9th Class Chemistry Unit 6 Equilibria Long Questions and Answers

1.How can you derive the reversible reaction at equilibrium? (a) in the forward direction (b) in the backward direction

(a) In the forward Direction:
To derive the reversible reaction in the forward direction at equilibrium, following step should be taken:
1. Concentration of reactants: Adding one or more reacting species will push the reaction towards forward direction producing more products.
2. Concentration of products: With drawing one or more product species will shift the equilibrium position to the right, favoring the formation of more products.
3. Temperature:
• If the reaction is exothermic, decreasing the temperature shift the equilibrium to forward direction producing more products.
• If the reaction is endothermic, increase the temperature shifts the equilibrium to forward direction producing more products.
4. Effect of Pressure & Volume:
• In gas phase reaction, if reaction proceeds with decrease in volume at equilibrium stage. Then further decrease in volume or increase in pressure at equilibrium will shift the equilibrium position in forward direction.
• In gas phase reaction if reaction proceeds with increase in volume at equilibrium stage. Then further increase in volume or decrease in pressure at equilibrium, will shift the equilibrium position in forward direction.

(b) In the backward Direction:
To derive the reversible reaction in the backward direction at equilibrium the following steps should be taken:
1. Concentration of reactants: By decreasing the concentration of reactants, it will shift the equilibrium towards backward direction producing more reactants.
2. Concentration of products: By increasing the concentration of products will push the reaction to backward direction producing more reactants.
3. Temperature:
• If the reaction is exothermic, increasing the temperature at equilibrium will shift the equilibrium position in the backward direction.
• If the reaction is endothermic, decreasing the temperature at equilibrium will shift the equilibrium position in the backward direction.
4. Effect of Pressure & Volume:
• In gas phase reaction if reaction proceeds with decrease in volume at equilibrium stage. Then further increase in volume or decrease in pressure at equilibrium will shift the equilibrium position in backward direction.

2.Explain how the forward and backward reactions change when the system approaches equilibrium.

When a system approaches equilibrium, the forward and backward reactions change in their rates until they become equal.

Forward Reaction If a reversible reaction is started by mixing the reactants, the reaction moves in the forward direction only.

N₂ + 3H₂ ⟶ 2NH₃

Backward Reaction After some time when enough concentrations of the products are built up, they react to give back the reactants in the reverse reaction.

2NH₃ ⟶ N₂ + 3H₂

Equilibrium State The reaction will keep on going in both the directions until the rate of forward reaction becomes equal to the rate of backward reaction. The number of reactant molecules which will disappear as a result of forward reaction becomes equal to the number of reactant molecules which will form as a result of the reverse reaction. The same will be true for the product molecules. At this stage, the reaction is said to be in a state of chemical equilibrium. It appears as if nothing is going on in the reaction vessel as the concentrations of both reactant and product molecules do not undergo any change at this stage. Since the reaction did not cease at this state of equilibrium, rather it keeps on going in both the directions, this state is called dynamic equilibrium. The concentrations of reacting species (reactants and products) remain constant at equilibrium.

N₂₍ₘ₎ + 3H₂₍ₘ₎ ⇌(400°C, 200 atm, Fe) 2NH₃₍ₘ₎

3.Describe the effect of a catalyst on the reversible reaction.

At the stage of dynamic equilibrium, the rates of both forward and backward reactions are equal.
A catalyst increases both the rates of forward and backward reactions of a reversible reaction. So if a reversible reaction is carried out in the presence of a catalyst it will decrease the time taken by the reaction to attain the state of equilibrium.

Example Consider the following Haber's process at equilibrium.

N₂₍ₘ₎ + 3H₂₍ₘ₎ ⇌(400°C, 200 atm, Fe) 2NH₃₍ₘ₎

The Haber's process is commonly used for ammonia synthesis in which iron catalyst is used to enhance the reaction rates. The catalyst helps increase the yield of ammonia without being consumed in the process. In this way, the yield of ammonia gas for production of urea fertilizer can be maximized.

Role of Catalyst in Haber's Process
For a dynamic equilibrium to be setup the rate of the forward reaction must be equal to the rate of backward reaction. This does not happen instantly and for very slow reaction, it may take years. Both the rates of formation and the decomposition of ammonia are reasonably fast at around 400°C in the presence of a catalyst. This reaction will reach the equilibrium state within minutes of the start of reaction.

4.How can a reversible reaction be forced to go to completion?

A reversible chemical can be forced to go to completion by disturbing in the following possible ways.
i. Adding one or more of the reacting species
ii. Withdrawing one or more of the product species
iii. Changing the temperature of the reaction
iv. Changing the pressure or volume of the reaction if it involves reactants and products in the gaseous state

Example Consider the following reversible reaction at equilibrium.

N₂₍ₘ₎ + 3H₂₍ₘ₎ ⇌(400°C, 200 atm, Fe) 2NH₃₍ₘ₎

(i) Adding N₂ or H₂ continuously in reaction mixture force this reversible reaction to completion.
(ii) Withdrawing of NH₃ force this reversible reaction to completion.
(iii) Decreasing temperature (up to 400°C) will force the reaction to completion.
(iv) Decreasing volume or increasing pressure at equilibrium will force the above reaction to completion.

5.How does change in temperature affect the reaction at equilibrium?

Effect of changing the temperature on the state of equilibrium:
The formation of ammonia is exothermic in the forward direction and hence this reaction will be endothermic in the reverse direction.

Heat
N₂₍ₘ₎ + 3H₂₍ₘ₎ ⇌ 2 NH₃₍ₘ₎ ΔH = - 92.4 KJ/mole

If this reaction is at equilibrium and its temperature is increased, the state of the equilibrium will be disturbed again. The ΔH of this reaction is negative. This means the total energy of the system containing N₂ and H₂ is higher than that of ammonia. The increase in temperature of this reaction at equilibrium will push the reaction in the backward direction i.e. the reactants side. Decreasing the temperature will derive the equilibrium to the forward direction.
If the reaction is exothermic, decreasing the temperature shift the equilibrium to forward direction producing more products. If the reaction is exothermic, increasing the temperature at equilibrium will shift the equilibrium position in the backward direction.
If the reaction is endothermic increase the temperature shifts the equilibrium to the forward direction producing more products can also help to derive the reaction forward. In gas phase reaction if reaction proceeds with increase in volume at equilibrium stage. Then Decrease in volume or increase in pressure at equilibrium, will shift the equilibrium position in Backward direction.

6.Study the effect of heat on hydrated CuSO₄. Why does this salt look coloured and why does it lose colour upon heating?

Copper sulphate pentahydrate (CuSO₄.5 H₂O) is an important salt which is blue in colour. When this salt is heated strongly, its colour changes to white. This physical change involves the following equilibrium.

CuSO₄ • 5 H₂O ⇌ CuSO₄ + 5 H₂O
Hydrated Copper Sulphate Anhydrous Copper Sulphate
Heat

When white anhydrous copper sulphate absorbs moisture from the atmosphere, it will turn blue again.

Presence of Blue Color The blue color of CuSO₄.5H₂O comes from the water molecules present in its structure. These water molecules interact with the copper ions in the compound, causing it to appear blue.

Absence of Blue Color The blue color of CuSO₄.5H₂O is lost upon heating because the water molecules within its structure are removed through dehydration. This removal of water causes the crystals to lose their color and appear white.

7.Synthesis of ammonia gas is very important industrially because it is used in the preparation of urea fertilizer. Explain the conditions you will use to get the maximum yield of ammonia.

Industrial production of ammonia in Haber Process is a very useful application of the phenomenon of chemical equilibrium. Ammonia gas leads to the formation of an important fertilizer urea. The ability of ammonia gas to be converted into its liquid form easily is used to drive the reaction to completion. In this way, practically 100% conversion of N₂ and H₂ to NH₃ is achieved.

N₂₍ₘ₎ + 3H₂₍ₘ₎ ⇌(400°C, 200 atm, Fe) 2NH₃₍ₘ₎

To achieve the maximum field of ammonia gas in the industrial synthesis process, the following condition are typically used:

i. High Pressure: Increasing the pressure shifts the equilibrium towards the formation of more ammonia molecules typically, pressures of around 200-300 atm are used to maximize the yield of ammonia.

ii. Low Temperature: Lower temperatures favor the exothermic reaction that produce ammonia. However, the reaction is faster at higher temperatures. Temperature of around 400-500°C is often used to balance the yield and reaction rate.

iii. Catalyst: The Haber's process is commonly used for ammonia synthesis in which iron catalyst is used to enhance the reaction rate. The catalyst increase the yield of ammonia without being consumed in the process. In this way, the yield of ammonia gas for production of urea fertilizer can be maximized.

Unit 6: Equilibria — Long Questions