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

Let’s solve each reaction one by one using the bond energy method.

The formula we use is:

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

Bonds broken → require energy (positive)
Bonds formed → release energy (negative)

We’ll go step by step for each reaction.

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1. H₂(g) + F₂(g) → 2 HF(g)

Break:
- 1 H–H bond = 432 kJ/mol
- 1 F–F bond = 154 kJ/mol
Total broken = 432 + 154 = 586 kJ

Form:
- 2 H–F bonds = 2 × 565 = 1130 kJ

ΔH = 586 – 1130 = –544 kJ

✔ Final Answer for #1: –544 kJ

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2. CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)

First, draw or imagine structures:

CH₄ has 4 C–H bonds
O₂ has O=O double bond (we have 2 molecules → 2 O=O bonds)
CO₂ has 2 C=O bonds (but note: table says “C=O (for CO₂)” = 799 kJ/mol — so each C=O in CO₂ is 799, and there are two → 2×799)
Each H₂O has 2 O–H bonds → 2 H₂O = 4 O–H bonds

Break:
- 4 C–H = 4 × 413 = 1652
- 2 O=O = 2 × 495 = 990
Total broken = 1652 + 990 = 2642 kJ

Form:
- 2 C=O (in CO₂) = 2 × 799 = 1598
- 4 O–H = 4 × 467 = 1868
Total formed = 1598 + 1868 = 3466 kJ

ΔH = 2642 – 3466 = –824 kJ

✔ Final Answer for #2: –824 kJ

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3. CO(g) + 2 H₂(g) → CH₃OH(l)

Note: CH₃OH is liquid, but bond energies are for gas phase. We’ll proceed as if all are gases unless told otherwise — but since product is liquid, this might be approximate. However, problem doesn’t give extra data, so we’ll ignore state change enthalpy and just use bond energies.

Structure:

CO(g): triple bond? But table gives C≡O = 1072 kJ/mol → that’s for carbon monoxide.

H₂: H–H bond

CH₃OH: has 3 C–H, 1 C–O, 1 O–H

Break:
- 1 C≡O = 1072
- 2 H–H = 2 × 432 = 864
Total broken = 1072 + 864 = 1936 kJ

Form:
In CH₃OH:
- 3 C–H = 3 × 413 = 1239
- 1 C–O = 358
- 1 O–H = 467
Total formed = 1239 + 358 + 467 = 2064 kJ

ΔH = 1936 – 2064 = –128 kJ

✔ Final Answer for #3: –128 kJ

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4. 2 H₂(g) + O₂(g) → 2 H₂O(g)

Break:
- 2 H–H = 2 × 432 = 864
- 1 O=O = 495
Total broken = 864 + 495 = 1359 kJ

Form:
Each H₂O has 2 O–H → 2 H₂O = 4 O–H = 4 × 467 = 1868 kJ

ΔH = 1359 – 1868 = –509 kJ

✔ Final Answer for #4: –509 kJ

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5. 2 H₂O(g) → 2 H₂(g) + O₂(g)

This is reverse of #4!

So ΔH should be opposite sign: +509 kJ

But let’s verify:

Break:
2 H₂O → break 4 O–H bonds = 4 × 467 = 1868 kJ

Form:
2 H–H = 2 × 432 = 864
1 O=O = 495
Total formed = 864 + 495 = 1359 kJ

ΔH = 1868 – 1359 = +509 kJ

✔ Final Answer for #5: +509 kJ

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6. H₂CCH₂(g) + Cl₂(g) → ClH₂CCH₂Cl(g)

H₂CCH₂ is ethene → has 1 C=C and 4 C–H bonds
Cl₂ has 1 Cl–Cl bond
Product: ClH₂C–CH₂Cl → now it’s single bond between carbons, each carbon has 2 H and 1 Cl → so bonds: 1 C–C, 4 C–H, 2 C–Cl

Break:
- 1 C=C = 614
- 4 C–H = 4 × 413 = 1652
- 1 Cl–Cl = 239
Total broken = 614 + 1652 + 239 = 2505 kJ

Form:
- 1 C–C = 347
- 4 C–H = 4 × 413 = 1652 (same number, but they’re reformed — still count them)
Wait — actually, in reactant, we had 4 C–H; in product, also 4 C–H — so net no change? But in bond energy calculation, we break ALL bonds in reactants and form ALL bonds in products.

So yes, we break 4 C–H and form 4 C–H — they cancel out numerically, but we still include them.

Actually, better to think: total bonds broken vs total bonds formed.

Reactants:
Ethene: 1 C=C, 4 C–H
Cl₂: 1 Cl–Cl
→ Total broken: C=C, 4 C–H, Cl–Cl

Products:
ClCH₂–CH₂Cl: 1 C–C, 4 C–H, 2 C–Cl
→ Total formed: C–C, 4 C–H, 2 C–Cl

So:

Broken:
C=C = 614
4 C–H = 1652
Cl–Cl = 239
Sum = 614 + 1652 + 239 = 2505

Formed:
C–C = 347
4 C–H = 1652
2 C–Cl = 2 × 339 = 678
Sum = 347 + 1652 + 678 = 2677

ΔH = 2505 – 2677 = –172 kJ

✔ Final Answer for #6: –172 kJ

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7. CH₄(g) + H₂O(g) → CO(g) + 3 H₂(g)

Break:
CH₄: 4 C–H = 4 × 413 = 1652
H₂O: 2 O–H = 2 × 467 = 934
Total broken = 1652 + 934 = 2586 kJ

Form:
CO: 1 C≡O = 1072
3 H₂: 3 H–H = 3 × 432 = 1296
Total formed = 1072 + 1296 = 2368 kJ

ΔH = 2586 – 2368 = +218 kJ

✔ Final Answer for #7: +218 kJ

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8. CH₃COOH + CH₃OH → CH₃COOCH₃ + H₂O

This is esterification.

Let’s write structural formulas to count bonds.

Acetic acid (CH₃COOH):
- CH₃– : 3 C–H
- –COOH: 1 C=O, 1 C–O, 1 O–H
Also, the central carbon is bonded to CH₃ and to OH and =O → so bonds:
Actually, standard structure:
Carbon 1 (methyl): 3 C–H, 1 C–C
Carbon 2 (carbonyl): 1 C=O, 1 C–O (to OH), and connected to methyl → so already counted C–C
Plus O–H in carboxylic acid.

Better to list all bonds in CH₃COOH:

- 3 C–H (from CH₃)
- 1 C–C (between CH₃ and COOH)
- 1 C=O
- 1 C–O
- 1 O–H

Methanol (CH₃OH):
- 3 C–H
- 1 C–O
- 1 O–H

Products:

Methyl acetate (CH₃COOCH₃):
Structure: CH₃–C(=O)–O–CH₃
Bonds:
- Left CH₃: 3 C–H, 1 C–C
- Carbonyl: 1 C=O
- Ester oxygen: 1 C–O (to left C), 1 C–O (to right CH₃)
- Right CH₃: 3 C–H, 1 C–O (already counted?) Wait — let's list unique bonds:

Actually, total bonds in CH₃COOCH₃:

- 6 C–H (two CH₃ groups)
- 2 C–C? No — only one C–C (between first CH₃ and carbonyl C)
Wait: atoms: C1 (methyl), C2 (carbonyl), O (ester), C3 (methyl)

Bonds:
C1–H x3
C1–C2
C2=O
C2–O
O–C3
C3–H x3

So:
- 6 C–H
- 1 C–C
- 1 C=O
- 2 C–O (one from C2–O, one from O–C3) — but both are C–O single bonds

Water (H₂O): 2 O–H

Now, let’s tabulate bonds broken and formed.

Reactants:

CH₃COOH:
- 3 C–H
- 1 C–C
- 1 C=O
- 1 C–O
- 1 O–H

CH₃OH:
- 3 C–H
- 1 C–O
- 1 O–H

Total broken:
C–H: 3 + 3 = 6
C–C: 1
C=O: 1
C–O: 1 (from acid) + 1 (from alcohol) = 2
O–H: 1 (acid) + 1 (alcohol) = 2

Products:

CH₃COOCH₃:
- 6 C–H
- 1 C–C
- 1 C=O
- 2 C–O (as explained)

H₂O:
- 2 O–H

Total formed:
C–H: 6
C–C: 1
C=O: 1
C–O: 2
O–H: 2

Wait — same bonds? That can’t be right. In reality, during esterification, we break the O–H from acid and the C–O from alcohol? Or rather, we break the C–O in acid and O–H in alcohol? Actually, mechanism involves breaking O–H of acid and C–O of alcohol? But in terms of net bond changes:

Actually, comparing reactants and products:

In reactants:
Acid has: C–O and O–H (carboxylic)
Alcohol has: C–O and O–H

In products:
Ester has: two C–O bonds (one from original acid’s C–O? Not exactly)
Water has: O–H bonds

Net change:
We lose one O–H from acid and one O–H from alcohol? But water has two O–H — so we form two new O–H? That doesn't make sense.

Better approach: identify which bonds are broken and which are formed specifically in the reaction.

Standard way for esterification:

Break:
- The O–H bond in carboxylic acid
- The C–O bond in alcohol (methanol) — because the OH from acid and H from alcohol form water, and the rest join.

Actually, more accurately:

In CH₃COOH, we break the C–OH bond (i.e., the C–O bond to the hydroxyl group)
In CH₃OH, we break the O–H bond
Then we form:
- A new C–O bond between carbonyl carbon and methoxy oxygen
- And form H₂O from the OH and H

But in bond energy terms, we consider all bonds.

Since the molecular formulas are the same on both sides? Let’s check atom count.

Reactants:
CH₃COOH = C₂H₄O₂
CH₃OH = CH₄O
Total: C₃H₈O₃

Products:
CH₃COOCH₃ = C₃H₆O₂
H₂O = H₂O
Total: C₃H₈O₃ — same atoms.

So, the difference is in bonding.

Specifically, in reactants:

- Carboxylic acid has a C=O, a C–O, and an O–H
- Alcohol has a C–O and an O–H

In products:

- Ester has a C=O, and two C–O bonds (no O–H)
- Water has two O–H bonds

So, compared to reactants, we have:

Lost:
- One O–H from acid
- One O–H from alcohol
Gained:
- Two O–H in water
But also, the C–O in acid and C–O in alcohol are replaced by two C–O in ester? Not exactly.

Actually, let’s count bond types:

Define:

In CH₃COOH:
Bonds:
- 3 C–H (methyl)
- 1 C–C
- 1 C=O
- 1 C–O (single, to OH)
- 1 O–H

In CH₃OH:
- 3 C–H
- 1 C–O
- 1 O–H

Total reactant bonds:
C–H: 6
C–C: 1
C=O: 1
C–O: 2 (one from acid, one from alcohol)
O–H: 2

Products:

CH₃COOCH₃:
As above:
- 6 C–H
- 1 C–C
- 1 C=O
- 2 C–O (both single bonds: one from carbonyl C to O, one from O to CH₃)

H₂O:
- 2 O–H

So total product bonds:
C–H: 6
C–C: 1
C=O: 1
C–O: 2
O–H: 2

Exactly the same as reactants! So ΔH = 0?

That can’t be right physically, but according to bond energies given, if the bond types and counts are identical, then yes.

But wait — in carboxylic acid, the C–O bond is part of a carboxyl group, and in ester, the C–O bonds are different? But the table doesn’t distinguish — it just gives "C–O" = 358 kJ/mol for any C–O single bond.

Similarly, O–H is 467 regardless.

So numerically, bonds broken and formed are identical.

Therefore, ΔH = 0 kJ?

But that seems odd. Let me double-check.

Perhaps I missed something.

In acetic acid, the carbon in COOH is bonded to:
- Carbon of CH₃ (C–C)
- Oxygen via double bond (C=O)
- Oxygen via single bond (C–O)
That oxygen is bonded to H (O–H)

In methanol:
Carbon bonded to three H and one O (C–O), that O bonded to H (O–H)

In methyl acetate:
Carbonyl carbon bonded to:
- CH₃ (C–C)
- O double bond (C=O)
- O single bond (C–O)
That O is bonded to CH₃ (so another C–O bond)

And the CH₃ of methanol is now attached via O, so its C–O bond is still there, but now it's C–O–C instead of C–O–H.

But in terms of bond types, we still have:

Same number of each bond type.

So yes, ΔH = 0 kJ based on given bond energies.

But let's calculate explicitly.

Break all bonds in reactants:

CH₃COOH:
3 C–H = 3*413 = 1239
1 C–C = 347
1 C=O = 745? Wait — here’s the issue!

In the table, there are two entries for C=O:
- C=O = 745 kJ/mol (probably for ketones/aldehydes)
- C=O (for CO₂) = 799

But for carboxylic acids and esters, what value to use?

Typically, in such problems, if not specified, we use the general C=O = 745 for organic compounds except CO₂.

In reaction #2, we used 799 for CO₂, which was specified.

For acetic acid and ester, we should use C=O = 745 kJ/mol.

Similarly, in CO(g), we used C≡O = 1072.

So for #8:

Reactants:

CH₃COOH:
- 3 C–H = 3*413 = 1239
- 1 C–C = 347
- 1 C=O = 745 (assuming general carbonyl)
- 1 C–O = 358
- 1 O–H = 467
Sum for acid = 1239+347+745+358+467 = let's compute:
1239+347=1586; +745=2331; +358=2689; +467=3156

CH₃OH:
- 3 C–H = 1239
- 1 C–O = 358
- 1 O–H = 467
Sum = 1239+358+467=2064

Total broken = 3156 + 2064 = 5220 kJ

Products:

CH₃COOCH₃:
- 6 C–H = 6*413 = 2478
- 1 C–C = 347
- 1 C=O = 745
- 2 C–O = 2*358 = 716
Sum = 2478+347=2825; +745=3570; +716=4286

H₂O:
- 2 O–H = 2*467 = 934

Total formed = 4286 + 934 = 5220 kJ

ΔH = 5220 – 5220 = 0 kJ

So indeed, with the given bond energies, it comes out to zero.

This makes sense because the bond types and quantities are conserved.

✔ Final Answer for #8: 0 kJ

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## Final Answers Summary:

1. –544 kJ
2. –824 kJ
3. –128 kJ
4. –509 kJ
5. +509 kJ
6. –172 kJ
7. +218 kJ
8. 0 kJ

Final Answer:
1. -544 kJ
2. -824 kJ
3. -128 kJ
4. -509 kJ
5. +509 kJ
6. -172 kJ
7. +218 kJ
8. 0 kJ