🌟 Equilibrium Constant & Reaction Quotient Numericals: Predictions for Direction of Reactions (Class 10, 11, MDCAT/ECAT) 📚🔬

Welcome to Inamjazbi Learn Chemistry! 🌟 Whether you're in Class 10 or 11, preparing for MDCAT or ECAT, or just passionate about chemistry, you’ve come to the right place. This post will guide you through Equilibrium Constant (Kc) and Reaction Quotient (Qc), with numericals and clear solutions to help you predict the direction of reactions. Get ready to unlock the secrets of Chemical Equilibrium and boost your exam scores! 🔥

🌟 Equilibrium Constant & Reaction Quotient Numericals: Predictions for Direction of Reactions (Class 10, 11, MDCAT/ECAT) 📚🔬

🔥🌿Formulas Used in Problems of Chemical Equilibrium

🔥🌿Writing Equilibrium Constant Expression for Reaction along with Rate of forward and reverse Reactions

Q1. Write an equilibrium equation of monoatomic carbon and a molecule of oxygen as reactant and carbon monoxide as product.

Answer

2C(s) + O₂(g) ⇌ 2CO(g)

The rate of forward reaction = Rf = Kf [O₂]

The rate of reverse reaction = Rr = Kr [CO]²

Q2. Write down the Equilibrium Constant expression for the reversible reaction of sulphur dioxide with oxygen to form sulphur trioxide.

Answer

For the reversible reaction of sulphur dioxide with oxygen to form sulphur trioxide, the Equilibrium Constant expression is derived as follows:

2SO₂ (g) + O₂ (g) ⇌ 2SO₃(g)

The rate of forward reaction = Rf = Kf [SO₂]²[O₂]

The rate of reverse reaction = Rr = Kr [SO₃]²

Q3. Write down the Equilibrium Constant expression for reversible reaction of nitrogen with oxygen to form nitrogen monoxide.

Answer

For the reversible reaction of nitrogen with oxygen to form nitrogen monoxide, the Equilibrium Constant expression is derived as follows:

N₂(g) + O₂(g) ⇌ 2NO(g)

The rate of forward reaction = Rf = Kf [N₂][O₂]

The rate of reverse reaction = Rr = Kr [NO]²

Q4. Write down the Equilibrium Constant expression for the reversible reaction of nitrogen with hydrogen to form ammonia,

Answer

For the reversible reaction of nitrogen with hydrogen to form ammonia, the Equilibrium Constant expression is derived as follows:

N₂(g) +3H₂(g) ⇌ 2NH₃(g)

The rate of forward reaction = Rf = Kf [N₂][H₂]³

The rate of reverse reaction = Rr = Kr [NH₃]²

Q5. Write down the Equilibrium Constant expression for the reversible reaction of combination of nitrogen dioxide into its dimer dinitrogen tetraoxide.

Answer

2NO₂(g) ⇌ N₂O₄(g)

The rate of forward reaction = Rf = Kf [NO₂]²

The rate of reverse reaction = Rr = Kr [N₂O₄]

🔥🌿Writing Equilibrium Constant Expression for Reaction

Q. Write down the equilibrium constant expression or Kc equation for given reactions

(i) H₂(g) + I₂(g) ⇌ 2HI(g)

(ii) S(s) + O₂(g) ⇌ SO₂(g)

(iii) 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

(iv) PCl₃ (g) + Cl₂(g) ⇌ PCl₅(g)

(v) N₂ + 2O₂(g) ⇌ 2NO₂(g)

(vi) N₂ + 3H₂(g) ⇌ 2NH₃ (g)

(vii) H₂ + Br₂(g) ⇌ 2HBr(g)

(viii) SO₂(g) + NO₂(g) ⇌ NO(g) + SO₃ (g)

(ix) CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)

(x) NH₄Cl(s) ⇌ NH₃ (g) + HCl(l)

Answer

🔥🌿Finding Out Coefficients for given Hypothetical and Real Reactions

Q. Figure out coefficients of each in the following hypothetical reactions

(i) 9X(g) + Y₃(g) ⇌ 3X₃Y(g)

(ii) 4X(g) ⇌ 2Y(g) + 3Z(g)

(iii) 2A(g) + 3B(g) ⇌ 4C(g) + 2D(g)

(iv) 2M(g) + 4N(g) ⇌ 5O(g)

General Consideration

In Chemistry, the coefficient is the number in front of the formulae or symbols to balance the chemical equations. The coefficient tells us how many molecules or atoms of a given formula are present. coefficient is a number used in chemical equation, just as a prefix of chemical formula or symbol to define the number of molecules or atoms reacting and producing in a reaction.

They multiply all the atoms in a formula.

e.g.

(i) Thus the symbol “2 NaHCO₃” indicates two units of sodium bicarbonate which contain 2 Na atoms 2 H atoms 2 C atoms and 6 O atoms (2 x 3 = 6 the coefficient times the subscript for O).

(ii) 2H₂O means we have 2 molecules of water

(iii)In the balanced chemical equation for Ammonia

N₂ + 3H₂ → 2NH₃

We know we have coefficients of

1 for N₂

3 for H₂

and

2 for NH₃

It takes 1 molecule of N₂ with 3 molecules of H₂ to produce 2 molecules of NH₃

Solution

(i) 9X(g) + Y₃(g) ⇌ 3X₃Y(g)

Coefficient of X = 9, Coefficient of Y₃ = 1, Coefficient of X₃Y = 3

(ii) 4X(g) ⇌ 2Y(g) + 3Z(g)

Coefficient of X = 4, Coefficient of Y = 2, Coefficient of Z = 3

(iii) 2A(g) + 3B(g) ⇌ 4C(g) + 2D(g)

Coefficient of A = 2, Coefficient of B = 3, Coefficient of C = 4, Coefficient of D = 2

(iv) 2M(g) + 4N(g) ⇌ 5O(g)

Coefficient of M = 2, Coefficient of N = 4, Coefficient of O = 5

🔥🌿Writing Forward and Reverse Reactions for reversible Reactions

Solution

🔥🌿Calculating Equilibrium Constant

Q1. Equilibrium occurs when nitrogen monoxide gas reacts with oxygen gas to form nitrogen dioxide gas

2NO(g) + O₂ (g) ⇌ 2NO₂(g)

At equilibrium at 230°C, the concentrations are measured to be

[NO] = 0.0542 mol dm⁻³, [O₂] = 0.127 mol dm⁻³ and [NO₂] = 15.5 mol dm⁻³

Calculate the equilibrium constant at this temperature. (Book problem 1; page 8)

Solution

Given

Given equilibrium concentrations are;

[NO] = 0.0542 mol dm⁻³

[O₂] = 0.127 mol dm⁻³

[NO₂] = 15.5 mol dm⁻³

Required

Equilibrium constant (Kc) = ?

Equilibrium constant (Kc) Expression

Calculation

Q2.A reaction takes place between iron and chloride ion as

Fe³⁺ + 4Cl⁻ ⇌ FeCl₄⁻

At equilibrium the concentrations are measured to be

Fe³⁺ = 0.2 mol dm⁻³, Cl⁻ = 0.28 mol dm⁻³ and FeCl₄⁻ = 0.95 x 10⁻⁴ mol dm⁻³

Calculate equilibrium constant Kc for given reaction. (Book problem 1; page 9)

Solution

Given

Given equilibrium concentrations are;

[Fe³⁺] = 0.2 mol dm⁻³

[Cl⁻] = 0.28 mol dm⁻³

[FeCl₄⁻] = 0.95 x 10⁻⁴ mol dm⁻³

Required

Equilibrium constant (Kc) = ?

Equilibrium constant (Kc) Expression

Q3. When hydrogen reacts with iodine at 25°C to form hydrogen iodide by a reversible reaction as follows:

H₂(g) + I₂ (g) ⇌ 2HI(g)

The equilibrium concentrations are:

[H₂] = 0.05 mol dm⁻³; [I₂] = 0.06 mol dm⁻³; and [HI] = 0.49 mol dm⁻³.

Calculate the equilibrium constant for this reaction.

Solution

Given

Given equilibrium concentrations are;

[H₂] = 0.05 mol dm⁻³

[I₂] = 0.06 mol dm⁻³

[HI] = 0.49 mol dm⁻³

Required

Equilibrium constant (Kc) = ?

Equilibrium constant (Kc) Expression

Q4.For the formation of ammonia by Haber’s process, hydrogen and nitrogen react reversibly at 500°C as follows

N₂(g) + 3H₂(g) ⇌2NH₃(g) , Kc = 3.0 x 10⁻⁹

The equilibrium concentrations of these gases are: nitrogen 0.602 mol dm⁻³; hydrogen 0.420 mol dm⁻³ and ammonia 0.113 mol dm⁻³. What is value of Kc.

Solution

Given

Given equilibrium concentrations are;

[NH₃] = 0.113 mol dm⁻³

[N₂] = 0.602 mol dm⁻³

[H₂] = 0.420 mol dm⁻³

Required

Equilibrium constant (Kc) = ?

Equilibrium constant (Kc) Expression

Q5.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 Kc at 250°C.

Solution

Given

[PCl₅] = 3.84 g

[PCl₃] = 9.14 g

[Cl₂] = 2.84 g

Required

Equilibrium constant (Kc) = ?

Calculation of Molar Equilibrium Concentration

Q6. In a reaction; H₂ + I₂ ⇌ 2HI, when equilibrium was attained, the concentrations of [H₂] = [I₂] = 4 mol/dm³ and [HI] = 6 mol/dm³. Calculate Kc.

Solution

Given

[H₂] = 4 mol dm⁻³

[I₂] = 4 mol dm⁻³

[HI] = 6 mol dm⁻³

Required

Equilibrium constant (Kc) = ?

Equilibrium constant (Kc) Expression

Q7. In a reaction; H₂ + I₂ ⇌ 2HI, when equilibrium was attained, the concentrations were [H₂] = [I₂] = [HI] = 4 mol/dm³. Calculate Kc.

Solution

Given

Given equilibrium concentrations are;

[H₂] = 4 mol dm⁻³

[I₂] = 4 mol dm⁻³

[HI] = 4 mol dm⁻³

Required

Equilibrium constant (Kc) = ?

Equilibrium constant (Kc) Expression

Q8. For the reaction:

N₂(g)+3H₂(g) ⇌ 2NH₃(g)

At equilibrium, the concentrations are:

[N₂]=0.2 mol/L

[H₂]=0.6 mol/L

[NH₃]=0.4 mol/L

Calculate the equilibrium constant, Kc

Solution:

The expression for Kc is

Kc=[NH₃]² /[N₂][H₂]³

Substitute the given concentrations:

Kc=(0.4)² /(0.2)(0.6)³ = 3.70

Q9. For the reaction:

2SO₂ (g)+O₂ (g)⇌2SO₃ (g)

At equilibrium, the concentrations are:

[SO₂]=0.4 mol/L

[O₂]=0.2 mol/L

[SO₃]=0.6 mol/L

Calculate the equilibrium constant, Kc

Solution:

The expression for Kc

Kc = [SO₃]²/[SO₂]² [O₂]

Substitute the concentrations:

Kc = (0.6)² /(0.4)2(0.2) =11.25

Q10. For the reaction:

A(g) + B(g) ⇌ C(g)

At equilibrium, the concentrations are:

[A] = 0.3 mol/L

[B] = 0.5 mol/L

[C] = 0.2 mol/L

Calculate the equilibrium constant, Kc

Solution:

The expression for Kc is:

Kc = [C]/[A][B]

Kc = 0.2/(0.3)(0.5) = 1.33

Q11. For the reaction:

PCl₅ (g)⇌PCl₃(g)+Cl₂ (g)

At equilibrium, the concentrations are:

[PCl₅] = 0.1 mol/L[PCl_5] = 0.1

[PCl₃] =0.2 mol/L

[Cl₂] = 0.2 mol/L

Calculate the equilibrium constant, Kc

Solution:

The expression for Kc

Kc = [PCl₃][Cl₂]/[PCl₅]

Kc = (0.2)(0.2)/0.1 = 0.4

Q12. For the reaction:

2NO(g) + O₂(g)⇌2NO₂(g)

At equilibrium, the concentrations of the gases are:

[NO] = 0.3 mol/L

[O₂] = 0.1 mol/L

[NO₂] = 0.4 mol/L

Calculate the equilibrium constant, Kc

Solution:

The expression for Kc

Kc=[NO₂]²/[NO]² [O₂]

Substitute the given concentrations:

Kc = (0.4)²/(0.3)² (0.1) = 17.78

🔥🌿Assignment Numericals on Kc

1. For the decomposition of dinitrogen oxide (N₂O) into nitrogen and oxygen, following reversible reaction takes place

2N₂O(g) ⇌ 2N₂(g) + O₂(g)

The concentration of N₂O, N₂ and O₂ are 1.1 mol dm⁻³, 3.90 mol dm⁻³ and 1.95 mol dm⁻³ respectively at equilibrium. Find out Kc for this reaction. (Ans: Kc = 20.25 mol dm⁻³)

2. Hydrogen iodide decomposes to form hydrogen and iodine. If the equilibrium concentration of HI is 0.078 mol dm⁻³, H₂ and I₂ is same 0.011 mol dm⁻³. Calculate the equilibrium constant value for this reversible reaction:

H₂(g) + I₂(g) ⇌ 2HI(g) (Ans: Kc = 50.28)

3. 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 Kc of reaction. PCl₅ ⇌ PCl₃ + Cl₂ (Ans: Kc = 0.041 mol dm⁻³)

4.PCl₅, PCl₃ and Cl₂ are at equilibrium at 500K in a closed container and their concentrations are 0.8 x 10⁻³ mol dm⁻³, 1.2 x 10⁻³ mol dm⁻³ and 1.2 x 10⁻³ mol dm⁻³ respectively. Calculate the value of Kc for the reaction along with unit. (Ans: Kc = 1.8 x 10⁻³ mol dm⁻³)

PCl₅(g) ⇌ PCl₃ (g) + Cl₂ (g)

5. 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 Kc. (Ans: Kc = 9.259)

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

(Ans: Kc = 0.666)

7. Calculate Kc 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: Kc = 3)

8. Dinitrogen tetraoxide (N₂O₄) decomposed into nitrogen dioxide (NO₂) in a reversible reaction. Derive equilibrium expression for the reaction of decomposition. Also interpret unit of Kc for balanced chemical reversible reaction.