Equilibrium Important Questions

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Very Short Answer Questions [1 mark questions]

Ques. Define Dynamic Equilibrium. 

Ans. Dynamic equilibrium is defined as the state when reactants present in a closed atmosphere react with each other to give rise to products, and end up entering a stage wherein the concentration of reactants start decreasing and products keep increasing, and no change is observed after some time. 

Ques. The boiling point of liquid depends on which factor? 

Ans. The boiling point of a liquid depends on the atmospheric pressure. It also depends upon the temperature and the vapour pressure of the liquid. Boiling starts when atmospheric pressure is equal to the vapor pressure of the liquid.

Ques. When does a reversible reaction become an irreversible reaction? 

Ans. A reversible reaction becomes an irreversible reaction when gas escapes through an open system.The process continues if we keep removing the products from the reversible reactions.

Ques. When will the reaction go farthest to completion, if K=10^-10, if K=10¹⁰ or if K=1? 

Ans. As the ratio of products/ reactants is greater when K=10¹⁰, the reaction will go farthest to completion in this condition.

Ques. The equilibrium constant for the reaction SO₃ (g) ⇌ SO₂ (g) + 1/2 O₂ (g) is 0.18 at 900 K. What will be the equilibrium constant for the reaction SO₂ (g) + 1/2 O₂ (g) ⇌ SO₃ (g)?

Ans. The equilibrium constant for the reaction SO₃ (g) ⇌ SO₂ (g) + 1/2 O₂ (g) is 0.18 at 900 K is 5.55. 

Ques. When will the ice water system be in equilibrium? 

Ans. When the temperature is at 273 K. In this case, the rates of transfer of molecules from ice into water and vice versa are equal at a temperature of 273 K.

Ques. Sequence the following in increasing order of pH. KNO₃ (aq), C₆H₅COONH₄ (aq), NH₄Cl (aq), CH₃COONa (aq)

Ans. NH₄Cl < C₆H₅COONH₄ < KNO₃ < CH₃COONa

Ques. What is the reaction quotient? 

Ans. Reaction quotient or Q is defined as the measure of amounts of products and reactants at any given time. 

Ques. What is the relation between Kc and Kp for the reaction: PCl₅ (g) ⇌ PCl₃ (g) + Cl₂ (g)

Ans. Δn (g) = 2 – 1 = 1

Therefore, K_p = K_c × (RT)^Δn = K × RT

Ques. What happens to the equilibrium constant when a catalyst is added? 

Ans. When a catalyst is added to an equilibrium solution then equilibrium constant remain constant. It increases the rate of reaction without getting consumed in the process.

Ques. What is an Ionic equilibrium? 

Ans. Ionic equilibrium is a type of equilibrium where ionic substance dissociates into its ions. Equilibrium is formed between ions in a solution and the unionized molecules, and it is formed in weak electrolytes.

Ques. What is heterogeneous chemical equilibrium?

Ans. Heterogeneous chemical equilibrium is a type of equilibrium where reactants and the product taking part in the chemical reaction are in different phases.

CO₂ (g) + C (s) ⇌ 2CO (g) 

Ques. How do you prepare ammonia by haber’s process?

Ans. In the haber process, ammonia is formed by combining nitrogen with hydrogen, mainly from natural gas (methane), in the presence of iron as a catalyst.

Ques. What is an example of equilibrium in real life?

Ans. The most common example of equilibrium is a book kept on a table at rest. In this case, the force exerted by the table on the book's mass and weight cancels out each other, keeping the book in a state of static equilibrium.

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Short Answer Type Questions [2 marks questions]

Ques. Explain why the pH of 0.1 molar solution of acetic acid is higher than 0.1 molar solution of HCl.

Ans. Acetic acid is a weak electrolyte, and gives less H⁺ ion concentration as it is not completely ionized. HCl is a strong acid, and it is completely ionized giving more H⁺ ions concentration. As pH = – log [H⁺], the lesser the no. of H⁺ ions, the more the value of pH. Therefore pH of 0.1 M acetic acid is more than that of 0.1 M HCl.

Ques. What is the pH of 0.002 M HBr? 

Ans. We can get the pH by using the following equation:

Ques. What is meant by physical equilibrium? Explain with an example. 

Ans. Physical equilibrium is the state occurring between two physical substances, an example of which is P=nVRT=CRT.  Phase equilibrium, Solute-solid equilibrium and Gas-liquid equilibrium are three types of physical equilibrium.

Ques. BF₃ does not have protons but upon reaction with NH₃, it still acts as an acid. Why? What is the bond forming between the two?

Ans. BF₃ does not have any protons and therefore acts as Lewis acid.: NH₃ has one lone pair, which can be donated to BF₃ and hence it can form a coordinate bond with NH₃, i.e. (NH₃: → BF₃).

Ques. How are Kc and Kp related in this particular reaction: N₂ (g) + O₂ (g) ⇌ 2NO (g)?

Ans. The relationship between Kc​ and Kp​ is  given by the equation Kp ​= Kc (RT)^Δn. As per analysis, the given reaction is given by Δn = (1+1) − 2 =0. In this particular reaction, Kc=Kp. 

Ques. On what factor does equilibrium depend?

Ans.The factors on which equilibrium depends are as follows:

• Change in concentration level.
• Change in pressure level.
• Effect of catalyst 
• Effect of inert gas
• Change in temperature level

Ques. What is forward reaction?

Ans. A forward reaction is a type of reversible reaction in which products are produced from reactants. It goes from left to right. The name is derived from Le Châtelier's Principle. In these types of reactions, the concentration of reactant increases. 

N₂ + O₂ ←→ 2NO

Ques. What happens after equilibrium is reached? 

Ans. The process of chemical balance never stops. Even after the equilibrium state has been achieved, the forward and reverse reactions keep happening. Reactant and product ratios, however, remain unchanged in this situation, meaning that during an equilibrium reaction, the relative concentrations of the reactants and products remain unchanged.

Ques. The equilibrium constant for the reaction H₂O + CO -----> H₂ + CO₂. Is 2.55 at 1260 K. What will be the value of the equilibrium constant for the reaction: 2H₂ + 2CO ---→ 2CO at 1260 K?

Ans: The value of equilibrium get reversed and also becomes double which can be shown as:

K_c = ( ½.55)²

K_c = 0.153

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Long Answer Questions [3 marks questions]

Ques. What does equilibrium mixture Kc tell us about the mixture? 

Ans. The equilibrium mixture Kc tell us following things about the mixture

• When Kc>10³ then the reaction will be dominated towards reactants and if the value of Kc is quite large it means that the reaction is going to complete.
• When Kc<10³ then reaction will be dominated towards the product and if the value is very small then reaction rarely proceeds.
• And when Kc is in between the range of 10⁻³ to 10³ then the concentration of both reactants as well as products are appreciated.

Ques. On the basis of the equation pH = – log [H⁺], the pH of the 10^-8 mol dm^-3 solution of HCI is 8. However, it is observed to be less than 7.0. Discuss the situation with a brief explanation?

Ans. The pH of an acidic solution cannot be more than 7.0. ehrneas the expected ph is 8. To explain this it is suggested that H⁺ or H₃O⁺ ions are released by the self ionisation of water. 

Since water is neytral, [H₃O⁺] = 10^-7 M.

therefore Total [H₃O ⁺] in solution = 10 ^– 8 M + 10^-7M

= 10 ^– 8 [1 + 10] = 11 x 10 ^– 8M

ph = – log [H₃O⁺] = – [11 x 10 ^– 8] = [8 – log 11]

ph = [8 – 1.0411] = 6.9586 = 6096

pH will be less than 7.0.

Ques. Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively?

Ans. pH of solution A = 6,

Therefore, the concentration of [H+] ion in solution A=10⁻⁶ mol / L-1

The pH of solution B = 4

And, the concentration of [H+] ion in solution B=10⁻⁴mol / L.

On mixing one litre of each solution, total volume = 1L + 1L = 2L.

Amount of H+ ions in 1L of solution A = concentration × Volume (V) =

Amount of H+ ions in 1L of solution A = 10⁻⁶ X 1 = 10⁻⁶

Amount of H+ ions in 1L of solution B =10⁻⁴ X 1 = 10⁻⁴

∴ The total amount of H⁺ ions in the solution formed by mixing solutions A and B is (10⁻⁵ mol + 10⁻⁴ mol)

This amount is present in 2L solution,

∴ Total [H+] = 10⁻⁴ (1 + 1.01) / 2 mol / L - 1

Total [H+] = 1.01 X 10⁻⁴ / 2 mol / L -1

Total [H+] = 0.5 X 10⁻⁴ mol / L -1

Total [H+] = 5 X 10⁻⁵ mol / L -1

pH = – log [H+]

pH = – log [5 X 10⁻⁵]

pH = – [log 5 – 5 log 10]

pH = – log 5 + 5

pH = 5 – log 5

pH = 5 – 0.6990

pH = 4.3010

pH = 4.3

And therefore, the pH = 4.3.

Ques. Tell if the solutions are acidic, basic or neutral? 

NaCl, KBr, NaCN, NaOH, H₂SO₄, NaNO₂, NH₄NO₃, KF

Ans. The analysis are as follows:

• Acidic solutions are H2SO4, NH4NO3, and KF. 
• Basic solutions are NaCN, NaOH and NaNO₂
• Neutral solutions are NaCl, and KBr

Ques. Name the three groups in which chemical equilibrium can be classified? 

Ans. The three groups are: 

• The reaction which proceeds quite into completion and has only negligible concentrations of the reactants left.
• The reaction in which only small amounts of products are formed and most of the reactants remain unchanged at equilibrium stage.
• The reaction in which concentrations of reactants and products are comparable, and the system is in equilibrium.

Ques. The constant KP for the reaction N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g) is 2.5 × 10^-4 atm^-2 at 400°C. What's going to be the constant of the chemical reaction at 500°C if the warmth of the reaction at this temperature varies from -25.14 kcal?

Ans. Equilibrium constants at completely different temperatures and the warmth of the reaction are connected by the equation,

• log KP₂ = -25140/2.303 × a pair of [773 – 673 / 773 × 673] + log 2.5 × 10^-4
• log KP₂ = -1.503
• KP₂ = 3.14 × 10^-5 atm^-2

Ques. The value for the term Kc for the reaction 2HI (g) ⇋ H₂ (g) + I₂ (g) is 1 x 10^-4. The composition of reaction mixture is [HI] = 2 x 10^-5 mol, [I₂ ] = 4 x 10^-5 mol. Determine the direction of the reaction?

Ans. Q = [H₂ ][I₂ ] / [HI]

• Q = 4 X 10^-5 X 4 X 10^-5 / (2 X 10^-5 )²
• Q = 8

Ques. The concentration of H⁺ in a soft drink is 5.8 x 10⁻³M. What is its pH?

Ans. Concentration of [H⁺] = 5.8 x 10⁻³M.

• We know that pH = -log[H⁺], put the value of H⁺
• pH = – log[5.8 x 10⁻³]
• - log 5.8 – log10⁻³
• 3 – 0.7634
• 2.2366

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Very Long Answer Questions[5 marks questions] 

Ques. Name as well as explain the factors influencing the equilibrium state.

Ans. The several factors influencing the equilibrium state are: 

• Concentration: Change in concentration influences the equilibrium state. When the concentration of the reactants is increased, the equilibrium shifts in a direction in which more products are formed. On the other hand, if the concentration of the products is increased, the equilibrium will shift in a direction in which more reactants are formed.
• Temperature: Like concentration, the change in temperature also affects the equilibrium state. An increase in temperature of the system will shift the equilibrium in a direction in which heat is absorbed (i.e. increase in the rate of endothermic reaction). Whereas, a decrease in temperature will shift the equilibrium in a direction in which heat is released (i.e., increase in the rate of exothermic reaction).
• Pressure: Pressure influences the equilibrium state only when the reaction proceeds with a change in volume. An increase in pressure shifts the equilibrium in a direction in which the volume of the system decreases. And a decrease in pressure shifts the equilibrium in a direction with an increase in the volume of the system.

To explain the effect of temperature, pressure, and concentration on the equilibrium state, we can also consider the combination of N₂ and H₂ to form NH3

N₂ + 3H₂ ⇔ 2NH₃     ΔH = – 92 kJ mol ^– 1

The reaction is exothermic, reversible and associated with a decrease in volume.

Effect of temperature: The formation of ammonia is accompanied by the release of heat, and is complemented by a decrease in temperature.

Effect of pressure: When going from left to right in the reaction above, there is a decrease in the volume or number of moles, and ammonia is formed with an increase in pressure.

Effect of concentration: The addition of N₂ or H2 favors the formation of ammonia.

Ques. Find the oxidation state of sulfur in the following compounds: 

H₂S, H₂SO₄, S₂O₄²⁻, S₂O₈²⁻ and HSO₃⁻

Ans. In H₂S, oxidation state of hydrogen is one then for sulfur oxidation state will be 

2+x=0; x=−2

In H₂SO₄, oxidation state of hydrogen is one and oxygen is -2 then sulfur will be

2+x−8=0; x=6

In S₂O₄²⁻, oxidation state of oxygen is -2 then oxidation state of sulfur will be

2x−8=−2; x=3

In S₂O₈²⁻, oxidation state of oxygen is -2 then oxidation state of sulfur will be

2x−16=−2;x=7

In HSO₃⁻, the oxidation state of hydrogen is one and oxygen is -2 then sulfur will be

1+x−6=−1;x=4.

Ques. Calculate the hydrolysis constant and degree of hydrolysis and pH of 0.10 M KCN solution at 25°C. Ka for HCN = 6.2 × 10^-10.

Ans. KCN is the salt of a weak acid HCN and strong base KOH

∴ Kh = Hydrolysis constant

Hydrolysis of KCN is 

Ques. What are the general characteristics of equilibria involving physical processes?

Ans. The general characteristics of equilibria involving physical processes are as follows:

• Solid-to-liquid equilibrium: Two phases can coexist at a single temperature, or one atm; if there is no heat exchange with the environment, the masses of the two phases stay constant.
• Liquid-vapor equilibrium: In this scenario, the pressure of the vapor at any temperature remains constant.
• Dissolution of solids in liquids: The solubility of solids in liquids is constant at all temperatures in this scenario.
• Gas dissolution in liquids: In this instance, the gas's concentration in the liquid is proportionate to its pressure above the liquid.

Ques. The solubility product of Al(OH)₃ is 6.7 x 10^-11. Calculate its solubility in g / L and also find out the pH of this solution. (Atomic mass of Al = 27 u)?

Ans. Let S be the solubility of Al(OH)₃

K_sp = [Al³⁺] [OH⁻]³

K_sp = (S) (3S)³

K_sp = 27S⁴

S⁴ = K_sp / 27

S⁴ = 6.7 × 10^-11 / 27 x 10

S⁴ = 0.24 x 10 ^-12 mol / L.

S = 0.24 x 10 ^-3 mol / L.

(i) Solubility of Al(OH)₃: Molar mass of Al(OH)₃ is 78g.

Therefore, the solubility of Al(OH)₃ in g / L = 0.24 ×10 ³ × 78 g / L

Solubility of Al(OH)₃ in g / L = 19.35 × 10⁻³ g / L

(ii) pH of the solution: S = 0.24 × 10⁻³ mol / L

[OH] = 3S = 3 × 0.24 × 10⁻³ = 0.72 × 10⁻³

pOH =3 − log 0.72

pH = 14 − pOH = 11 + 0.1426 = 11.1426

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