Termokimia (5) | Menentukan Perubahan Entalpi berdasarkan data Energi Ikatan | Kimia Kelas 11

Main Ideas / Concepts Taught

• Bond energy (energi ikatan) is defined as:

  • The amount of energy/heat required to break 1 mole of a bond between atoms in the gaseous state.
  • If you break a bond: energy is absorbed, so it effectively counts as positive in enthalpy accounting.
  • If you form (connect) a bond: energy is released, so it contributes as negative in enthalpy accounting.

• Enthalpy change (ΔH) using bond energies:

  • The method computes:
    • ΔH = (sum of energies of broken bonds) − (sum of energies of formed bonds)
  • Broken bonds are those on the left (reactants) side of the equation.
  • Formed bonds are those on the right (products) side.

• Structural formula construction rules (for solving examples):

  • C forms 4 bonds
  • H forms 1 bond
  • O forms 2 bonds
  • Group 7A elements (e.g., F, Cl, Br, I) form 1 bond

• Counting bonds carefully:

  • Count how many of each bond type appears on each side, such as C–H, C–Cl, H–Cl, O–H, etc.
  • If bond counts are the same on both sides, they can be simplified by subtraction before substituting values.

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Methodology / Step-by-Step Instructions Shown

A) General procedure to determine ΔH using bond energies

1. Write the reaction clearly.
2. Draw/construct structural formulas for all reactants and products.
3. Count bonds on the reactant side (left):
   • Identify each type of bond (e.g., C–H, C–Cl).
   • Sum the corresponding bond energies → this becomes the broken bonds sum.
4. Count bonds on the product side (right):
   • Identify and sum bond energies → this becomes the formed bonds sum.
5. Apply the formula:
   • ΔH = Σ(bond energies left) − Σ(bond energies right)
6. Interpret the sign:
   • Negative ΔH → heat is released (exothermic).

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B) Example 1: CH₄ + Cl₂ → CH₃Cl + HCl

Given bond energies (kJ/mol)

• C–H = 414
• Cl–Cl = 244
• H–Cl = 432
• C–Cl = 326

Bond counting concept

• Left (reactants):
  • CH₄ contains 4 C–H bonds
  • Cl₂ contains 1 Cl–Cl bond
• Right (products):
  • CH₃Cl contains 3 C–H bonds and 1 C–Cl bond
  • HCl contains 1 H–Cl bond
• Simplification mentioned:
  • The difference in C–H bonds is treated as 4 − 3 = 1 C–H effectively remaining to be counted.

Compute ΔH (as carried out)

• Use: ΔH = (C–H + Cl–Cl) − (C–Cl + H–Cl) after simplification
• Result:
  • ΔH = 658 − 758 = −100 kJ/mol

Conclusion

• 100 kJ/mol of heat is released, because ΔH is negative.

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C) Example 2: Using enthalpy of formation of H₂O

Given

• ΔHf(H₂O) = −242.5 kJ/mol
• Bond energies:
  • H–H = 436 kJ/mol
  • O–O = 495 kJ/mol
• Goal (subtitle unclear/garbled in the source), but the computation targets:
  • Bond energy of O–H (for 1 mole of O–H bonds)

Reaction context

• Formation of water from elements:
  • H₂ + ½ O₂ → H₂O (balanced via coefficients)

Bond counting idea

• Left:
  • H₂ contributes H–H
  • O₂ contributes O–O, but because of the ½ coefficient, the O–O term is multiplied by ½
• Right:
  • H₂O contains 2 O–H bonds

Key algebra/sign handling

• Start from:
  • ΔH (formation) = (energies to break reactant bonds) − (energies to form product bonds)
• The video rearranges the equation to solve for the unknown bond energy.

Computation shown

• The total for 2(O–H) bonds is found to be:
  • 2(O–H) = 926 kJ/mol
• Therefore, for 1 mole of O–H bonds:
  • O–H = 926 / 2 = 463 kJ/mol

Conclusion

• O–H bond energy = 463 kJ/mol.

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Speakers / Sources Featured

• Brother (the main presenter/teacher)
• “Hello little brothers and sisters” (same presenter addressing the audience)
• No other named speakers or external sources are featured.