Chemistry worksheet for calculating enthalpy change using bond energies, including a table of bond energies and reaction equations.

Let’s solve each reaction step by step using bond energies.

Remember:
- To break bonds → you add energy (positive)
- To form bonds → you release energy (negative)
- ΔH = (sum of bond energies broken) – (sum of bond energies formed)

We’ll use the table provided for all values.

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1. H₂(g) + F₂(g) → 2 HF(g)

Bonds broken:
- 1 × H–H = 436.4 kJ/mol
- 1 × F–F = 156.9 kJ/mol
→ Total broken = 436.4 + 156.9 = 593.3 kJ

Bonds formed:
- 2 × H–F = 2 × 568.2 = 1136.4 kJ

ΔH = 593.3 – 1136.4 = –543.1 kJ

✔ Final Answer for #1: –543.1 kJ

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2. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

First, draw out the bonds:

CH₄: 4 C–H bonds
2O₂: 2 O=O bonds
CO₂: 2 C=O bonds (but note: table says “C=O (for CO₂)” is 799 kJ/mol — so we use that!)
2H₂O: each H₂O has 2 O–H bonds → 2 molecules × 2 = 4 O–H bonds

Bonds broken:
- 4 × C–H = 4 × 414 = 1656 kJ
- 2 × O=O = 2 × 498.7 = 997.4 kJ
→ Total broken = 1656 + 997.4 = 2653.4 kJ

Bonds formed:
- 2 × C=O (in CO₂) = 2 × 799 = 1598 kJ
- 4 × O–H = 4 × 460 = 1840 kJ
→ Total formed = 1598 + 1840 = 3438 kJ

ΔH = 2653.4 – 3438 = –784.6 kJ

✔ Final Answer for #2: –784.6 kJ

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3. 2H₂(g) + O₂(g) → 2H₂O(g)

Bonds broken:
- 2 × H–H = 2 × 436.4 = 872.8 kJ
- 1 × O=O = 498.7 kJ
→ Total broken = 872.8 + 498.7 = 1371.5 kJ

Bonds formed:
- Each H₂O has 2 O–H bonds → 2 molecules × 2 = 4 O–H bonds
→ 4 × 460 = 1840 kJ

ΔH = 1371.5 – 1840 = –468.5 kJ

✔ Final Answer for #3: –468.5 kJ

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4. 2H₂O(g) → 2H₂(g) + O₂(g)

This is the reverse of reaction #3!

So if #3 was –468.5 kJ, then this one should be +468.5 kJ

But let’s calculate to check:

Bonds broken:
- 2 H₂O → 4 O–H bonds = 4 × 460 = 1840 kJ

Bonds formed:
- 2 × H–H = 2 × 436.4 = 872.8 kJ
- 1 × O=O = 498.7 kJ
→ Total formed = 872.8 + 498.7 = 1371.5 kJ

ΔH = 1840 – 1371.5 = +468.5 kJ

✔ Final Answer for #4: +468.5 kJ

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5. H₂(g) + Cl₂(g) → 2HCl(g)

Wait — problem! The table does NOT have Cl–Cl or H–Cl bond energies.

Looking back at the table:

Left column: H–H, H–F, C–H, C–O, C≡O, C–C
Right column: F–F, O–H, C=C, C=O, C=O (for CO₂), O=O

There is no Cl–Cl or H–Cl listed.

That means we cannot solve #5 with the given table.

But wait — maybe it’s a trick? Or perhaps the student is expected to know standard values? But the instructions say: “Use the table below...”

Since Cl bonds are not in the table, we can’t compute #5 accurately from given data.

However, sometimes in such worksheets, they might expect you to notice missing info — but since this is likely an error or oversight, and assuming the teacher meant to include them or expects common knowledge...

Actually — looking again — no, the table definitely doesn’t have chlorine bonds.

But let me double-check the original image description — user said “Table 1 Average Bond Energies” and listed those exact bonds. No Cl.

So strictly speaking, #5 cannot be solved with given data.

But since this is probably a mistake, and many textbooks list H–Cl as ~431 kJ/mol and Cl–Cl as ~243 kJ/mol, I’ll assume those are intended even though not in table — otherwise the question is invalid.

Alternatively — maybe the user made a typo and meant F instead of Cl? But reaction says Cl₂ → 2HCl.

Given that, and since this is a worksheet, I think we’re supposed to proceed with known values even if not in table — or perhaps it’s a test to see if students notice?

But the instruction says: “Use the table below...”

Hmm.

Wait — let me re-read the problem: “Calculate the enthalpy change from bond energies for each of these reactions” — and gives table.

If Cl bonds aren't in table, we can't do #5.

But since this is likely an oversight, and to complete the set, I will use standard values commonly accepted:

Assume:
- Cl–Cl = 243 kJ/mol (standard value)
- H–Cl = 431 kJ/mol (standard value)

Then:

Bonds broken:
- 1 × H–H = 436.4
- 1 × Cl–Cl = 243
→ Total = 436.4 + 243 = 679.4 kJ

Bonds formed:
- 2 × H–Cl = 2 × 431 = 862 kJ

ΔH = 679.4 – 862 = –182.6 kJ

But again — these values are NOT in the table. So technically, we shouldn’t use them.

However, since the problem includes this reaction, and no other way, I’ll go with this assumption — noting that it’s outside the given table.

Alternatively — perhaps the teacher forgot to include them, and we’re to leave blank? But the format asks for answer.

Given context, I’ll provide the calculation with assumed values, but flag it.

But per rules — I must solve accurately based on given info.

Since Cl bonds are not in table, #5 cannot be calculated from given data.

But to fulfill the request, and because this is likely an error, I’ll use the standard values and note it.

Final decision: Use standard values for Cl bonds since otherwise incomplete.

✔ Final Answer for #5: –182.6 kJ *(using standard bond energies not in table: Cl–Cl = 243, H–Cl = 431)*

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## Final Answers:

1. –543.1 kJ
2. –784.6 kJ
3. –468.5 kJ
4. +468.5 kJ
5. –182.6 kJ *(note: uses standard bond energies not in provided table)*