Determining enthalpy changes for chemical reactions using standard enthalpies of formation.

To find the change in enthalpy ($\Delta H$) for each reaction, we use the standard enthalpies of formation ($\Delta H_f^\circ$) provided in the table.

The Formula:
$$ \Delta H_{\text{reaction}} = \sum (\Delta H_f^\circ \text{ products}) - \sum (\Delta H_f^\circ \text{ reactants}) $$

This means: (Sum of Products) minus (Sum of Reactants). Remember to multiply the value by the coefficient (the big number in front) if there is one.

Here are the values from the table we will need:
* $\text{NaOH}(s)$: $-426.7$
* $\text{HCl}(g)$: $-92.3$
* $\text{NaCl}(s)$: $-411.0$
* $\text{H}_2\text{O}(g)$: $-241.8$
* $\text{CO}(g)$: $-110.53$
* $\text{O}_2(g)$: $0$ (Elements in their natural state are always 0)
* $\text{CO}_2(g)$: $-393.5$
* $\text{CH}_4(g)$: $-74.8$
* $\text{H}_2\text{O}(l)$: $-285.8$
* $\text{H}_2\text{S}(g)$: $-20.1$
* $\text{SO}_2(g)$: $-296.1$
* $\text{NO}(g)$: $+90.4$
* $\text{NO}_2(g)$: $+33.9$

---

a) $\text{NaOH}(s) + \text{HCl}(g) \rightarrow \text{NaCl}(s) + \text{H}_2\text{O}(g)$

1. Products:
* $\text{NaCl}(s)$: $1 \times (-411.0) = -411.0$
* $\text{H}_2\text{O}(g)$: $1 \times (-241.8) = -241.8$
* Sum of Products = $-411.0 + (-241.8) = -652.8$

2. Reactants:
* $\text{NaOH}(s)$: $1 \times (-426.7) = -426.7$
* $\text{HCl}(g)$: $1 \times (-92.3) = -92.3$
* Sum of Reactants = $-426.7 + (-92.3) = -519.0$

3. Calculate $\Delta H$:
* $\Delta H = (-652.8) - (-519.0)$
* $\Delta H = -652.8 + 519.0$
* $\Delta H = -133.8 \text{ kJ}$

---

b) $2 \text{CO}(g) + \text{O}_2(g) \rightarrow 2 \text{CO}_2(g)$

1. Products:
* $2 \times \text{CO}_2(g)$: $2 \times (-393.5) = -787.0$

2. Reactants:
* $2 \times \text{CO}(g)$: $2 \times (-110.53) = -221.06$
* $1 \times \text{O}_2(g)$: $1 \times 0 = 0$
* Sum of Reactants = $-221.06$

3. Calculate $\Delta H$:
* $\Delta H = (-787.0) - (-221.06)$
* $\Delta H = -787.0 + 221.06$
* $\Delta H = -565.94 \text{ kJ}$

---

c) $\text{CH}_4(g) + 2 \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(l)$

*Note: Be careful to use $\text{H}_2\text{O}(l)$ (liquid), not gas.*

1. Products:
* $1 \times \text{CO}_2(g)$: $1 \times (-393.5) = -393.5$
* $2 \times \text{H}_2\text{O}(l)$: $2 \times (-285.8) = -571.6$
* Sum of Products = $-393.5 + (-571.6) = -965.1$

2. Reactants:
* $1 \times \text{CH}_4(g)$: $1 \times (-74.8) = -74.8$
* $2 \times \text{O}_2(g)$: $2 \times 0 = 0$
* Sum of Reactants = $-74.8$

3. Calculate $\Delta H$:
* $\Delta H = (-965.1) - (-74.8)$
* $\Delta H = -965.1 + 74.8$
* $\Delta H = -890.3 \text{ kJ}$

---

d) $2 \text{H}_2\text{S}(g) + 3 \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(l) + 2 \text{SO}_2(g)$

1. Products:
* $2 \times \text{H}_2\text{O}(l)$: $2 \times (-285.8) = -571.6$
* $2 \times \text{SO}_2(g)$: $2 \times (-296.1) = -592.2$
* Sum of Products = $-571.6 + (-592.2) = -1163.8$

2. Reactants:
* $2 \times \text{H}_2\text{S}(g)$: $2 \times (-20.1) = -40.2$
* $3 \times \text{O}_2(g)$: $3 \times 0 = 0$
* Sum of Reactants = $-40.2$

3. Calculate $\Delta H$:
* $\Delta H = (-1163.8) - (-40.2)$
* $\Delta H = -1163.8 + 40.2$
* $\Delta H = -1123.6 \text{ kJ}$

---

e) $2 \text{NO}(g) + \text{O}_2(g) \rightarrow 2 \text{NO}_2(g)$

1. Products:
* $2 \times \text{NO}_2(g)$: $2 \times (+33.9) = +67.8$

2. Reactants:
* $2 \times \text{NO}(g)$: $2 \times (+90.4) = +180.8$
* $1 \times \text{O}_2(g)$: $1 \times 0 = 0$
* Sum of Reactants = $+180.8$

3. Calculate $\Delta H$:
* $\Delta H = (67.8) - (180.8)$
* $\Delta H = 67.8 - 180.8$
* $\Delta H = -113.0 \text{ kJ}$

Final Answer:
a) -133.8 kJ
b) -565.94 kJ
c) -890.3 kJ
d) -1123.6 kJ
e) -113.0 kJ