Question

using enthalpies of formation, determine the enthalpy of the following reaction:
2 c₆h₆ (l) + 15 o₂ (g) → 12 co₂ (g) + 6 h₂o (l)
note that the enthalpies of formation can be found in the tables of chemical data posted on canvas. report the answer in kilojoules to 4 sig figs, but do not include units in your answer.

question 8
0.5 pts
using enthalpies of formation, determine the balanced thermal chemical equation for the following reaction:
3 co₃o₄ (s) + 8 al (s) → 9 co (s) + 4 al₂o₃ (s)
how much heat is given off or absorbed when 44.88 g of co is formed. the reaction occurs at 25.0°c at a constant pressure of 1.00 atm. note that the enthalpies of formation can be found in the tables of chemical data posted on canvas. report the answer in kilojoules to 4 sig figs, but do not include units in your answer.

Explanation:

First Reaction (Benzene Combustion)

Step1: Recall Enthalpy of Formation Formula

The enthalpy change of a reaction ($\Delta H_{rxn}$) is calculated by:
$$\Delta H_{rxn} = \sum n \Delta H_f^{\circ}(\text{products}) - \sum m \Delta H_f^{\circ}(\text{reactants})$$
where $n$ and $m$ are stoichiometric coefficients, and $\Delta H_f^{\circ}$ is the standard enthalpy of formation.

Step2: Identify $\Delta H_f^{\circ}$ Values (from standard tables)

• $\Delta H_f^{\circ}(\text{C}_6\text{H}_6(l)) = 49.0\ \text{kJ/mol}$
• $\Delta H_f^{\circ}(\text{O}_2(g)) = 0\ \text{kJ/mol}$ (element in standard state)
• $\Delta H_f^{\circ}(\text{CO}_2(g)) = -393.5\ \text{kJ/mol}$
• $\Delta H_f^{\circ}(\text{H}_2\text{O}(l)) = -285.8\ \text{kJ/mol}$

Step3: Calculate $\sum n \Delta H_f^{\circ}(\text{products})$

For products: $12\ \text{mol}\ \text{CO}_2$ and $6\ \text{mol}\ \text{H}_2\text{O}(l)$:
$$\sum n \Delta H_f^{\circ}(\text{products}) = 12 \times (-393.5) + 6 \times (-285.8)$$
$$= -4722 - 1714.8 = -6436.8\ \text{kJ}$$

Step4: Calculate $\sum m \Delta H_f^{\circ}(\text{reactants})$

For reactants: $2\ \text{mol}\ \text{C}_6\text{H}_6(l)$ and $15\ \text{mol}\ \text{O}_2(g)$:
$$\sum m \Delta H_f^{\circ}(\text{reactants}) = 2 \times 49.0 + 15 \times 0 = 98.0\ \text{kJ}$$

Step5: Compute $\Delta H_{rxn}$

$$\Delta H_{rxn} = -6436.8 - 98.0 = -6534.8\ \text{kJ}$$
(Adjust to 4 sig figs: $-6535$)

Second Reaction (Aluminum + Cobalt Oxide)

Step1: Find $\Delta H_{rxn}$ for the Balanced Reaction

First, calculate $\Delta H_{rxn}$ for $3\ \text{Co}_3\text{O}_4(s) + 8\ \text{Al}(s)
ightarrow 9\ \text{Co}(s) + 4\ \text{Al}_2\text{O}_3(s)$:

$\Delta H_f^{\circ}$ values (standard tables):

• $\Delta H_f^{\circ}(\text{Co}_3\text{O}_4(s)) = -1776\ \text{kJ/mol}$
• $\Delta H_f^{\circ}(\text{Al}(s)) = 0\ \text{kJ/mol}$ (element in standard state)
• $\Delta H_f^{\circ}(\text{Co}(s)) = 0\ \text{kJ/mol}$ (element in standard state)
• $\Delta H_f^{\circ}(\text{Al}_2\text{O}_3(s)) = -1675.7\ \text{kJ/mol}$

$$\Delta H_{rxn} = \sum n \Delta H_f^{\circ}(\text{products}) - \sum m \Delta H_f^{\circ}(\text{reactants})$$
$$= [9 \times 0 + 4 \times (-1675.7)] - [3 \times (-1776) + 8 \times 0]$$
$$= -6702.8 + 5328 = -1374.8\ \text{kJ}$$ (for 9 mol Co formed)

Step2: Moles of Co Formed

Molar mass of Co = $58.93\ \text{g/mol}$. Moles of Co:
$$n(\text{Co}) = \frac{44.88\ \text{g}}{58.93\ \text{g/mol}} \approx 0.7616\ \text{mol}$$

Step3: Heat for 0.7616 mol Co

From the reaction, 9 mol Co releases $1374.8\ \text{kJ}$ (since $\Delta H_{rxn}$ is negative, heat is given off). For $0.7616\ \text{mol}$ Co:
$$\text{Heat} = \frac{-1374.8\ \text{kJ}}{9\ \text{mol}} \times 0.7616\ \text{mol} \approx -117.0\ \text{kJ}$$ (4 sig figs: $-117.0$)

Answer:

s:
First reaction enthalpy: $\boldsymbol{-6535}$
Second reaction heat: $\boldsymbol{-117.0}$