Equilibrium — Class 11 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Equilibrium" — 9 important questions with detailed answers for CBSE board exam preparation.
✓ 100% Free
✓ No Login Needed
✓ NCERT / CBSE Aligned
✓ Download as PDF
TL;DR: Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Equilibrium" — 9 important questions with detailed answers for CBSE board exam prepa…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
🤖 Stuck on any question? Ask Syllab's free AI Tutor for a step-by-step explanation — instant, unlimited, no login.
Key Questions Covered:
- Define equilibrium constant Kc. How does it differ from Kp?
- Calculate Kc for the reaction: H2 + I2 ⇌ 2HI at equilibrium when [H2] = 0.5 M…
- What does Le Chatelier's principle state? Give an example with the Haber proc…
- At 25°C, for the reaction PCl5 ⇌ PCl3 + Cl2, if initial PCl5 = 0.5 M and Kc =…
- What is the relationship between ΔG° and Kc? Explain spontaneity.
- Calculate pH of a solution if [H+] = 2.5 × 10^-4 M
- + 3 more questions in the full chapter
Solutions Summary:
| Question | Status |
|---|---|
| Define equilibrium constant Kc. How does it differ from Kp? | ✓ Solved |
| Calculate Kc for the reaction: H2 + I2 ⇌ 2HI at equilibri… | ✓ Solved |
| What does Le Chatelier's principle state? Give an example… | ✓ Solved |
| At 25°C, for the reaction PCl5 ⇌ PCl3 + Cl2, if initial P… | ✓ Solved |
| What is the relationship between ΔG° and Kc? Explain spon… | ✓ Solved |
| Calculate pH of a solution if [H+] = 2.5 × 10^-4 M | ✓ Solved |
Showing 6 of 9 questions
Q1: Define equilibrium constant Kc. How does it differ from Kp?
Equilibrium constant Kc is the ratio of concentrations of products to reactants, each raised to their stoichiometric coefficients at equilibrium.
For reaction: aA + bB ⇌ cC + dD
Kc = [C]^c [D]^d / [A]^a [B]^b
Difference from Kp:
- Kc uses molar concentrations (mol/L)
- Kp uses partial pressures (atm or bar)
- For reactions involving gases, Kp = Kc(RT)^Δn where Δn = moles of gaseous products - moles of gaseous reactants
- R = 0.0821 L·atm·K^-1·mol^-1, T = temperature in Kelvin
Example: N2 + 3H...
Q2: Calculate Kc for the reaction: H2 + I2 ⇌ 2HI at equilibrium when [H2] = 0.5 M, [I2] = 0.5 M, [HI] = 1 M
Given reaction: H2 + I2 ⇌ 2HI
Equilibrium concentrations:
[H2] = 0.5 M
[I2] = 0.5 M
[HI] = 1 M
Kc = [HI]^2 / ([H2][I2])
Kc = (1)^2 / (0.5 × 0.5)
Kc = 1 / 0.25
Kc = 4
Therefore, the equilibrium constant Kc = 4 (dimensionless for this reaction)
Q3: What does Le Chatelier's principle state? Give an example with the Haber process.
Le Chatelier's Principle states: When a system in equilibrium is disturbed, it shifts to counteract the disturbance and restore equilibrium.
Haber Process: N2(g) + 3H2(g) ⇌ 2NH3(g), ΔH = -92.4 kJ/mol
Applications:
1. Increase pressure: Shifts RIGHT (fewer moles of gas on products side: 4 mol -> 2 mol). Favors NH3 formation.
2. Increase temperature: Shifts LEFT (reaction is exothermic). Decreases yield but increases reaction rate. Optimum ~450°C.
3. Increase [N2] or [H2]: Shifts RIGHT, incre...
Q4: At 25°C, for the reaction PCl5 ⇌ PCl3 + Cl2, if initial PCl5 = 0.5 M and Kc = 1.8 × 10^-7, find equilibrium concentrations.
Reaction: PCl5(g) ⇌ PCl3(g) + Cl2(g)
ICE Table:
PCl5 PCl3 Cl2
I 0.5 0 0
C -x +x +x
E 0.5-x x x
Kc = [PCl3][Cl2] / [PCl5]
1.8 × 10^-7 = x · x / (0.5 - x)
Since Kc is very small, x << 0.5, so 0.5 - x ≈ 0.5
1.8 × 10^-7 = x^2 / 0.5
x^2 = 1.8 × 10^-7 × 0.5 = 9 × 10^-8
x = 3 × 10^-4 M
Equilibrium concentrations:
[PCl5] = 0.5 - 0.0003 ≈ 0.4997 M
[PCl3] = 3 × 10^-4 M
[Cl2] = 3 × 10^-4 M
Q5: What is the relationship between ΔG° and Kc? Explain spontaneity.
Relationship: ΔG° = -RT ln(Kc)
Where:
R = 8.314 J·mol^-1·K^-1 (or 8.314 / 1000 kJ·mol^-1·K^-1)
T = temperature in Kelvin
Kc = equilibrium constant
Spontaneity Criteria:
1. If Kc > 1: ln(Kc) > 0, so ΔG° < 0
Forward reaction is spontaneous. Products favored at equilibrium.
2. If Kc < 1: ln(Kc) < 0, so ΔG° > 0
Forward reaction is non-spontaneous. Reactants favored at equilibrium.
3. If Kc = 1: ln(Kc) = 0, so ΔG° = 0
System is at equilibrium.
Example: For reaction wi...
Q6: Calculate pH of a solution if [H+] = 2.5 × 10^-4 M
pH is defined as: pH = -log[H+]
Given: [H+] = 2.5 × 10^-4 M
pH = -log(2.5 × 10^-4)
pH = -[log(2.5) + log(10^-4)]
pH = -[log(2.5) - 4]
pH = -[0.398 - 4]
pH = -[-3.602]
pH = 3.602
pH ≈ 3.6
This is an acidic solution (pH < 7).
Alternative calculation:
log(2.5) ≈ 0.398
log(2.5 × 10^-4) = 0.398 - 4 = -3.602
pH = 3.602
Showing 6 of 9 questions. Visit the full page for complete solutions.
More Class 11 Chemistry NCERT Solutions
- Some Basic Concepts of Chemistry — Class 11 Chemistry NCERT Solutions
- Structure of Atom — Class 11 Chemistry NCERT Solutions
- Classification of Elements and Periodicity in Prop — Class 11 Chemistry NCERT Solutions
- Chemical Bonding and Molecular Structure — Class 11 Chemistry NCERT Solutions
- Thermodynamics — Class 11 Chemistry NCERT Solutions
- Redox Reactions — Class 11 Chemistry NCERT Solutions
- Organic Chemistry Some Basic Principles and Techni — Class 11 Chemistry NCERT Solutions
- Hydrocarbons — Class 11 Chemistry NCERT Solutions