Test: Law of Chemical Equilibrium and Equilibrium Constant (3 July) - JEE MCQ

Concept:
Van't Hoff equation -The Van't Hoff equation gives the relationship between the standard Gibbs free energy change and the equilibrium constant.
It is represented by the equation -
-ΔG° = RT log_eK_p
where, R = gas constant 
T = temperature 
K_p = equilibrium constant
or 

Thus, the equilibrium constant K is temperature-dependent.
Explanation:
The given reaction is -
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)
Given condition - the total pressure at which the equilibrium is established, is increased without changing the temperature or at a constant temperature.
According to Van't Hoff equation, the equilibrium constant K is temperature-dependent. 
If the temperature is constant, the reaction is neither exothermic nor endothermic.
Therefore, K will remain same as temperature is constant.
Conclusion:
Hence, if the total pressure at which the equilibrium is established, is increased without changing the temperature the equilibrium constant K for the reaction 
N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g), remains same.
 Hence, the correct answer is option 1.

Given Reaction are

we need to final K for the following Reaction

We know

Dividing K₂ by K₁,

Concept:

Relationship between K_p and K_c - For a reaction, aA + bB  ⇋ xX + yY

• K_p - It is the equilibrium constant when concentration is expressed in terms of partial pressure(generally when reactant and products are gaseous).
  
• K_c - It is the equilibrium constant when concentration is expressed in terms of molarity.
  • 

Relationship between K_p and K_c:

Where 

• Δn is the change in number of moles of gaseous substance.
• R is gas constant.
• T is temperature

Calculation:
The chemical reaction is -
C₁₀H₈(s) ⇋ C₁₀H₈(v) 
The vapour pressure above the balls was found to be 0.10 mm Hg.
∴ K_p =  0.10 mm Hg.
Relation between K_p and K_c is -

Where 

• Δn is the change in number of moles.
• R is gas constant.
• T is temperature

Given,  Kp =  0.10 mm Hg or 0.1/760 atm
R = 0.0821 L(atm) mol^-1K^-1.
T = 273 + 27 = 300 K
Δn = 1

= 5.36 × 10-6
Conclusion:
Therefore, the value of Kc(sublimation) = 5.36 × 10^-6

Concept:
Relation between K_c and Q_c - 

1. The comparison between K_c and Q_c determines the direction of the reaction to take place.
2. If K_c > Q_c the reaction proceed to form products or to the right.
3. K_c = Q_c this is condition of equillibrium
4. K_c < Q_c the reaction proceed in the reverse direction or to the left.

Explanation:
For the reaction, 2A ⇋ B + C, reaction quaotient(Q_c) is given by -

Given, the reaction mixture has [A] = [B] = [C] = 3 × 10^-3 M.

Given, K_c = 2 × 10^-3 or  0.002 
As (Q_c) >  (K_c)
So, the reaction will proceed in the reverse manner or to the left.
Conclusion:
For a reaction 2A ⇋ B + C, K_c is 2 × 10^-3. At a given time, the reaction mixture has [A] = [B] = [C] = 3 × 10^-3 M,  the reaction will proceed in the reverse manner or to the left.

Concept:

• The constant which defines the relationship between the molar concentration of the reactants and the rate of the chemical reaction is known as the rate constant.
• The rate constant of a reaction depends on the following factors :
• Nature of the reactants. 
• Temperature of the reaction. As the temperature increases, the velocity constant (rate constant) increases. 
• The conditions of the reactions like the presence of the catalyst, solvent, pH, etc. 
• It does not depend on the concentration of the reactants. But if one or more substances are in excess concentration, then the order of the reaction is independent of them. 

Explanation:
2SO₂ + O₂ ⇋ 2SO₃
Equilibrium constant for this reaction

Equilibrium constant for this reaction

On squaring both side of equation (ii), we get

Conclusion:
Thus, the equilibrium constant for the reaction SO₃ ⇋ SO₂ + (1/2)O₂ at the same temperature is 1/8.

Concept:-

• Equilibrium Constant (K): The equilibrium constant, often denoted by K, is a value that expresses the balance between reactants and products of a reaction at equilibrium. In general terms, it's equal to the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their stoichiometric coefficient.
• Direction of Reaction & Equilibrium Constant: The equilibrium constant can help to predict the direction of a reaction to reach equilibrium if non-equilibrium concentrations of reactants and products are provided. A comparison between the reaction quotient (Q), calculated the same way as K but from initial concentrations, and K determines the direction of the reaction.
• Dependence of K on Initial Concentrations: The value of the equilibrium constant for a given reaction at a given temperature is a constant value, independent of the initial concentrations of reactants and products. However, it is worth noting that the individual equilibrium concentrations of reactants and products depend on initial concentrations.

Explanation:-
K_equ = kf/kb
K_equ= [C]^c [D]^d/[A]^a [B]^b = K_c
The equilibrium constant can indeed be used for calculating the equilibrium concentrations given initial concentrations and other equilibrium constants, so this statement is correct.
The direction of a reaction can be predicted by comparing the reaction quotient (Q) to the equilibrium constant.

• If Q > K, the reaction will proceed in the direction of the reactants.
• If Q < K, the reaction will proceed in the direction of the products.
• If Q = K, the reaction is at equilibrium.
• So, this statement is also correct.​

​The value of the equilibrium constant is independent of initial concentrations of reactants and products but is a function of temperature. Therefore, this statement is incorrect.
So, the correct answer is 4) More than one of the above, as both statements 1 and 2 are correct.

Concept:

• Reversible reactions involve the condition where the rate of forward reaction becomes equal to the rate of backward reaction.
• It is termed an equilibrium condition.

Explanation:

• Consider a hypothetical reversible reaction: aA(g)+bB(g)⇌cC(g)+dD(g)
• The expression for the rate constant of the forward reaction is shown below:
• 
• The expression for the rate constant of the backward reaction is shown below:
• 
• Thus, from (1) and (2) the relation between K_f and K_b is shown below:
• 

Concept:

• Effect of pressure on equilibrium: When a system at equilibrium is subjected to a change in the pressure of gaseous reactants and if pressure is increased, then equilibrium shifts towards the fewer moles of the gaseous reactants.
• Since pressure and volume are mutually inversely proportional to each other in the case of gaseous substrates.

Explanation:
The given equilibrium reaction is shown below:
N₂(g)+3H₂(g) ⇌ 2NH₃(g)
 The pressure at the equilibrium position is increased.

• When pressure is increased, then equilibrium will move in such a direction that it opposes the change according to LeChatlier's principle.
• The total number of moles of gaseous substrates on the left side of the equilibrium is 4 mol.
• The total number of moles of gaseous substrates on the left side of the equilibrium is 2 mol.
• An increase in pressure favours the decrease in the volume of gases at a constant temperature according to Boyle's law.
• Hence, an increase in pressure at this equilibrium, allows the equilibrium to move towards a lesser number of moles.

Hence, the formation of ammonia NH₃ gas will be there.

Concept:
The equilibrium constants for a perfect gaseous mixture are K_P and K_C.
K_P: When gas phase equilibrium concentrations are stated in terms of atmospheric pressure, the equilibrium constant is designated by K_p.
K_C: When equilibrium concentrations are expressed in molarity, the equilibrium constant is expressed by Kc.
The relationship between K_P and K_C is:

Where  is the number of products - number of moles of reactants
Explanation:
The given reaction is:

The equilibrium constant of the reaction is expressed as:
The product and reactant both are in the gaseous phase, so the concentration of both comes under the equilibrium equation:

Calculating the value of 

Conclusion:
For the reaction,   value is 

Concept :
Relation between K_p and K_c - 

• K_p and K_c - are the equilibrium constant of ideal gas mixture consider under reversible reaction.
• K_p is the equilibrium constant wrt pressure.
• And K_c is the equilibrium constant wrt concentration. 
• K_p and K_c are related as - K_p = K_c(RT)^Δn ​

Calculation:
K_p and K_c are related as - 
K_p = K_c(RT)^Δn 
Given, (K_c) = 4 × 10^-4 ,T = 1000 K
Δn = no. of moles of product - no. of moles of reactant
Δn = 3-2 = 1
R = 8.314 
we have to calculate K_p at 800K
K_p = 4 × 10^-4 ×(8.314 × 800)¹
K_p= 0.026
Conclusion:
Therefore, K_p for the reaction at 800K is 0.026.