[Solved] Le-Chateliers Principle And Its Application MCQ [Free PDF] - Objective Question Answer for Le-Chateliers Principle And Its Application Quiz - Download Now! (2024)

1. Zero
2. 1/1
3. 2/2
4. More than one of the above
5. None of the above

The correct answer is 1/1

Concept:

• Relation Between Kp and Kc : For a given reaction, the equilibrium constants Kp (in terms of partial pressures) and Kc (in terms of concentrations) are related by the equation:

  \( K_p = K_c (RT)^{\Delta n}\)

• Δn: Δn is the change in the number of moles of gas between the products and reactants, calculated as:

  Δn = (moles of gaseous products) - (moles of gaseous reactants)

Explanation:

For the given reaction:

N₂(g) + O₂(g) ⇌ 2NO(g)

• Number of moles of gaseous reactants = 1 (N₂) + 1 (O₂) = 2
• Number of moles of gaseous products = 2 (NO)

So, Δn = 2 - 2 = 0

Therefore:

\(K_p = K_c (RT)^{\Delta n}\)

Since Δn = 0:

\(K_p = K_c (RT)^0 = K_c \cdot 1 = K_c\)

Thus, the ratio\( \frac{K_p}{K_c} \) is 1.

So, the correct answer is 1/1.

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1. Zero
2. 1/1
3. 2/2
4. -1/-1

The correct answer is 1/1

Concept:

• Relation Between Kp and Kc : For a given reaction, the equilibrium constants Kp (in terms of partial pressures) and Kc (in terms of concentrations) are related by the equation:

  \( K_p = K_c (RT)^{\Delta n}\)

• Δn: Δn is the change in the number of moles of gas between the products and reactants, calculated as:

  Δn = (moles of gaseous products) - (moles of gaseous reactants)

Explanation:

For the given reaction:

N₂(g) + O₂(g) ⇌ 2NO(g)

• Number of moles of gaseous reactants = 1 (N₂) + 1 (O₂) = 2
• Number of moles of gaseous products = 2 (NO)

So, Δn = 2 - 2 = 0

Therefore:

\(K_p = K_c (RT)^{\Delta n}\)

Since Δn = 0:

\(K_p = K_c (RT)^0 = K_c \cdot 1 = K_c\)

Thus, the ratio\( \frac{K_p}{K_c} \) is 1.

So, the correct answer is 1/1.

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1. Zero
2. 1/1
3. 2/2
4. More than one of the above
5. None of the above

The correct answer is 1/1

Concept:

• Relation Between Kp and Kc : For a given reaction, the equilibrium constants Kp (in terms of partial pressures) and Kc (in terms of concentrations) are related by the equation:

  \( K_p = K_c (RT)^{\Delta n}\)

• Δn: Δn is the change in the number of moles of gas between the products and reactants, calculated as:

  Δn = (moles of gaseous products) - (moles of gaseous reactants)

Explanation:

For the given reaction:

N₂(g) + O₂(g) ⇌ 2NO(g)

• Number of moles of gaseous reactants = 1 (N₂) + 1 (O₂) = 2
• Number of moles of gaseous products = 2 (NO)

So, Δn = 2 - 2 = 0

Therefore:

\(K_p = K_c (RT)^{\Delta n}\)

Since Δn = 0:

\(K_p = K_c (RT)^0 = K_c \cdot 1 = K_c\)

Thus, the ratio\( \frac{K_p}{K_c} \) is 1.

So, the correct answer is 1/1.

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1. Zero
2. 1/1
3. 2/2
4. -1/-1

The correct answer is 1/1

Concept:

• Relation Between Kp and Kc : For a given reaction, the equilibrium constants Kp (in terms of partial pressures) and Kc (in terms of concentrations) are related by the equation:

  \( K_p = K_c (RT)^{\Delta n}\)

• Δn: Δn is the change in the number of moles of gas between the products and reactants, calculated as:

  Δn = (moles of gaseous products) - (moles of gaseous reactants)

Explanation:

For the given reaction:

N₂(g) + O₂(g) ⇌ 2NO(g)

• Number of moles of gaseous reactants = 1 (N₂) + 1 (O₂) = 2
• Number of moles of gaseous products = 2 (NO)

So, Δn = 2 - 2 = 0

Therefore:

\(K_p = K_c (RT)^{\Delta n}\)

Since Δn = 0:

\(K_p = K_c (RT)^0 = K_c \cdot 1 = K_c\)

Thus, the ratio\( \frac{K_p}{K_c} \) is 1.

So, the correct answer is 1/1.

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1. The reaction is endothermic
2. The reaction is exothermic
3. If NO₂ (g) and N₂O₄ (g) are mixed at 400 K at partial pressures 20 bar and 2 bar respectively, more N₂O₄ (g) will be formed.
4. The entropy of the system increases.

Concept :

• Let a reaction is A + B\(\rightleftharpoons\) C +D
  • Then the equilibrium constant (K) = [C][D] / [A][B]
  • Where [A],[B],[C],[D] are the concentration of A,B,C,D.
  • If the gaseous product is there then partial pressure is taken instead of concentration
• Exothermic reaction: The reaction in which heat is evolved
• Endothermic reaction: the reaction in which heat is absorbed

For a reaction that is in equilibrium then if the forward reaction is exothermic then the backward reaction must be endothermic and vice-versa

• At a constant temperature forreaction equilibrium constant is a constant
• According to the la Chatelier principle-
  • With increasing temperature the reaction goes in a direction in which direction the heat is absorbedeg - with increasing temperature exothermic reaction will go backward direction and an endothermic reaction will go in the forward direction
  • Entropy is the measure of randomness. with increasing the no. of molecules in a reaction the entropy increases.

Explanation :

For the reaction it is given that K = 40 at 400 and 1700 at 500

So increasing the temperature value K increases which means the concentration of the product increases.

That means the reaction is going forward.

With increasing temperature, the reaction goes forward so the reaction is endothermic.

Now given that at 400 K p_N2O4 = 2 bar and p_NO2 = 20 bar

Then K =\(\frac{pNO2}{pN2O4}\) = ²⁰/₂= 10

So at 400 K value of K is 10 but it already said that the value of Kis 50 at 400 K

At a temperature value of K is constant

Thus to be 50 value ofpNO2has to be increased

So more NO₂ has to be formed.

Again from 1 N₂O₄ molecule, 2 NO₂ is produced

So for dissociation of N₂O₄, the no.of molecules increased in the reaction mixture

As the no. of molecules increased entropy also increased.

Conclusion:

So the reaction is endothermic and by the dissociation entropy of the reaction increases.

the correct option is 1 and 4.

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1. H₂ (g) + I₂ (g)\(\rightleftharpoons\) 2HI (g)
2. PCl₅ (g)\(\rightleftharpoons\) PCl₃ (g) + Cl₂ (g)
3. N₂ (g) + 3H₂ (g) \(\rightleftharpoons\) 2NH₃ (g)
4. The equilibrium will remain unaffected in all the three cases.

Concept:

Addition of inert gas at constant volume-

• Inert gas is unreactive but can act as a catalyst which increases the rate of the reaction without being utilized in the reaction.
• Inert gas at constant volume when added to a system in equilibrium, the total pressure will increase.
• But the concentration of reactant and product remains the same or does not change.
• Therefore, the equilibrium constant remains unaffected when an inert gas is added to equilibrium at constant volume.

Explanation:

Noble gas or inert gas does not react itself due to the extremely low reactivity of noble gas with other substances.

As Ar is a noble gas, it is also unreactive but it decreases the activation energy in both forward as well as backward directions by the same amount of reversible reaction.

When Ar is added at constant volume, the equilibrium will not change as there is no change in the no. of moles of the reaction (forward and backward)of either reactant or products.

⇒H2(g) + I2(g)\(\rightleftharpoons\)2HI (g)

• No. of moles for forward reaction =2 for reactant and 2 for product
• No. of moles for backwardreaction =2for reactant and 2 for product

⇒PCl5(g)\(\rightleftharpoons\)PCl3(g) + Cl2(g)

• No. of moles for forward reaction =2 for reactant and 1 for product
• No. of moles for backward reaction =1 for reactant and 2 for product

⇒N2(g) + 3H2(g)\(\rightleftharpoons\)2NH3(g)

• No. of moles for forward reaction =4 for reactant and 2 for product
• No. of moles for backwardreaction =2for reactant and 4 for product

Hence, in all cases equilibrium remains unaffected.

Conclusion:

Therefore, the equilibrium will remain unaffected in all three cases.

Hence, the correct answer is option4.

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