Thermodynamics Solved Examples (Class 11 Chemistry)

Thermodynamics numericals in chemistry cover enthalpy, entropy, Hess's law, and spontaneity. These problems connect to reaction mechanisms and equilibrium

Thermodynamics Solved Examples (Class 11 Chemistry)

Thermodynamics numericals in chemistry cover enthalpy, entropy, Hess's law, and spontaneity. These problems connect to reaction mechanisms and equilibrium

✓ 100% Free ✓ No Login Needed ✓ NCERT / CBSE Aligned ✓ Download as PDF

TL;DR: Thermodynamics numericals in chemistry cover enthalpy, entropy, Hess's law, and spontaneity. These problems connect to reaction mechanisms and equilib…

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.

Thermodynamics numericals in chemistry cover enthalpy, entropy, Hess's law, and spontaneity. These problems connect to reaction mechanisms and equilibrium

Thermodynamics — Solved Numerical Examples (Step by Step)

Example 1: The enthalpy of combustion of carbon is -394 kJ/mol. Calculate energy released when 6 g of carbon burns. (Atomic mass of C = 12)

Solution: Molar mass of C = 12 g/mol. Moles of C = 6/12 = 0.5 mol. Energy released = 0.5 × 394 = 197 kJ.

Example 2: For reaction A → B, ΔH = -100 kJ/mol. If 2 mol of A reacts, calculate heat released.

Solution: For 1 mol of A, heat released = 100 kJ. For 2 mol of A, heat released = 2 × 100 = 200 kJ.

Example 3: Using Hess's law, find ΔH for: C + O₂ → CO₂. Given: (1) C + 0.5O₂ → CO, ΔH = -111 kJ. (2) CO + 0.5O₂ → CO₂, ΔH = -283 kJ.

Solution: Adding equations (1) and (2): C + 0.5O₂ → CO (ΔH = -111) + CO + 0.5O₂ → CO₂ (ΔH = -283). Result: C + O₂ → CO₂, ΔH = -111 + (-283) = -394 kJ.

Example 4: Calculate the heat of neutralization if 100 mL of 1 M HCl neutralizes 100 mL of 1 M NaOH with temperature rise of 5.76°C. (Assume density = 1 g/mL, specific heat = 4.18 J/g°C)

Solution: Mass of solution = 200 g. Heat absorbed by solution q = m × c × ΔT = 200 × 4.18 × 5.76 = 4815.36 J = 4.815 kJ. Moles of HCl = 0.1 mol. Heat of neutralization = 4.815 / 0.1 = 48.15 kJ/mol.

Example 5: If ΔH = -100 kJ and ΔS = 50 J/K, is the reaction spontaneous at 300 K? (Calculate ΔG)

Solution: Using ΔG = ΔH - T × ΔS. ΔG = -100,000 - 300 × 50 = -100,000 - 15,000 = -115,000 J = -115 kJ. Since ΔG < 0, the reaction is spontaneous.

Tips

  • Hess's law: ΔH of a reaction equals sum of ΔH of its steps, independent of path.
  • ΔG = ΔH - TΔS. Reaction is spontaneous if ΔG < 0.
  • Exothermic reactions (ΔH < 0) release heat; endothermic reactions (ΔH > 0) absorb heat.

Frequently Asked Questions

Why is ΔH negative for exothermic reactions?

Exothermic reactions release energy to surroundings. The system loses energy, so ΔH = Hproducts - Hreactants < 0.

What determines if a reaction is spontaneous?

A reaction is spontaneous if ΔG < 0. Both ΔH and ΔS contribute: ΔG = ΔH - TΔS. Temperature affects spontaneity via the entropy term.

More Chemistry Solved Examples

🤖 Stuck on any of these? Ask Syllab's free AI Tutor to explain step by step →

More free resources for this chapter