Worksheet for calculating enthalpy changes using standard enthalpies of formation.

To solve this problem, we will use the standard enthalpy of formation (ΔH°f) values provided in the table to calculate the enthalpy change (ΔH°rxn) for each reaction using the following formula:

$$
\Delta H^\circ_{\text{rxn}} = \sum \Delta H^\circ_f (\text{products}) - \sum \Delta H^\circ_f (\text{reactants})
$$

We’ll go through each reaction one by one.

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a) NaOH(s) + HCl(g) → NaCl(s) + H₂O(l)

Step 1: Identify ΔH°f values from the table

- NaOH(s): -426.7 kJ/mol
- HCl(g): -92.3 kJ/mol
- NaCl(s): -411.0 kJ/mol
- H₂O(l): -285.8 kJ/mol

Step 2: Apply the formula

$$
\Delta H^\circ_{\text{rxn}} = [\Delta H^\circ_f(\text{NaCl}) + \Delta H^\circ_f(\text{H}_2\text{O})] - [\Delta H^\circ_f(\text{NaOH}) + \Delta H^\circ_f(\text{HCl})]
$$

$$
= [(-411.0) + (-285.8)] - [(-426.7) + (-92.3)]
$$

$$
= (-696.8) - (-519.0) = -696.8 + 519.0 = -177.8 \text{ kJ/mol}
$$

✔ Answer: ΔH° = -177.8 kJ/mol

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b) 2 CO(g) + O₂(g) → 2 CO₂(g)

Step 1: Find ΔH°f values

- CO(g): -110.53 kJ/mol
- O₂(g): 0 kJ/mol (element in standard state)
- CO₂(g): -393.5 kJ/mol

Step 2: Apply formula

$$
\Delta H^\circ_{\text{rxn}} = [2 \times \Delta H^\circ_f(\text{CO}_2)] - [2 \times \Delta H^\circ_f(\text{CO}) + \Delta H^\circ_f(\text{O}_2)]
$$

$$
= [2 \times (-393.5)] - [2 \times (-110.53) + 0]
$$

$$
= (-787.0) - (-221.06) = -787.0 + 221.06 = -565.94 \text{ kJ/mol}
$$

✔ Answer: ΔH° = -565.94 kJ/mol

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c) CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l)

Step 1: Find ΔH°f values

- CH₄(g): -74.8 kJ/mol
- O₂(g): 0 kJ/mol
- CO₂(g): -393.5 kJ/mol
- H₂O(l): -285.8 kJ/mol

Step 2: Apply formula

$$
\Delta H^\circ_{\text{rxn}} = [\Delta H^\circ_f(\text{CO}_2) + 2 \times \Delta H^\circ_f(\text{H}_2\text{O})] - [\Delta H^\circ_f(\text{CH}_4) + 2 \times \Delta H^\circ_f(\text{O}_2)]
$$

$$
= [(-393.5) + 2 \times (-285.8)] - [(-74.8) + 2 \times 0]
$$

$$
= (-393.5 - 571.6) - (-74.8) = (-965.1) + 74.8 = -890.3 \text{ kJ/mol}
$$

✔ Answer: ΔH° = -890.3 kJ/mol

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d) 2 H₂S(g) + 3 O₂(g) → 2 H₂O(l) + 2 SO₂(g)

Step 1: Find ΔH°f values

- H₂S(g): -20.1 kJ/mol
- O₂(g): 0 kJ/mol
- H₂O(l): -285.8 kJ/mol
- SO₂(g): -296.1 kJ/mol

Step 2: Apply formula

$$
\Delta H^\circ_{\text{rxn}} = [2 \times \Delta H^\circ_f(\text{H}_2\text{O}) + 2 \times \Delta H^\circ_f(\text{SO}_2)] - [2 \times \Delta H^\circ_f(\text{H}_2\text{S}) + 3 \times \Delta H^\circ_f(\text{O}_2)]
$$

$$
= [2 \times (-285.8) + 2 \times (-296.1)] - [2 \times (-20.1) + 0]
$$

$$
= [-571.6 - 592.2] - [-40.2] = (-1163.8) + 40.2 = -1123.6 \text{ kJ/mol}
$$

✔ Answer: ΔH° = -1123.6 kJ/mol

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e) 2 NO(g) + O₂(g) → 2 NO₂(g)

Step 1: Find ΔH°f values

- NO(g): +90.4 kJ/mol
- O₂(g): 0 kJ/mol
- NO₂(g): +33.9 kJ/mol

Step 2: Apply formula

$$
\Delta H^\circ_{\text{rxn}} = [2 \times \Delta H^\circ_f(\text{NO}_2)] - [2 \times \Delta H^\circ_f(\text{NO}) + \Delta H^\circ_f(\text{O}_2)]
$$

$$
= [2 \times 33.9] - [2 \times 90.4 + 0]
$$

$$
= 67.8 - 180.8 = -113.0 \text{ kJ/mol}
$$

✔ Answer: ΔH° = -113.0 kJ/mol

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✔ Final Answers Summary:

| Reaction | ΔH° (kJ/mol) |
|--------|----------------|
| a) NaOH(s) + HCl(g) → NaCl(s) + H₂O(l) | -177.8 |
| b) 2 CO(g) + O₂(g) → 2 CO₂(g) | -565.94 |
| c) CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) | -890.3 |
| d) 2 H₂S(g) + 3 O₂(g) → 2 H₂O(l) + 2 SO₂(g) | -1123.6 |
| e) 2 NO(g) + O₂(g) → 2 NO₂(g) | -113.0 |

All reactions are exothermic (negative ΔH), meaning they release heat.

Let me know if you'd like a visual explanation or help interpreting these results!