Bond Energy Calculations Worksheet 1 | PDF - Free Printable

To calculate the enthalpy change ($\Delta H$) for each reaction using bond energies, we follow these steps:

1. Identify the bonds broken and formed in the reaction.
2. Use the bond energies from the table to calculate the energy required to break the bonds and the energy released when new bonds are formed.
3. Calculate $\Delta H$ using the formula:
\[
\Delta H = \text{(Energy to break bonds)} - \text{(Energy to form bonds)}
\]

Let's solve each reaction step by step.

---

Reaction 1: \( \text{H}_2(g) + \text{F}_2(g) \rightarrow 2 \text{HF}(g) \)

#### Bonds Broken:
- 1 H-H bond: \( 432 \, \text{kJ/mol} \)
- 1 F-F bond: \( 154 \, \text{kJ/mol} \)

#### Bonds Formed:
- 2 H-F bonds: \( 2 \times 565 = 1130 \, \text{kJ/mol} \)

#### Calculate $\Delta H$:
\[
\Delta H = (\text{Energy to break bonds}) - (\text{Energy to form bonds})
\]
\[
\Delta H = (432 + 154) - 1130
\]
\[
\Delta H = 586 - 1130 = -544 \, \text{kJ/mol}
\]

Answer:
\[
\boxed{-544}
\]

---

Reaction 2: \( \text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \)

#### Bonds Broken:
- 4 C-H bonds in CH₄: \( 4 \times 413 = 1652 \, \text{kJ/mol} \)
- 4 O=O bonds in 2 O₂: \( 4 \times 495 = 1980 \, \text{kJ/mol} \)

#### Bonds Formed:
- 2 C=O bonds in CO₂: \( 2 \times 799 = 1598 \, \text{kJ/mol} \)
- 4 O-H bonds in 2 H₂O: \( 4 \times 467 = 1868 \, \text{kJ/mol} \)

#### Calculate $\Delta H$:
\[
\Delta H = (\text{Energy to break bonds}) - (\text{Energy to form bonds})
\]
\[
\Delta H = (1652 + 1980) - (1598 + 1868)
\]
\[
\Delta H = 3632 - 3466 = -166 \, \text{kJ/mol}
\]

Answer:
\[
\boxed{-166}
\]

---

Reaction 3: \( \text{CO}(g) + 2 \text{H}_2(g) \rightarrow \text{CH}_3\text{OH}(l) \)

#### Bonds Broken:
- 1 C≡O bond in CO: \( 1072 \, \text{kJ/mol} \)
- 4 H-H bonds in 2 H₂: \( 4 \times 432 = 1728 \, \text{kJ/mol} \)

#### Bonds Formed:
- 3 C-H bonds in CH₃OH: \( 3 \times 413 = 1239 \, \text{kJ/mol} \)
- 1 C-O bond in CH₃OH: \( 358 \, \text{kJ/mol} \)
- 1 O-H bond in CH₃OH: \( 467 \, \text{kJ/mol} \)

#### Calculate $\Delta H$:
\[
\Delta H = (\text{Energy to break bonds}) - (\text{Energy to form bonds})
\]
\[
\Delta H = (1072 + 1728) - (1239 + 358 + 467)
\]
\[
\Delta H = 2800 - 2064 = 736 \, \text{kJ/mol}
\]

Answer:
\[
\boxed{-736}
\]

---

Reaction 4: \( 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(g) \)

#### Bonds Broken:
- 4 H-H bonds in 2 H₂: \( 4 \times 432 = 1728 \, \text{kJ/mol} \)
- 1 O=O bond in O₂: \( 495 \, \text{kJ/mol} \)

#### Bonds Formed:
- 4 O-H bonds in 2 H₂O: \( 4 \times 467 = 1868 \, \text{kJ/mol} \)

#### Calculate $\Delta H$:
\[
\Delta H = (\text{Energy to break bonds}) - (\text{Energy to form bonds})
\]
\[
\Delta H = (1728 + 495) - 1868
\]
\[
\Delta H = 2223 - 1868 = -355 \, \text{kJ/mol}
\]

Answer:
\[
\boxed{-355}
\]

---

Reaction 5: \( 2 \text{H}_2\text{O}(g) \rightarrow 2 \text{H}_2(g) + \text{O}_2(g) \)

This is the reverse of Reaction 4. The enthalpy change will be the opposite sign but the same magnitude.

Answer:
\[
\boxed{+355}
\]

---

Reaction 6: \( \text{H}_2\text{CCH}_2(g) + \text{Cl}_2(g) \rightarrow \text{CH}_3\text{CCl}_2\text{Cl}(g) \)

#### Bonds Broken:
- 1 C=C bond in H₂CCH₂: \( 614 \, \text{kJ/mol} \)
- 1 Cl-Cl bond in Cl₂: \( 239 \, \text{kJ/mol} \)

#### Bonds Formed:
- 1 C-Cl bond: \( 339 \, \text{kJ/mol} \)
- 1 C-Cl bond: \( 339 \, \text{kJ/mol} \)
- 1 C-H bond: \( 413 \, \text{kJ/mol} \)

#### Calculate $\Delta H$:
\[
\Delta H = (\text{Energy to break bonds}) - (\text{Energy to form bonds})
\]
\[
\Delta H = (614 + 239) - (339 + 339 + 413)
\]
\[
\Delta H = 853 - 1091 = -238 \, \text{kJ/mol}
\]

Answer:
\[
\boxed{-238}
\]

---

Reaction 7: \( \text{CH}_4(g) + \text{H}_2\text{O}(g) \rightarrow \text{CO}(g) + 3 \text{H}_2(g) \)

#### Bonds Broken:
- 4 C-H bonds in CH₄: \( 4 \times 413 = 1652 \, \text{kJ/mol} \)
- 2 O-H bonds in H₂O: \( 2 \times 467 = 934 \, \text{kJ/mol} \)

#### Bonds Formed:
- 1 C≡O bond in CO: \( 1072 \, \text{kJ/mol} \)
- 6 H-H bonds in 3 H₂: \( 6 \times 432 = 2592 \, \text{kJ/mol} \)

#### Calculate $\Delta H$:
\[
\Delta H = (\text{Energy to break bonds}) - (\text{Energy to form bonds})
\]
\[
\Delta H = (1652 + 934) - (1072 + 2592)
\]
\[
\Delta H = 2586 - 3664 = -1078 \, \text{kJ/mol}
\]

Answer:
\[
\boxed{-1078}
\]

---

Reaction 8: \( \text{CH}_3\text{COOH} + \text{CH}_3\text{OH} \rightarrow \text{CH}_3\text{COOCH}_3 + \text{H}_2\text{O} \)

#### Bonds Broken:
- 1 C=O bond in CH₃COOH: \( 799 \, \text{kJ/mol} \)
- 1 O-H bond in CH₃OH: \( 467 \, \text{kJ/mol} \)
- 1 C-O bond in CH₃OH: \( 358 \, \text{kJ/mol} \)

#### Bonds Formed:
- 1 C=O bond in CH₃COOCH₃: \( 799 \, \text{kJ/mol} \)
- 1 O-H bond in H₂O: \( 467 \, \text{kJ/mol} \)
- 1 C-O bond in CH₃COOCH₃: \( 358 \, \text{kJ/mol} \)

#### Calculate $\Delta H$:
\[
\Delta H = (\text{Energy to break bonds}) - (\text{Energy to form bonds})
\]
\[
\Delta H = (799 + 467 + 358) - (799 + 467 + 358)
\]
\[
\Delta H = 1624 - 1624 = 0 \, \text{kJ/mol}
\]

Answer:
\[
\boxed{0}
\]

---

Final Answers:

1. \(\boxed{-544}\)
2. \(\boxed{-166}\)
3. \(\boxed{-736}\)
4. \(\boxed{-355}\)
5. \(\boxed{+355}\)
6. \(\boxed{-238}\)
7. \(\boxed{-1078}\)
8. \(\boxed{0}\)