Chapter No.1: Chemical Equilibrium -Chemistry For Class X (2022 and Onward) - Numericals (Section "D")

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Chemical Equilibrium
Numericals

SECTION- D: Numericals

Numerical No. 1:
Dinitrogen tetra oxide N₂O₄, decomposed into nitrogen dioxide NO₂, in a reversible reaction. Derive equilibrium constant expression for the reaction of decomposition. Also interpret unit of K_c for balanced chemical reversible reaction.
Solution:
Equation:

N₂O₄ ⇌ 2NO₂

Derivation Of Equation For Equilibrium Constant:
Let us apply law of mass action on above reversible reaction.
Forward Reaction:
The rate of forward reaction according to law of mass action is:

R_f ∝ [N₂O₄]

R_f = k_f [N₂O₄] ...(i)

Where k_f is the rate constant for forward reaction.

Reverse reaction:
The rate of reverse reaction according to law of mass action is:

R_r ∝ [NO₂]²

R_r = k_r [NO₂]² ... (ii)

Where k_r is the rate constant for reverse reaction.

At Equilibrium:
We know that, at equilibrium rate of forward and reverse reaction becomes equal. So,

R_f = R_r ... (iii)

Putting the values of R_f and R_r from eq. (i) & (ii) in eq. (iii), we have:

k_f [N₂O₄] = k_r [NO₂]²

By taking constants on L.H.S and variables on R.H.S, we have

Where K_c is called equilibrium constant.

Unit Of K_c for given reaction:

Ans: Hence Unit of K_c for the given reaction is [mol.dm^-3].

Numerical No. 2:
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, 12 x 10^-3 mol.dm^-3 and 1.2 x 10^-3 mol.dm^-3 respectively. Calculate the value of K_c for the reaction along with unit.
Solution:
Given:

• [PCl₅] = 0.8 x 10^-3 mol.dm^-3
• [PCl₃] = 12 x 10^-3 mol.dm^-3
• [Cl₂] = 1.2 x 10^-3 mol.dm^-3
• K_c at 500K = ?

Equation:

PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)

Therefore

Ans: The value of K_c at 500K is 1.8 x 10^-3 mol.dm-3

Numerical No.3:
The value of K_c for the reaction is 1 x 10^-4
2HI_(g) ⇌ H_2(g) + I_2(g)
At a given temperature, the molar concentration of reaction mixture is HI = 2 X 10^-5 mol.dm^-3, H₂ = 1 X 10^-5 mol.dm^-3 and I₂ = 1 x 10^-5 mol.dm^-3. Predict the direction of the reaction.
Solution:
Given:

• ^{K_c = 1 x 10-4}
• ^{[HI] = 2 X 10-5 mol.dm-3}
• ^{[H₂] = 1 X 10-5 mol.dm-3}
• ^{[I₂] = 1 x 10-5 mol.dm-3}
• ^{Direction of reaction = Q_c = ?}

Working:
At a given temperature the reaction quotient Q_c for the reaction will be given by the expression: As the value of the reaction quotient is greater than the value of K_c i.e 0.25 > 1 x 10^-4

Ans: Hence Q_c > K_c, the reaction will be proceed in reverse direction.

Test Yourself

Numerical No. 4:
The value of K_c for the following reaction at 717K is 48.

H_2(g) + I_2(g) ⇌ 2HI_(g)

At a particular instant, the concentration of H₂, I₂ and HI are found to be 0.2 molL^-1, 0.2 molL^-1, and 0.6 molL^-1 respectively. Calculate reaction quotient for given reaction. Also predict direction of reaction.

Solution:
Given:

• K_c = 48 at 717K
• [H₂] = 0.2 molL^-1
• [I₂] = 0.2 molL^-10.2 molL^-1
  
• [HI] = 0.6 molL^-1

Working Formula:
At a given time the reaction quotient Q_c for given reaction will be given by the expression:

Therefore Q_c < K_c
Ans: Q_c < K_c, so reaction will proceed in forward direction.

Numericals - Examples From Text Book

Numerical 01:
Equilibrium occurs when nitrogen monoxide gas reacts with oxygen gas to form nitrogen dioxide gas.

2NO_(g) + O_2(g) ⇄ 2NO_2(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.
Solution:
Given equilibrium concentrations of reactants and product are:

• [NO] = 0.0542 mol.dm^-3
• [O₂] = 0.127 mol.dm^-3
• [NO₂] = 15.5 mol.dm^-3

Write equilibrium expression as:

Now put equilibrium concentration values in equilibrium constant expression

Ans: K_c = 6.44 X 10⁵ mol^-1.dm³

Numerical 02:
A reaction takes place between iron ion and chloride ion as:

Fe⁺³_(aq) + 4Cl^-_(aq) ⇌ FeCl₄^-_(aq)

At equilibrium, the concentrations are measured to be (Fe⁺³) is 0.2 mol.dm^-3, Cl^- is 0.28 mol.dm^-3 and FeCl₄^- is 0.95x10^-4 mol.dm^-3. Calculate equilibium constant K_c for given reaction.
Solution:
Given equilibrium concentrations of reactants and product are:

• [Fe⁺³] = 0.2 mol.dm^-3
• [C1^- = 0.28 mol.dm^-3
• [FeCl₄^-] = 0.95x 10^-4 mol.dm^-3

Write equilibrium expression as:

Now put equilibrium concentration values in equilibrium constant expression:

Ans: K_c = 7.72 x 10^-2 mol^-4.dm¹²

Numerical 03:
Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperature. At 2000 °C, the value of the equilibrium constant for the given reaction is 4.1 x 10^-4

N_2(g) + O_2(g) ⇌ 2NO_(g)

Find the concentration of NO in an equilibrium mixture at 1 atm pressure at 2000°C. In air, [N₂ = 0.036 mol/L and [O₂] = 0.0089 mol/L
Solution:
Given: equilibrium concentrations of reactants and product are:

• [N₂] = 0.036 mol/L
• [O₂] = 0.0089 mol/L
• [NO] = ?

We are given all of the equilibrium concentrations except that of NO. Thus, we can solve for the missing

Taking square root on both the sides, we have

√[NO₂] = √(4.1 x 10^-4 mol/L)(0.036 mol/L) (0.0089 mol/L)

Ans: [NO] = 3.6x10^-1 mol/L