Equilibrium

NCERT Solution

Part 3

Question 31: Dihydrogen gas used in Haber's process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction

CO (g) + H₂O ⇌ CO₂ (g) + H₂ (g)

If a reaction vessel at 400° C is charged with an equimolar mixture of CO and steam such that p_CO = p_H2O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium? K_p = 10.1 at 400°C

`K_p=(pC\O_2xx\pH_2)/(pC\O\xx\H_2O)`

Or, `10.1=(p^2)/(4-p)^2`

Or, ` p/(4-p)= sqrt(10.1)=3.178`

Or, `p=3.178(4-p)`

Or,`p=12.702-3.178p`

Or, `4.178p=12.702`

Or, `p=(12.702)/(4.178)=3.04`

Question 32: Predict which of the following reactions will have appreciable concentration of reactants and products:

1. Cl₂ (g) ⇌ 2Cl (g) K_c = 5 × 10^-39
2. Cl₂ (g) + 2NO (g) ⇌ 2NOCl (g) K_c 3.7 × 10⁸
3. Cl₂ (g) + 2NO₂ (g) ⇌ 2NO₂Cl (g) K_c = 1.8

Answer: Reaction (a) has the least concentration of products because K_c very small, (b) has the highest concentration of products because K_c is very high, and (c) has both products and reactants in appreciable concentration because K_c is neither too large nor too small.

Question 33: The value of K_c for the reaction 3O₂ (g) ⇌ 2O₃ (g) is 2.0 × 10^-50 at 25°C. If the equilibrium concentration of O₂ in air at 25°C is 1.6 × 10^-2, what is the concentration of O₃?

Answer: `K_c=([O_3]^2)/([O_2]^3)`

Or, `2xx10^(-50)=([O_3]^2)/((2xx10^(-2))^3)`

Or, `[O_3]^2=2xx2^3xx10^(-56)`

Or, `[O_3]=4xx10^(-28)`

Question 34: The reaction, CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g) is at equilibrium at 1300 K in a 1 L flask. It also contains 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant K_c

Answer: `K_c=([CH_4][H_2O])/([CO][H_2]^3)`

Or, `3.9=([CH_4]xx0.02)/(0.3xx0.1^3)`

Or, `[CH_4]=(3.9xx0.3xx0.001)/(0.02)=0.0585` M L^-1

Question 35: What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:

HNO₂, CN^-, HClO₄, F^-, OH^-, CO₃^2- and S^2-

Answer: Acid-base pair which differs by 1 proton only is called conjugate acid base pair.

HA → H⁺ + A^-

Question 36: Which of the following are Lewis acids? H₂O, BF₃, H⁺ and NH₄⁺

Answer: Acid which can accept a pair of electrons is called Lewis acid. H₂ is not a Lewis acid, but all others are Lewis acid.

Question 37: What will be the conjugate bases for the Bronsted acids: HF, H₂SO₄ and HCO₃^-?

Question 38: Write the conjugate acids for the following Bronsted bases: NH₂^-, NH₃ and HCOO^-.

Answer:

Question 39: The species: H₂O, HCO₃^-, HSO₄^- and NH₃ can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.

Answer:

Question 40: Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH^- (b) F^- (c) H⁺ (d) BCl₃

Answer: (a) and (b) are Lewis base because they can donate a pair of electrons. (c) and (d) are Lewis acids because they can accept a pair of electrons.

Question 41: The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^-3 M. What is its pH?

Answer: pH = -log [H⁺]

= - log 3.8 + 3 = 3 – 0.5798 = 2.42

Question 42: The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen iron in it.

Answer: pH = - log [H⁺]

Or, log [H⁺] = - pH

Or, [H⁺] = antilog (- pH)

= antilog (- 3.76) = 1.74 × 10^-4

Question 43: The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 × 10^-4, 1.8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base.

Answer: In case of HF, for F^-:

`K_b=(K_w)/(K_a)`

`=(10^(-14))/(6.8xx10^(-4))=1.5xx10^(-11)`

For HCOO^-:

`K_b=(10^(-14))/(1.8xx10^(-4)=5.6xx10^(-11)`

For CN^-:

`K_b=(10^(-14))/(4.8xx10^(-9))=2.08xx10^(-6)`

Question 44: The ionization constant of phenol is 1.0 × 10^-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?

Answer:

So,`K_b=(x^2)/(0.05-x)=10xx10^(-10)`

Or, `x^2=5xx10^(-12)`

Or, `x=2.2xx10^(-6)` M

Let us assume that in presence of 0.01 C₆H₅ONa, amount of dissociated phenol is y

Then , `K_a=(0.01y)/(0.05)=1.0xx10^(-10)`

Or, `y=5xx10^(-10)`

Or, `α=(5xx10^(-10))/(5xx10^(-2))=10^(-8)`

Question 45: The first ionization constant of H₂S is 9.1 × 10^-8. Calculate the concentration of HS^- ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H₂ is 1.2 × 10^-13, calculate the concentration of S^2- under both conditions.

Answer: Calculation of [HS^-]

Let us consider 0.1 – x ∼ 0.1

`K_s=(x^2)/(0.1)=9.1xx10^(-8)`

Or, `x^2=9.1xx10^(-9)`

Or, `x=9.54xx10^(-5)`

Let us now assume that in presence of 0.1 M HCl, dissociated H₂S is y

`K_a=(0.1xx\y)/(0.10)=9.1xx10^(-8)`(given)

Or, `y=9.1xx10^(-8)`

Calculation of [S^2-]

H₂S ⇌ H⁺ + HS^-

HS^- ⇌ H⁺ + S^2-

Overall Reaction: H₂S ⇌ 2H⁺ + S^2-

If K_a1 is the first ionization constant and K_a2 is the second ionization constant

`K_a=K_(a1)xx\K_(a2)`

= 9.2 × 10^-8 × 1.2 × 10^-13

= 1.092 × 10^-20

We know: `K_a=([H^+][S^(2-)])/([H_2S])`

In the absence of 0.1 M HCL:

[H⁺] = 2[S^2-]

So, if [S^2-] = x then [H⁺] = 2x

So, `((2x^2)x)/(0.1)=1.092xx10^(-20)`

Or, `4x^3=1.092xx10^(-21)=273xx10^(-24)`

Or, 3 log x = log 273 – 24 = 2.4362 – 24

Or, log x = 0.8127 – 8 = 8.8121

Or, x = Antilog 8.8127 = 273 × 10^-24 = 6.5 × 10^-8 M

In presence of 0.1 M HCl

Suppose [S^2-] = y

Then [H₂S] = 0.1 – y ∼ 0.1 M

[H⁺] = 0.1 + y ∼ 0.1 M

`K_a=(0.1^2xx\y)/(0.1)=1.09xx10^(-20)`

Or, `y=1.09xx10^(-19)` M

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