JEE Important Chapter - Equilibrium

The state of a process in which the properties of the system, such as temperature, pressure, and concentration, do not change throughout time is called equilibrium. All processes that reach equilibrium entail two opposing processes. When the concentrations of the two conflicting processes are equal, it approaches equilibrium.

Because the opposing mechanisms only demand physical alterations, the equilibrium is called Physical Equilibrium. Chemical Equilibrium occurs when chemical reactions are in opposition to one another.

JEE Main Chemistry Chapter-wise Solutions 2022-23 

Important Topics for Equilibrium

• Equilibrium

• Law of Mass Action

• Equilibrium Constant

• Le-Chatelier's Principle

• Free Energy

• Ionic Equilibrium

Important Concepts for Equilibrium

Reversible and Irreversible Reactions

• Reversible Reactions: Reversible reactions are ones in which the reactants are not completely transformed into products.

A reversible reaction is one in which an acid and a base, one or both of which are weak, neutralise each other.

This is the reaction that occurs when a weak acid (CH3COOH) reacts with a strong base (NaOH).
CH3COOH + NaOH ⇌ CH3COONa + H2O

• Irreversible Reactions: Irreversible reactions are ones in which the entire quantity of reactants is transformed into products.

Irreversible reactions include the neutralisation reaction between a strong acid (HCl) and a strong base (NaOH).
NaOH + HCl → NaCl + H2O.

Equilibrium and Its Dynamic Character

• The state of equilibrium is when the concentrations of reactants and products do not change over time i.e. the reactant and product concentrations become constant.

• A chemical equilibrium is dynamic in nature, which implies that reactions continue to occur at the same rate in both the forward and backward directions. 

• As a result, the amount of product created reacts backwards to give reactants, leaving no change in reactant or product concentration with the passage of time.

Law of Mass Action and Equilibrium Constant

• “At any given moment, the product of the molar concentrations of the reactants at a fixed temperature determines the pace of a chemical reaction."

• Take a look at a simple reversible reaction, at a particular temperature.
  aA + bB ⇌ cC + dD.

• Therefore at equilibrium ,
  Forward reaction rate = Backward reaction rate
  $k_{f}[A]^{a}[B]^{b}=k_{b}[C]^{c}[D]^{d}$
  $\dfrac{k_{f}}{k_{b}}= K_{c} = \dfrac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}$; Kc stands for equilibrium constant.
  
  

Relation Between the Three Constants - Kp, Kc, and Kx

• The values of Kp can be obtained as follows:
  Kp = Kc (RT)Δn
  Kp = Kc (P)Δn
  Δn = In a chemical equation, the number of moles of gaseous products equals the number of moles of gaseous reactants.

Relationship Between Equilibrium Constant and G°

• The relation relates ΔG for a reaction under any situation to ΔG°,
  ΔG = ΔG° + 2.303 RT log(Q)

• At temperature T, the standard free energy change of a reaction and its equilibrium constant are related by the relation,
  ΔG° = - 2.303 RT logK

• For any general reaction,
  aA + bB ⇌ cC + dD;
  The rate constant is given as,
  K = ((aC)c(aD)d)/((aA)a(aB)b); The activity of the reactants and products is represented by the letter a. It has a lower unit count.

• For pure solids and liquids, use the following formula:
  a = 1

• For gases, use:
  a = gas pressure measured in atm.

• For components in solution: a = molar concentration.

Le-Chatelier's Principle

• Changes in any of the components that determine a system's equilibrium conditions will cause the equilibrium to shift in such a way that the effect of the change is reduced or neutralised.

• Application of Le-Chatelier's Principle:
  The Le-Chatelier's concept is extremely important in chemical, physical, and everyday life systems that are in a state of equilibrium.

Applications to the Chemical Equilibrium

Solved Examples From the Chapter

Question 1: A critical stage in the industrial manufacture of sulfuric acid is the reaction between gaseous sulphur dioxide and oxygen:

2SO2(g) + O2(g) → 2SO3(g)

The system was kept at 800 K with a mixture of SO2 and O2 until it reached equilibrium. The equilibrium mixture was made up of 5.0 × 10−2 M SO3, 3.5 × 10−3 M O2, and 3.0 × 10−3 M SO2. At this temperature, calculate K and Kp.

Solution: 

• In the equilibrium constant expression, substitute the relevant equilibrium concentrations,

• K = [SO3]2/([SO2]2[O2])

K = (5.0 × 10−2)2/(3.0 × 10−3)2.(3.5 × 10−3) 

∴ K = 7.9 × 104.

• We use the relationship between K and Kp to solve for Kp, where
  Kp = K(RT)Δn

Kp = 7.9 × 104((0.08206 L⋅atm/mol⋅K)(800 K))−1 

∴ Kp = 1.2 × 103

Key Points to Remember: The relationship between K and Kp is Kp = K(RT)Δn.

Question 2: When 0.1 M CH3COOH (50 ml) and 0.1 M NaOH (50 ml) are mixed, calculate the pH of the solution [Ka (CH3COOH) = 10-5].

Solution: 

• The three dissociation into ions in given chemical reactions are given as:
  CH3COOH ⇌ CH3COO- + H+,
  NaOH → Na+ + OH-, and H+ + OH- ⇌ H2O.

• For the reaction,
  CH3COOH + OH- ⇌ CH3COO- + H2O
  Keq = Ka/Kw.

• Concentration of H2O thus remains constant,
  109 = x/(0.05-x)2; 

• Because the value of equilibrium constant is very high, x >> 0.05

• ∴ Let 0.05 - x = a

109 = 0.05/a2

a = 7.07 x 10-6

pOH = 6 - log(7.07)

pOH = 6 - 0.85

• pH = 14 - (6 - 0.85) = 8.85

Hence, the pH is 8.85 which is basic.

Key Points to Remember: The relation between equilibrium constant and acid dissociation constant is Keq = Ka/Kw.

Solved Examples from Previous Year Question Papers

Question 1: For the reaction, 2SO2(g) + O2(g) ⇌ 2SO3(g), ∆H = - 57.2 kJ mol–1 and Kc = 1.7 × 1016. Which of the following statements is incorrect?

(a) The equilibrium will shift in the forward direction as the pressure increases.

(b) The addition of inert gas at constant volume will not affect the equilibrium constant.

(c) The equilibrium constant is large, suggestive of reaction going to completion and so no catalyst is required.

(d) The equilibrium constant decreases as the temperature increases.

Solution: 

• The big value of Kc indicates that the reaction is nearly complete. SO2 oxidation to SO3 is a slow process. 

• As a result, adding a catalyst speeds up the process. 

• As a result, option (c) is the correct answer.

Question 2: 20 mL of 0.1 M H2SO4 solution is added to 30 mL of 0.2 M NH4OH solution. The pH of the resultant mixture is [pKb of NH4OH = 4.7]

(a) 9.4

(b) 9.0

(c) 5.0

(d) 5.2

Solution: 

• Given that NH4OH has a pKb of 4.7,

• 0.1 M H2SO4 (20 ml) ⇒ nH+ = 4.

• nNH4OH = 6 from 30 ml 0.2 M NH4OH.

• The chemical reaction is given as:
  NH4OH + H+ ⇌ NH4+ + H2O
      6           4            0          0
      2           0            4          4  

• Basic buffer is the solution.

• pOH = pKb + log [NH4+]/[NH4OH] = 4.7 + log (4/2) pOH = pKb + log [NH4+]/[NH4OH] = 4.7 + log (4/2) pOH = pK

• pOH = log 2 + 4.7

pOH = 4.7 + 0.3

 pOH = 5

• 14 - pOH, is the pH value.

14 - 5 = 9

• As a result, option (b) is the correct answer.

Question 3: ​​The increase of pressure on the ice water system at constant temperature will lead to

(a) no effect on that equilibrium.

(b) a decrease in the entropy of the system.

(c) a shift of the equilibrium in the forward direction.

(d) an increase in the Gibbs energy of the system.

Solution: When the pressure on this system in equilibrium is increased, the equilibrium tends to shift in the direction of volume reduction, i.e., forward.

As a result, option (c) is the correct answer.

Practice Questions

Question 1: The reaction between hydrogen gas and iodine produces hydrogen iodide:
H2(g)+ I2(g) → 2HI(g)

At 740 K, a combination of H2 and I2 was kept until the system reached equilibrium. The equilibrium mixture was made up of
1.37 × 10−2 M HI, 6.47 × 10−3 M H2, and 5.94 × 10−4 M I2.
For this reaction, calculate K and Kp.

Answer: K = 4.88, and Kp = 48.8.

Question 2: For a chemical reaction in equilibrium, which of the following is true?

(a) The forward reaction is the only thing that comes to a halt.

(b) Only the opposite reaction is halted.

(c) The forward and reverse reactions have the same rate constants.

(d) The forward and reverse reactions have the same rates.

Answer: (d) The forward and reverse reactions have the same rates. 

Conclusion

When the observable attributes of a process, such as colour, temperature, pressure, concentration, and so on, do not vary, it is said to be in equilibrium.

The word equilibrium implies 'balance,' implying that the reactants and products involved in a chemical reaction are in a state of equilibrium. The equilibrium condition can also be seen in physical processes like the melting point of ice at 0°C, when both ice and water are present at the same time.

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