation, all products given in the balanced equation are placed on numerator while reactants on denominator in enclosed parenthesis separately. Each partial pressure term of given substance is raised to the power of its stoichiometric coefficient in the balanced chemical equation.

Q1. Write down the K_c and K_p expression for the following reversible reactions

(i)  2NO_2(g)  + 7H_2(g)  ⇌  2NH_3(g) + 4H₂O_(g)

 (ii)2H₂S_(g)  + 3O_2(g)  ⇌  2SO_2(g)  + 2H₂O_(g)

Answer 

Q2. Write down the equilibrium constant expression or K_c equation for given reactions

(i)        H_2(g)      +  I_2(g)    ⇌   2HI_(g)

(ii)         CO_(g)     +  3H_2(g)  ⇌  CH_4(g)  +  H₂O_(g)

(iii)       2NO_2(g)       ⇌ N₂O_4(g)

(iv)        PCl_3(g)+  Cl_2(g)   ⇌  PCl_5(g)

(v)         N_2(g)       +  2O_2(g) ⇌ 2NO_2(g)

(vi)        N_2(g)       +  3H_2(g) ⇌  2NH_3(g)

(vii)      H_2(g)       +  Br_2(g)   ⇌   2HBr_(g)

(viii)     SO_2(g) +  NO_2(g) ⇌  NO_(g)     + SO_3(g)

(ix)        N₂O_(g) ⇌ N_2(g) + ½ O_2(g)

Answer 

Q3. Write down the equilibrium constant expression or K_c equation for given reactions

Answer 

Q1.  An essential step in Contact process is the oxidation of SO₂ to SO₃  shown as        2SO_2(g) + O_2(g) ⇌ 2SO_3(g)

If an experiment, there are 5 moles of SO₂, 4 moles of O₂ and 2.8 moles of SO₃ present at  equilibrium state in a 2dm³ flask. Calculate K_c.  (Example 7.3, Page 143)

Solution

Conversion of moles into molar concentration

volume of flask = 2 dm³

Molar equilibrium concentration of SO₂ = [SO₂] = moles/volume = 5/2    = 2.5 mol/dm³

Molar equilibrium concentration of O₂   = [O₂]   = moles/volume = 4/2     = 2.0 mol/dm³

Molar equilibrium concentration of SO₃ = [SO₃] = moles/volume = 2.8/2 = 1.4 mol/dm³

ICE Table (Initial, Change and Equilibrium Concentration Table)

Q2. Ethyl acetate is an ester of ethanol and acetic acid commonly used as an organic solvent; 

CH₃COOH_(l)  + C₂H₅OH_(l) ⇌ CH₃COOC₂H_5(l)  + H₂O_(l)

 In an esterification process, 180 g of acetic acid and 138 g of ethanol were mixed at 298K and allowed to start reaction under necessary conditions. After equilibrium is established,    60 g of unused acid were present in the reaction mixture. Calculate K_c.

(Example 7.4, Page 143)

Solution

Conversion of Mass into moles

Suppose volume of container = 1 dm³

Moles of CH₃COOH   = mass/molar mass = 180/60 = 3 mol

Moles of C₂H₅OH     = mass/molar mass = 138/46 = 3 mol

Moles of unused acid = mass/molar mass = 60/60  = 1 mol

Equilibrium concentration of acid = Initial mole – x = 3 – x

Moles of unused acid = 1

Therefore

3 – x  = 1 or x = 3 – 1 = 2

 

ICE Table (Initial, Change and Equilibrium Concentration Table)

Q3. At 444^oC reaction of hydrogen and iodine is performed in a sealed 1 dm³ steel vessel

H_2(g) + I_2(g)  ⇌  2HI_(g)

If equilibrium mixture contains 1 mole of H₂, 1 mole of I₂ and 7 moles of HI, calculate

(a) Equilibrium constant (K_c)                             

(b) Initial concentration of H₂ and I₂

(Example , Page 143)

Solution

Q8. In the synthesis of nitric acid by Ostwald process, one of the important reactions is the oxidation of nitric oxide to nitrogen dioxide.                                                                                                                                                                           (Example 7.6; Page # 155)

                  2NO_(g)  + O_2(g)  ⇌ 2NO₂   ∆H^o = − 114 kJ/mol

Use Le-Chatelier’s principle to predict the direction of reaction when the equilibrium is disturbed by

(a) Increasing the pressure                                                

(b) Increasing the temperature

(c) Adding O₂                                                                                                 

(d) Removing NO

 Answer

Predicting effect of Increasing Pressure using Le-Chatelier’s Principle

Since reaction occurs by decrease in moles or volume from 3 moles of gaseous reactants to 2 moles of gaseous products, an increasing pressure shifts the equilibrium to the side where there is less no. of moles i.e. to the right thus more product (NO₂) will be formed.

Predicting effect of increasing temperature using Le-Chatelier’s Principle

Since the forward reaction is exothermic as shown by the negative sign of ∆H, so increasing temperature shifts the equilibrium to the side where heat is absorbed i.e. to the left thus more reactants (NO and O₂) will be formed.

Predicting effect of adding more O₂ (reactant)

Adding more reactant here O₂ gas in the reaction mixture shifts the reaction to the right giving  more products (NO₂).

Predicting effect of removing NO gas (product)

Removing product here NO gas from the reaction mixture shifts the system to the left producing more reactants. 

Q1. 9.2 g ethyl alcohol and 12 g of acetic acid kept at constant temperature until equilibrium was established. 4.0 g of acid was left unused. Calculate Kc for the reaction.                                         [K.B- 2021]

Solution

Conversion of Mass into moles

Moles of CH₃COOH = mass/molar mass = 12/60  = 0.2 mol

Moles of C₂H₅OH  = mass/molar mass = 9.2/46 = 0.2 mol

Moles of unused acid = mass/molar mass = 4.0/60  = 0.067 mol

Suppose volume of container = V dm³

Suppose

No. of equilibrium moles of reactant = Initial mole – x

No. of equilibrium moles of acid   = 0.2 – x

No. of equilibrium moles of alcohol = 0.2 – x

Therefore

0.2 – x = 0.067 or x = 0.1 – 0.067 = 0.1333 mol

Q2. 6.0 g CH₃COOH and 4.6 g C₂H₅OH were heated together and allowed to attain equilibrium state. Calculate K_c for the reaction, if 2.0 g of CH₃COOH is found unconsumed at equilibrium state. [K.B- 2019]

Solution

Conversion of Mass into moles

Moles of CH₃COOH = mass/molar mass = 6/60 = 0.1 mol

Moles of C₂H₅OH  = mass/molar mass = 4.6/46 = 0.1 mol

Moles of unused acid = mass/molar mass = 2/60 = 0.033 mol

Suppose volume of container = V dm³

Suppose

No. of equilibrium moles of reactant  = Initial mole – x

No. of equilibrium moles of acid   = 0.1 – x

No. of equilibrium moles of alcohol = 0.1 – x

Therefore

0.1 – x = 0.033 or x = 0.1 – 0.033 = 0.067 mol 

Q3. One mole of PCl₅ was introduced in a vessel of 10 dm³ capacity at constant temperature. A equilibrium, 0.465 moles of Cl₂ gas were present. Calculate K_C.  PCl₅ ⇌ PCl₃ +  Cl₂

Solution

ICE Table (Initial, Change and Equilibrium Concentration Table)

Q4. 1.5 moles of acetic acid and 1.5 moles of ethyl alcohol were reacted at a certain temperature. At equilibrium, 1 mole of ethyl acetate was present in one litre of the equilibrium mixture. Calculate K_C.

Solution

ICE Table (Initial, Change and Equilibrium Concentration Table)

Q5. 0.2 mole of A and 0.4 mole of B were reacted at a certain temperature and were allowed to come to an equilibrium. The equilibrium mixture contained 0.1 mole of A and 0.2 mole of AB; find K_c if the volume of the container was 2 dm³.

Solution

ICE Table (Initial, Change and Equilibrium Concentration Table)

Q6. 1 mole of HI is introduced into a vessel held at constant temperature when equilibrium is reached, it is found that 0.1 mole of I₂ have been formed.  Calculate K_C.

Solution

ICE Table (Initial, Change and Equilibrium Concentration Table)

Suppose volume of container is V:

Q7. Phosphorus pentachloride decomposes in a gas phase reaction at 250°C as follows:

                                             PCl₅       ⇌  PCl₃ +  Cl₂

An equilibrium mixture in a 5 liter container is found to have 3.84 g PCl₅, 9.14 g PCl₃ and 2.84 g of Cl₂. Evaluate K_c at 250°C. (K.B- 2017)

Solution                                                                                       

Conversion of Mass into moles

Molar mass of PCl₅ = 31 + 5(35.5) = 208.5 gmol⁻¹

Molar mass of PCl₃ = 31 + 3(35.5) = 137.5 gmol⁻¹

Molar mass of Cl₂   = 2(35.5)  = 71 gmol⁻¹

Moles of PCl₅ = mass/molar mass = 3.84/208.5 = 0.0184 mol

Moles of PCl₃ = mass/molar mass = 9.14/137.5 = 0.0665 mol

Moles of Cl₂ = mass/molar mass  = 2.84/71= 0.04 mol

volume of container = 5 liter

Q8.    164.4 g N₂ and 518.4 g O₂ gases are mixed and heated at 2300^oC until the equilibrium is established. 338.4 g nitric oxide (NO) is formed. Calculate Kc.

               N_2(g) + O₂ ⇌ 2NO

Solution                                                                                       

Conversion of Mass into moles

Molar mass of N₂  = 2(14) = 28 gmol⁻¹

Molar mass of O₂  = 2(16) = 32 gmol⁻¹

Molar mass of NO = 14+16 = 30 gmol⁻¹

Moles of N₂   = mass/molar mass = 164.4 /28 = 5.87 mol

Moles of O₂  = mass/molar mass = 518.4 /32 = 16.2 mol

Moles of NO = mass/molar mass = 338.4 /30 = 11.28 mol

Let, volume of container = V liter

Q9.  In a reaction;  H₂  +  I₂ ⇌  2HI, when equilibrium was attained, the concentrations were [H₂]  =  [I₂]   = [HI]  =  4 mol/dm³.  Calculate K_c and also the initial concentration of H₂ and I₂. (K.B. 2015)

Solution

Q10. In a reaction;  H₂  +  I₂ ⇌ 2HI, when equilibrium was attained, the concentrations of [H₂] = [I₂] = 4 mol/dm³ and [HI] = 6 mol/dm³. Calculate K_c and also the initial concentration of H₂ and I₂

Solution

Q11.  In a reaction;  A  +  B  ⇌ 2C, when equilibrium is attained, the concentrations of B  =  4 mol/dm³ and C = 6 mol/dm³. Calculate the initial and equilibrium concentrations of A.  (K_c  =  2.25).

Solution

Q12. When the equilibrium was attained for the reaction;  A  +  B ⇌ 2C, the concentrations of [A] = [B] was 4 mol/dm³ and that of [C] was 6 mol/dm³. Calculate K_c and the initial concentrations of A and B. (K.B. 2009)

Solution          

Q13. In a reaction; A + B ⇌ 2C, 7 mol/dm³ of A and 7 mol/dm³ of B were mixed and allowed to attain equilibrium. If K_c = 2.25, find out the concentration of A, B and C at equilibrium state. (K.B. 2008)

Solution

1.  At equilibrium, a 12 litre flask contains 0.21 mole of PCl₅, 0.32 mole of PCl₃ and 0.32 mole of Cl₂ at 250°C. Find the K_c of reaction.     PCl₅       ⇌  PCl₃  +  Cl₂                                                              

        (Ans: K_c = 0.041)          

2. The equilibrium of N₂ + 3H₂ ⇌ 2NH₃ at 300°C in a 5 litre container has 1.0 mole of NH₃, 0.1 mole of N₂ and 3.0 moles of H₂.  Find K_c.

        (Ans: K_c = 9.259)                       

3. For the reaction  x + 3y ⇌ 2z, the equilibrium concentration are, x = 0.3 M, y = 0.2 M, z = 0.04 M.  Calculate K_c.

        (Ans: K_c = 0.666)

4. Calculate K_c for the reaction 2A + B ⇌ C + D. If equilibrium concentration are A = 0.4 formula weight/litre, B = 0.5 formula weight/litre, C = 0.3 formula weight/litre, D = 0.8 formula weight/litre.   

        (Ans: K_c = 3)                   

5. 1 mole of CH₃COOH and 1 mole of C₂H₅OH in 1 dm³ of solution were reacted at a certain temperature.  At equilibrium, 0.667 mole of water was present; calculate K_c.                 

{Hint:  [Water] = 0.667, [Ester] = 0.667,  [Acid] = 1 – 0.667,  [Alcohol] = 1 – 0.667 }

    (Ans: K_c = 4)                       

6. When 60 g of acetic acid and 46 g of alcohol was heated to give an equilibrium mixture of 12 g water and 58.7 g ethyl acetate, find K_c.                                                               (Ans: K_c = 4)

7.     92 g of ethyl alcohol was mixed with 120 g of acetic acid to give an equilibrium mixture of 1.33 moles of each of ester and water.  Find K_c.                                                                                                 

        (Ans: K_c = 4)

8.  4 mole of HI is introduced into a vessel held at constant temperature. When equilibrium is reached, it is found that 0.4 mole of I₂ have been formed.  Calculate K_c.                                             (Ans: K_c = 0.0156)

               2HI        ⇌  H₂  +  I₂                                                                        

9. In a reversible reaction PCl₅ ⇌ PCl₃ + Cl₂, 2 moles of PCl₅ are heated in a two dm³ flask. When the equilibrium is obtained, 40% PCl₅ is dissociated into PCl₃ and Cl₂.Calculate K_c.

                                                                  (Ans: K_c = 0.266)

Q1. Calculate K_c and K_p for the given reaction N₂O_4(g) ⇌ 2NO_2(g) at 295K, if the equilibrium concentrations are 

[N₂O₄] = 0.75M and [NO₂] = 0.062M      

Solution

Calculating K_c

[N₂O₄] = 0.75M 

[NO₂]    = 0.062M

N₂O_4(g) ⇌ 2NO_2(g)

Calculating K_p

T = 295 K

R = 0.0821 atm.dm³.K⁻¹.mol⁻¹

K_c = 0.00512

Moles of Reactants = n_R = 1

Moles of Products   = n_p = 2

∆n = n_p - n_R = 2 – 1 = 1

K_p = K_c(RT)^Δn      

⇒ K_p = 0.00512 (0.0821 x 295)¹                  

⇒ K_p = 0.00512 (24.23)¹ 

⇒ K_p = 0.00512 (6740.41)
K_p = 0.124

Q2. For the Haber process, K_c = 9.60 at 300°C. Calculate K_p for this reaction at this temperature.

                N_2(g)  +  3H_2(g) ⇌ 2NH_3(g)

Solution

Calculating K_p

T = 300°C  ⇒ 300 +273 = 573 K

R = 0.0821 atm.dm³.K⁻¹.mol⁻¹

K_c = 9.60

Moles of Reactants = n_R = 1 + 3 = 4

Moles of Products   = n_p = 2

∆n = n_p − n_R = 2 – 4 = –2

K_p = K_c(RT)^Δn      

⇒ K_p = 9.60 (0.0821 x 573)⁻²     

⇒ K_p = 9.60 (12.03)⁻² 

⇒ K_p = 9.60 (144.72)⁻ 

⇒ K_p = 9.60/144.72
K_p = 0.0663

Q3. For the reaction: N₂(g)+3H₂ (g)⇌2NH₃(g)

If K_c = 0.04 at 500 K, find K_p.

Solution

Calculating K_p

T = 500 K

R = 0.0821 atm.dm³.K⁻¹.mol⁻¹

K_c = 0.04

Moles of Reactants = n_R = 1 + 3 = 4

Moles of Products   = n_p = 2

∆n = n_p − n_R = 2 – 4 = –2

K_p = K_c(RT)^Δn      

⇒ K_p = 0.04 (0.0821×500)^⁻²  

⇒ K_p = 0.04 (41.05)^⁻²   

⇒ K_p = 0.04/(1685.1)²  

⇒ K_p = 2.37×10^⁻⁵

Q4. For the reaction:2SO₂ (g)+O₂ (g)⇌2SO₃ (g)

       If K_p = 2.5 × 10⁴ at 700 K, find K_c.

Solution

Calculating K_p

T = 700 K

R = 0.0821 atm.dm³.K⁻¹.mol⁻¹

K_p = 2.5 × 10⁴

Moles of Reactants = n_R = 2 + 1 = 3

Moles of Products   = n_p = 2

∆n = n_p − n_R = 2 – 3 = –1

K_p = K_c(RT)^Δn      

⇒ K_c = K_p/(RT)^⁻¹  

⇒ K_c = K_p(RT) 

⇒ K_c = 2.5 × 10⁴(0.0821 × 700) 

K_c = 2.5 × 10⁴(0.0821 × 57.47) 

K_c =1.44 × 10⁶ mol⁻¹·L

💡 Tips & Tricks

Always find Δn carefully. Count only gaseous moles.

Convert temperature to Kelvin before using the formula.

Choose correct R value (0.0821 if P in atm, 8.314 if in J).

For Δn = 0 → K_p = K_c (a common trick question).

Memorize the pattern:

Δn positive → K_p > K_c

Δn negative → K_p < K_c

Q1. Suppose a gaseous mixture consists of 0.02 M CO, 0.02 M H₂, 0.001 M CH₄ and 0.001 M H₂O. If the mixtureis passed over a catalyst at 1200 K, the following reaction occurs

    CO_(g) + 3H_2(g)  ⇌ CH_4(g) + H₂O_(g); K_c = 3.0 × 10^-2

predict the direction in which the reaction moves so as to achieve the equilibrium.       

(Reaction Quotient, Q_c = 6.25 > K_c = 3.0 × 10^-2, hence the reaction will proceed in the reverse direction).

Q2. 0.035 mole of SO₂, 0.500 mole of SO₂Cl₂, and 0.080 mole of Cl₂ are combined in an evacuated 5.00 L flask and heated to 100^oC.  What is Q_c before the reaction begins?  Which direction will the reaction proceed in order      to establish equilibrium?

  SO₂Cl_2(g) ⇌ SO_2(g) + Cl_2(g       K_c = 0.078 at 100^oC

(Reaction Quotient, Q_c = 0.011 < K_c = 0.078, hence the reaction will proceed in the forward direction).

Q3. The reaction between ammonia and oxygen in the gas phase, produces nitric oxide and water. Using K_c = 0.6 and the           concentrations of reactants and products listed below, which direction does the reaction favour?

      4NH_3(g) + 5O2_(g) ⇌ 2NO_(g) + 6H₂O_(g)

 [NH₃] = 0.1 M, [O₂] = 0.2 M, [NO] = 0.2 M, [H₂O] = 0.1M

(Reaction Quotient, Q_c = 0.05 < K_c = 0.6, hence the reaction will proceed in the forward direction).

Q4. Consider the reaction and its equilibrium constant:

 N₂O_4(g) ⇌ 2NO_2(g)            K_c = 5.85 x 10^-3 (at some temperature)

A reaction mixture contains 0.0255 M NO₂ and 0.0331 M N₂O₄. Calculate Q_c and determine the direction in which the  reaction will proceed.

(Reaction Quotient, Q_c = 0.0196 > K_c = 5.85 x 10^-3, hence the reaction will proceed in the reverse direction). 

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