Class 11 Chemistry – Chapter 8: Chemical Equilibrium (FBISE)
This section provides complete, exam-oriented notes for Class 11 Chemistry Chapter 8 – Chemical Equilibrium strictly according to the Federal Board (FBISE) syllabus. The chapter builds a strong foundation for understanding reversible reactions and equilibrium systems.
Major topics include reversible and irreversible reactions, law of mass action, equilibrium constant (Kc & Kp), reaction quotient (Q), Le Chatelier’s principle, factors affecting equilibrium, and industrial applications. Numerical problems and exam-focused concepts are explained step by step.
For better conceptual clarity and exam preparation, students can access video lectures, MCQs, numericals practice, test series, and live revision classes on our YouTube channel and stay updated through our WhatsApp channel.
8.2 Reversible Reactions and Dynamic Equilibrium
A reversible reaction is one in which the products can react back to form the original reactants. These reactions never reach 100% completion; instead, they exist in a state where both forward and backward reactions occur simultaneously.
Key Concepts
- Forward Reaction: Reactants reacting to form products.
- Reverse Reaction: Products reacting to form reactants.
- Dynamic Equilibrium: A state where the rate of the forward reaction equals the rate of the reverse reaction. While the concentrations of reactants and products remain constant, the molecules continue to react at the microscopic level.
Example Reaction
Consider the reaction between steam and carbon monoxide:$$\text{H}_2\text{O}_{(g)} + \text{CO}_{(g)} \rightleftharpoons \text{H}_{2(g)} + \text{CO}_{2(g)}$$
8.3 The Law of Mass Action
Proposed by Guldberg and Waage in 1864, this law states that the rate of a chemical reaction is proportional to the product of the active masses (molar concentrations) of the reacting substances.
Deriving the Equilibrium Constant ($K_c$)
For a general reaction: $$a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$$
- Rate of forward reaction ($R_f$): $R_f = k_f [\text{A}]^a [\text{B}]^b$
- Rate of reverse reaction ($R_r$): $R_r = k_r [\text{C}]^c [\text{D}]^d$
- At equilibrium ($R_f = R_r$): $k_f [\text{A}]^a [\text{B}]^b = k_r [\text{C}]^c [\text{D}]^d$
The equilibrium constant expression is:$$K_c = \frac{k_f}{k_r} = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b}$$
Questions and Answers
Q1: Why do the lines in a concentration-time graph become parallel at equilibrium?
A: The lines become parallel (horizontal) because the concentrations of reactants and products have become constant. This happens because the rate at which substances are being consumed is exactly equal to the rate at which they are being produced.
Q2: What is meant by the term "active mass"?
A: Active mass refers to the molar concentration of a substance, usually expressed in units of $\text{mol dm}^{-3}$ for dilute solutions, denoted by square brackets $[ ]$.
Q3: Why is equilibrium called "dynamic" rather than "static"?
A: It is called dynamic because the reaction has not actually stopped. At the microscopic level, reactants are still forming products and products are still forming reactants, but since the rates are equal, no net change is observed macroscopically.
Q4: How do you write a $K_c$ expression for a specific reaction?
A: Place the product concentrations in the numerator and the reactant concentrations in the denominator. Each concentration is raised to the power of its stoichiometric coefficient from the balanced equation.
Chemical Equilibrium Notes
1. Key Features of Equilibrium Constant ($K_c$)
- State: $K_c$ applies only when the reaction has reached the equilibrium state.
- Dependency: It does not depend on initial concentrations. It is only dependent on temperature.
- Expression: Concentrations of products are placed in the numerator, and reactants in the denominator. Each is raised to a power equal to its stoichiometric coefficient in the balanced equation.
- Magnitude:
- If $K_c < 1$: Concentration of reactants is greater than products.
- If $K_c > 1$: Concentration of products is greater than reactants.
2. Various Forms of Equilibrium Constants
The equilibrium constant can be expressed based on the units of the substances involved:
| Symbol | Basis | Relationship with $K_p$ |
|---|---|---|
| $K_c$ | Molar Concentration ($\text{mol/dm}^3$) | $K_p = K_c(RT)^{\Delta n}$ |
| $K_p$ | Partial Pressures | - |
| $K_n$ | Number of Moles | $K_p = K_n(\frac{P}{n})^{\Delta n}$ |
| $K_x$ | Mole Fractions | $K_p = K_x(P)^{\Delta n}$ |
Note: $\Delta n$ is the difference between the total number of moles of gaseous products and gaseous reactants.
Questions and Answers
Q1. Write the $K_c$ expression for the following reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Answer: Following the rule (products over reactants), the expression is:$$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$$
Q2. Give the balanced equation that corresponds to the following expression: $K_c = \frac{[CH_3OH]}{[CO][H_2]^2}$
Answer: By reversing the $K_c$ logic, the numerator is the product and the denominators are reactants. The powers become coefficients:$$CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_3OH_{(g)}$$
Q3. Balance and write the $K_c$ expression for: $SO_{2(g)} + O_{2(g)} \rightleftharpoons SO_{3(g)}$
Answer:Balanced Equation: $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$Expression: $$K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$$
Q4. At $500^\circ C$, the equilibrium concentrations for $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$ are $[A_2] = 0.399M$, $[B_2] = 1.197M$, and $[AB] = 0.203M$. Calculate $K_c$.
Answer:Formula: $K_c = \frac{[AB]^2}{[A_2][B_2]}$Calculation: $$K_c = \frac{(0.203)^2}{(0.399)(1.197)}$$$$K_c = \frac{0.041209}{0.477603} \approx 0.086$$
Chemistry Notes: Equilibrium Constants $K_p$ and $K_c$
1. Core Definitions
- $K_p$: The equilibrium constant calculated using the partial pressures of gaseous reactants and products.
- $K_c$: The equilibrium constant calculated using the molar concentrations ($mol/dm^3$) of reactants and products.
2. The Relationship Equation
To convert between $K_p$ and $K_c$, we use the following formula:
$$K_p = K_c(RT)^{\Delta n}$$
Variables Defined:
- $R$: Ideal gas constant ($0.0821 \text{ dm}^3 \cdot \text{atm} / \text{K} \cdot \text{mol}$)
- $T$: Absolute temperature in Kelvin ($T(K) = ^\circ\text{C} + 273$)
- $\Delta n$: Change in moles of gas ($\text{moles of gaseous products} - \text{moles of gaseous reactants}$)
3. Problem Solving Strategy
- Write the Expression: Set up the $K_p$ or $K_c$ expression based on the balanced chemical equation (Products / Reactants).
- Substitute Values: Plug in the given equilibrium partial pressures or concentrations.
- Calculate $\Delta n$: Sum the coefficients of gaseous products and subtract the sum of gaseous reactants.
- Convert: Use the $K_p = K_c(RT)^{\Delta n}$ relationship to find the requested constant.
4. Concept Assessment & Q&A
Question 1: Calculating $K_p$ and $K_c$
Reaction: $2\text{NO}_{(g)} + \text{Cl}_{2(g)} \rightleftharpoons 2\text{NOCl}_{(g)}$ at $25^\circ\text{C}$
Given: $P_{\text{NOCl}} = 1.2 \text{ atm}$, $P_{\text{NO}} = 5.0 \times 10^{-2} \text{ atm}$, $P_{\text{Cl}_2} = 3.0 \times 10^{-1} \text{ atm}$
Answer:
Step 1: Calculate $K_p$
$$K_p = \frac{(P_{\text{NOCl}})^2}{(P_{\text{NO}})^2(P_{\text{Cl}_2})} = \frac{(1.2)^2}{(5.0 \times 10^{-2})^2(3.0 \times 10^{-1})} = 1.9 \times 10^3$$
Step 2: Calculate $\Delta n$
$$\Delta n = 2 - (2 + 1) = -1$$
Step 3: Calculate $K_c$
Using $K_p = K_c(RT)^{\Delta n} \Rightarrow K_c = \frac{K_p}{(RT)^{\Delta n}}$
Since $\Delta n = -1$, $K_c = K_p \times (RT)^1$
$K_c = (1.9 \times 10^3) \times (0.0821 \times 298) = 4.65 \times 10^4$
Question 2: Finding $K_p$ from Concentrations
Reaction: $2\text{SO}_{2(g)} + \text{O}_{2(g)} \rightleftharpoons 2\text{SO}_{3(g)}$ at $450^\circ\text{C}$
Given: $[\text{SO}_2] = 0.59\text{M}$, $[\text{O}_2] = 0.05\text{M}$, $[\text{SO}_3] = 0.259\text{M}$
Answer:
Step 1: Calculate $K_c$ first
$$K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} = \frac{(0.259)^2}{(0.59)^2(0.05)} \approx 3.85$$
Step 2: Solve for $K_p$
$\Delta n = 2 - (2 + 1) = -1$
$T = 450 + 273 = 723\text{K}$
$$K_p = K_c(RT)^{-1} = \frac{3.85}{(0.0821 \times 723)} \approx 0.0649 \text{ atm}^{-1}$$
1. Types of Equilibrium
Chemical equilibrium is classified based on the physical states of the reactants and products:
- Homogeneous Equilibria: All reactants and products are in the same phase (e.g., all gases or all liquids).Example: $$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$$
- Heterogeneous Equilibria: Reactants and products involve more than one phase.Example: $$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$$
Note: In heterogeneous equilibria, the concentrations of pure solids and pure liquids are excluded from the equilibrium constant expression ($$K_c$$and$$K_p$$) because their concentrations remain constant regardless of the amount present.
2. Ways to Recognize Equilibrium
Equilibrium can be identified when the macroscopic properties of a system become constant over time.
a) Physical Method (Spectrometric Method)
Used when a reactant or product absorbs specific light (UV, Visible, or IR). For example, in the reaction:
$$N_2O_{4(g)} \text{ (colourless)} \rightleftharpoons 2NO_{2(g)} \text{ (reddish brown)}$$
The intensity of the reddish-brown color (measured as absorbance) increases as $$NO_2$$ forms. When the absorbance becomes constant, the system has reached equilibrium.
b) Chemical Method
Involves periodic titration to determine concentration. For the reaction between acetic acid and ethanol:
$$CH_3COOH_{(l)} + C_2H_5OH_{(l)} \rightleftharpoons CH_3COOC_2H_{5(l)} + H_2O_{(l)}$$
Samples are withdrawn at regular intervals and titrated against a standard $$NaOH$$ solution. When the concentration of acetic acid remains constant, equilibrium is attained.
3. Practice Exercises & Solutions
Concept Assessment Exercise 8.3
Write $$K_c$$and$$K_p$$ expressions for the following:
| Reaction | Equilibrium Expressions |
|---|---|
| (i) $$FeO_{(s)} + CO_{(g)} \rightleftharpoons Fe_{(s)} + CO_{2(g)}$$ | $$K_c = \frac{[CO_2]}{[CO]}$$$$K_p = \frac{P_{CO_2}}{P_{CO}}$$ |
| (ii) $$P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)}$$ | $$K_c = \frac{1}{[O_2]^5}$$$$K_p = \frac{1}{P_{O_2}^5}$$ |
| (iii) $$CH_{4(g)} + 4Cl_{2(g)} \rightleftharpoons CCl_{4(l)} + 4HCl_{(g)}$$ | $$K_c = \frac{[HCl]^4}{[CH_4][Cl_2]^4}$$$$K_p = \frac{P_{HCl}^4}{P_{CH_4} \cdot P_{Cl_2}^4}$$ |
4. Relevant Questions & Answers
Q1: Why are pure solids omitted from the $$K_c$$ expression?A: The concentration (density) of a pure solid is an intrinsic property and does not change significantly during the reaction, regardless of how much solid is present. Therefore, it is treated as a constant and incorporated into the value of $$K_c$$.
Q2: How does a spectrometer help in recognizing equilibrium?A: A spectrometer measures the absorbance of light by a substance. Since absorbance is proportional to concentration, a constant absorbance reading over time indicates that the concentrations of the species are no longer changing, signifying equilibrium.
Q3: In the titration of acetic acid, what indicates that equilibrium has been reached?A: Equilibrium is reached when the volume of $$NaOH$$ required to neutralize the withdrawn samples remains constant in successive titrations, showing that the concentration of acetic acid is no longer decreasing.
8.6 Factors Affecting Equilibrium (Le Chatelier's Principle)
Le Chatelier's Principle: States that if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.
8.6.1 The Effect of a Change in Concentration
- Adding Reactants: Equilibrium shifts to the right (product side) to consume the added substance.
- Adding Products: Equilibrium shifts to the left (reactant side).
- Removing Reactants: Equilibrium shifts to the left to restore the lost substance.
- Removing Products: Equilibrium shifts to the right to produce more of the removed substance.
Key Example Walkthrough: Decomposition of ICl
Reaction: $$2\text{ICl}_{(g)} \rightleftharpoons \text{Cl}_{2(g)} + \text{I}_{2(g)}$$
Given $$K_c = 1.0 \times 10^{-3}$$ at **230°C**.
Example 8.5 (Addition of Reactant)
If **1 mole** of $$\text{ICl}$$is added to an equilibrium mixture containing 1.6 moles$$\text{ICl}$$, 0.05 mole $$\text{I}_2$$, and 0.05 mole $$\text{Cl}_2$$ in a **2 dm³** container:
- New initial moles of $$\text{ICl} = 1.6 + 1 = 2.6$$ moles.
- Let $$x$$be the change in concentration for products. Due to stoichiometry,$$\text{ICl}$$decreases by$$2x$$.
- Equilibrium concentrations:
- $$[\text{ICl}] = \frac{2.6 - 2x}{2}$$
- $$[\text{I}_2] = [\text{Cl}_2] = \frac{0.05 + x}{2}$$
- Substitute into $$K_c$$expression:$$1.0 \times 10^{-3} = \frac{(\frac{0.05+x}{2})(\frac{0.05+x}{2})}{(\frac{2.6-2x}{2})^2}$$
- Solving for $$x$$gives$$0.029 \text{ moles dm}^{-3}$$.
Concept Assessment Exercise 8.4: Questions & Answers
Q1: Determine the equilibrium concentrations of $$\text{I}_2, \text{Cl}_2$$, and $$\text{ICl}$$when the equilibrium is restored after the removal of one mole of$$\text{ICl}$$.
Answer:
If we remove 1 mole of $$\text{ICl}$$, the system will shift to the left (reactant side) to replace the lost reactant.
| Species | Initial Moles (after removal) | Change ($$x$$) | Equilibrium Concentration ($$M$$) |
|---|---|---|---|
| $$\text{ICl}$$ | $$1.6 - 1.0 = 0.6$$ | $$+2x$$ | $$\frac{0.6 + 2x}{2}$$ |
| $$\text{Cl}_2$$ | $$0.05$$ | $$-x$$ | $$\frac{0.05 - x}{2}$$ |
| $$\text{I}_2$$ | $$0.05$$ | $$-x$$ | $$\frac{0.05 - x}{2}$$ |
Applying the $$K_c$$ formula:
$$1.0 \times 10^{-3} = \frac{(\frac{0.05 - x}{2})^2}{(\frac{0.6 + 2x}{2})^2}$$
Taking the square root of both sides:
$$3.16 \times 10^{-2} = \frac{0.05 - x}{0.6 + 2x}$$
Solving for $$x$$:
$$3.16 \times 10^{-2}(0.6 + 2x) = 0.05 - x$$
$$0.01896 + 0.0632x = 0.05 - x$$
$$1.0632x = 0.03104$$
$$x \approx 0.0292$$
Final Concentrations:
- $$[\text{I}_2] = [\text{Cl}_2] = \frac{0.05 - 0.0292}{2} = 0.0104 \text{ mol/dm}^3 \text{ (Total concentration is } 0.0208M \text{ based on book answer key context)}$$
- $$[\text{ICl}] = \frac{0.6 + 2(0.0292)}{2} = 0.329 \text{ mol/dm}^3 \text{ (Total concentration is } 0.658M \text{ based on book answer key context)}$$
Note: The provided answer in the image (0.0208M, 0.658M) represents the molarity values calculated without the volume division error in the final step.
Q2: Why does an increase in reactant concentration increase the rate of the forward reaction?
According to the collision theory, an increase in concentration means there are more molecules in a given volume. This increases the frequency of effective collisions between reacting molecules, thereby accelerating the forward reaction rate until a new equilibrium is established.
8.6 Chemical Equilibrium and Le Chatelier’s Principle
Le Chatelier's Principle states that if a system at equilibrium is disturbed by changing the conditions, the system shifts its equilibrium position to counteract the disturbance.
1. The Effect of Pressure Change
Pressure changes primarily affect equilibrium systems containing gases. The effect depends on the total number of moles of gas on each side of the balanced equation.
- Increase in Pressure: The system shifts toward the side with the fewer number of gaseous molecules to reduce volume and relieve the pressure.
- Decrease in Pressure: The system shifts toward the side with the larger number of gaseous molecules to increase volume.
- No Change: If the number of moles of gas is equal on both sides (e.g., $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$), changing pressure has no effect on the equilibrium position.
2. The Effect of Temperature Change
To predict shifts, treat "heat" as a reactant or product based on the enthalpy change ($\Delta H$):
- Exothermic Reactions ($\Delta H = -ve$): Heat is a product.
- Increasing temperature shifts the equilibrium to the left (reactants).
- Decreasing temperature shifts the equilibrium to the right (products).
- Endothermic Reactions ($\Delta H = +ve$): Heat is a reactant.
- Increasing temperature shifts the equilibrium to the right (products).
- Decreasing temperature shifts the equilibrium to the left (reactants).
Note: Temperature is the only factor that changes the value of the equilibrium constant ($K_c$).
3. The Effect of Addition of Catalyst
- A catalyst speeds up both the forward and reverse reactions to the same degree.
- It has no effect on the equilibrium concentrations or the value of $K_c$.
- It only helps the system reach equilibrium faster.
Concept Assessment Exercises
Exercise 8.5: Formation of Methanol
Reaction: $CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_3OH_{(g)}$
Question: A student decreases pressure to increase the yield of methanol. Is this approach reasonable?
Answer: No. On the reactant side, there are 3 moles of gas ($1 CO + 2 H_2$), while on the product side, there is only 1 mole ($CH_3OH$). Decreasing pressure shifts the equilibrium to the side with more molecules (the left). Therefore, decreasing pressure will actually decrease the yield of methanol.
Exercise 8.6: Temperature Effects
Q1: $2I_{(g)} \rightleftharpoons I_{2(g)}$ (Bond formation is exothermic). Effect of decreasing temperature?
Answer: Since the forward reaction is exothermic ($2I_{(g)} \rightleftharpoons I_{2(g)} + heat$), decreasing temperature will shift the equilibrium to the right (forward direction) to produce more heat.
Q2: $CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_3OH_{(g)}$ ($\Delta H = -ve$). Effect of increasing temperature on product amount?
Answer: This is an exothermic reaction. Increasing temperature adds "heat" to the product side, causing the equilibrium to shift to the left. The amount of product ($CH_3OH$) will decrease.
8.7 Industrial Application of Le Chatelier's Principle
In industrial chemistry, maximizing product yield while minimizing costs is essential. Le Chatelier's Principle and reaction kinetics are used to determine the optimum conditions (temperature, pressure, and concentration) for large-scale production.
8.7.1 Synthesis of Ammonia by Haber's Process
The chemical equation for the synthesis of ammonia is:
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \quad \Delta H^\circ = -92.46kJ$
Key Factors for Maximum Yield:
- Low Temperature: Since the forward reaction is exothermic ($\Delta H$ is negative), lower temperatures favor the formation of $NH_3$. An optimum temperature of $400^\circ\text{C}$ is typically used to balance yield and speed.
- High Pressure: The reaction proceeds with a decrease in the number of moles (4 moles of reactants produce 2 moles of product). High pressure ($200 - 300 \text{ atm}$) shifts the equilibrium to the right.
- Catalyst: At low temperatures, the reaction is too slow. A piece of iron with other metal oxides is used as a catalyst to increase the reaction rate.
- Continual Removal of Ammonia: Ammonia is removed by liquefying it at $-33.4^\circ\text{C}$. This constant "stress" shifts the equilibrium forward, allowing for nearly $100\%$ conversion.
8.7.2 Synthesis of Sulphuric Acid by Contact Process
The production involves the oxidation of sulphur dioxide:
$2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \quad \Delta H^\circ = -198kJ$
Optimum Conditions:
- Temperature: The reaction is exothermic. A compromised temperature of $450 - 500^\circ\text{C}$ is used to maintain a practical reaction speed.
- Pressure: Fewer moles are produced in the forward reaction, so high pressure favors yield. However, because high pressure is expensive, a compromised pressure of $2 \text{ atm}$ is used.
- Catalyst: Vanadium pentaoxide ($V_2O_5$) is used to speed up the oxidation of $SO_2$ to $SO_3$.
- Concentration: Adding excess $O_2$ shifts the equilibrium to the right, though a $1:1$ ratio of $SO_2$ and $O_2$ is often used to avoid slowing the rate by over-dilution.
Questions & Answers
Q1: Why is a catalyst used in the Haber's process if it doesn't affect the equilibrium position?A: Although a catalyst does not change the yield (equilibrium position), the reaction at the ideal low temperature ($400^\circ\text{C}$) is too slow for industrial use. The catalyst increases the rate of reaction, making the process economically viable.
Q2: In the Contact process, why is a pressure of only $2 \text{ atm}$ used despite Le Chatelier's principle favoring higher pressure?A: While higher pressure increases the yield of $SO_3$, maintaining high-pressure equipment is very expensive. $2 \text{ atm}$ is a "compromised pressure" that provides a sufficient yield at a lower operational cost.
Q3: How does the removal of ammonia affect the equilibrium in Haber's process?A: According to Le Chatelier's principle, removing a product ($NH_3$) creates a stress that the system tries to counteract by producing more of that product. By continually liquefying and removing $NH_3$, the reaction is forced to move in the forward direction.
Q4: What is the enthalpy change ($\Delta H^\circ$) for the synthesis of $SO_3$, and what does it signify?A: The $\Delta H^\circ$ is $-198kJ$. The negative sign signifies that the reaction is exothermic, meaning it releases heat to the surroundings.