Quiz worksheet on calculating bond enthalpy for a reaction between chlorine and ethane.

Let’s solve Question 1 step by step.

We are given a chemical reaction:

Ethane (C₂H₆) + Chlorine (Cl₂) → Chloroethane (C₂H₅Cl) + Hydrogen chloride (HCl)

The structural diagram shows:
- In ethane: 6 C–H bonds and 1 C–C bond
- In Cl₂: 1 Cl–Cl bond
- In chloroethane: 5 C–H bonds, 1 C–C bond, 1 C–Cl bond
- In HCl: 1 H–Cl bond

So, to find the enthalpy change of reaction using bond enthalpies, we use:

ΔH = Σ (bond energies of bonds broken) – Σ (bond energies of bonds formed)

Why? Because breaking bonds requires energy (positive), forming bonds releases energy (negative). So we subtract the energy released from the energy absorbed.

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Step 1: Identify which bonds are broken.

In reactants:
- Ethane has 6 C–H bonds and 1 C–C bond — but wait! Looking at the products, in chloroethane, one H is replaced by Cl. That means one C–H bond is broken, not all six.
Actually, let’s look carefully at what changes between reactants and products.

Reactants:
- Ethane: C₂H₆ → has 6 C–H, 1 C–C
- Cl₂: 1 Cl–Cl

Products:
- Chloroethane: C₂H₅Cl → has 5 C–H, 1 C–C, 1 C–Cl
- HCl: 1 H–Cl

So, comparing:

Bonds broken:
- 1 C–H bond (because we go from 6 to 5 C–H)
- 1 Cl–Cl bond

Bonds formed:
- 1 C–Cl bond
- 1 H–Cl bond

Note: The C–C bond stays the same (still 1 in both sides), so it doesn’t contribute to ΔH.

So:

Energy required to break bonds:
= (1 × C–H) + (1 × Cl–Cl)
= (1 × 413) + (1 × 239)
= 413 + 239 = 652 kJ/mol

Energy released when bonds form:
= (1 × C–Cl) + (1 × H–Cl)
= (1 × 339) + (1 × 427)
= 339 + 427 = 766 kJ/mol

Now,

ΔH = Energy absorbed – Energy released
= 652 – 766
= -114 kJ/mol

That matches one of the options.

Let me double-check:

Broken: C–H (413) + Cl–Cl (239) = 652
Formed: C–Cl (339) + H–Cl (427) = 766
652 - 766 = -114 ✔️

Yes, correct.

Final Answer:
-114 kJ/mol