53 1 Calculating Enthalpy Change from Bond | StudyX - Free Printable

To calculate the enthalpy change ($\Delta H$) for each reaction using bond energies, we use the following formula:

\[
\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}
\]

We will go through each reaction step by step.

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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: \( 154 \, \text{kJ/mol} \)

Total energy to break bonds:
\[
432 + 154 = 586 \, \text{kJ/mol}
\]

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

Total energy to form bonds:
\[
1130 \, \text{kJ/mol}
\]

#### Enthalpy Change:
\[
\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}
\]
\[
\Delta H = 586 - 1130 = -544 \, \text{kJ/mol}
\]

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

---

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 molecules: \( 4 \times 495 = 1980 \, \text{kJ/mol} \)

Total energy to break bonds:
\[
1652 + 1980 = 3632 \, \text{kJ/mol}
\]

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

Total energy to form bonds:
\[
1598 + 1868 = 3466 \, \text{kJ/mol}
\]

#### Enthalpy Change:
\[
\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}
\]
\[
\Delta H = 3632 - 3466 = 166 \, \text{kJ/mol}
\]

Answer:
\[
\boxed{166}
\]

---

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 molecules: \( 4 \times 432 = 1728 \, \text{kJ/mol} \)

Total energy to break bonds:
\[
1072 + 1728 = 2800 \, \text{kJ/mol}
\]

#### 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: \( 467 \, \text{kJ/mol} \)

Total energy to form bonds:
\[
1239 + 358 + 467 = 2064 \, \text{kJ/mol}
\]

#### Enthalpy Change:
\[
\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}
\]
\[
\Delta H = 2800 - 2064 = 736 \, \text{kJ/mol}
\]

Answer:
\[
\boxed{736}
\]

---

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 molecules: \( 4 \times 432 = 1728 \, \text{kJ/mol} \)
- 1 O=O bond in O₂: \( 495 \, \text{kJ/mol} \)

Total energy to break bonds:
\[
1728 + 495 = 2223 \, \text{kJ/mol}
\]

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

Total energy to form bonds:
\[
1868 \, \text{kJ/mol}
\]

#### Enthalpy Change:
\[
\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}
\]
\[
\Delta H = 2223 - 1868 = 355 \, \text{kJ/mol}
\]

Answer:
\[
\boxed{355}
\]

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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 negative of reaction 4.

\[
\Delta H = -355 \, \text{kJ/mol}
\]

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

---

6. \( \text{H}_2\text{CCH}_2(g) + \text{Cl}_2(g) \rightarrow \text{ClH}_2\text{CCH}_2\text{Cl}(g) \)

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

Total energy to break bonds:
\[
614 + 239 = 853 \, \text{kJ/mol}
\]

#### Bonds Formed:
- 2 C-Cl bonds in ClH₂CCH₂Cl: \( 2 \times 339 = 678 \, \text{kJ/mol} \)
- 1 C-C bond in ClH₂CCH₂Cl: \( 347 \, \text{kJ/mol} \)

Total energy to form bonds:
\[
678 + 347 = 1025 \, \text{kJ/mol}
\]

#### Enthalpy Change:
\[
\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}
\]
\[
\Delta H = 853 - 1025 = -172 \, \text{kJ/mol}
\]

Answer:
\[
\boxed{-172}
\]

---

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: \( 2 \times 467 = 934 \, \text{kJ/mol} \)

Total energy to break bonds:
\[
1652 + 934 = 2586 \, \text{kJ/mol}
\]

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

Total energy to form bonds:
\[
1072 + 2592 = 3664 \, \text{kJ/mol}
\]

#### Enthalpy Change:
\[
\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}
\]
\[
\Delta H = 2586 - 3664 = -1078 \, \text{kJ/mol}
\]

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

---

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: \( 745 \, \text{kJ/mol} \)
- 1 O-H bond in CH₃OH: \( 467 \, \text{kJ/mol} \)
- 1 C-O bond in CH₃COOH: \( 358 \, \text{kJ/mol} \)

Total energy to break bonds:
\[
745 + 467 + 358 = 1570 \, \text{kJ/mol}
\]

#### Bonds Formed:
- 1 C=O bond in CH₃COOCH₃: \( 745 \, \text{kJ/mol} \)
- 1 O-H bond in H₂O: \( 467 \, \text{kJ/mol} \)
- 1 C-O bond in CH₃COOCH₃: \( 358 \, \text{kJ/mol} \)

Total energy to form bonds:
\[
745 + 467 + 358 = 1570 \, \text{kJ/mol}
\]

#### Enthalpy Change:
\[
\Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed}
\]
\[
\Delta H = 1570 - 1570 = 0 \, \text{kJ/mol}
\]

Answer:
\[
\boxed{0}
\]

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Final Answers:

1. \(\boxed{-544}\)
2. \(\boxed{166}\)
3. \(\boxed{736}\)
4. \(\boxed{355}\)
5. \(\boxed{-355}\)
6. \(\boxed{-172}\)
7. \(\boxed{-1078}\)
8. \(\boxed{0}\)