14.4 - 1 standard enthalpies of formation and $delta h_{f}^{circ}$\nuse the table below to answer the…

14.4 - 1 standard enthalpies of formation and $delta h_{f}^{circ}$\nuse the table below to answer the following questions.\ntable 2 standard enthalpies of formation (kj/mol)\n| compound | $delta h_{f}^{circ}$ (kj/mol) | compound | $delta h_{f}^{circ}$ (kj/mol) |\n| ---- | ---- | ---- | ---- |\n| $ch_{3}oh(l)$ | - 239.1 | $fe_{2}o_{3}(s)$ | - 824.2 |\n| $ch_{4}(g)$ | - 74.4 | $h_{2}o(l)$ | - 285.8 |\n| $c_{2}h_{2}(g)$ | + 228.2 | $h_{2}o(g)$ | - 241.8 |\n| $c_{8}h_{18}(l)$ | - 250.1 | $nh_{3}(g)$ | - 45.9 |\n| $co(g)$ | - 110.5 | $no(g)$ | + 90.2 |\n| $co_{2}(g)$ | - 393.5 | $no_{2}(g)$ | + 33.2 |\ndetermine the enthalpy change of reaction from standard enthalpies of formation for each of these reactions. (for reactions 6 and 7, first write the chemical equation using a coefficient of 1 for the hydrocarbon reactant.)\n1. $2fe(s)+3o_{2}(g)\rightarrow fe_{2}o_{3}(s)$ $delta h_{r}^{circ}=-824.2kj/mol$\n2. $co(g)+2h_{2}(g)\rightarrow ch_{3}oh(l)$ $delta h_{r}^{circ}=-128.6kj/mol$\n3. $2h_{2}(g)+o_{2}(g)\rightarrow 2h_{2}o(l)$ $delta h_{r}^{circ}=-571.6kj/mol$\n4. $4nh_{3}(g)+5o_{2}(g)\rightarrow 4no(g)+6h_{2}o(g)$ $delta h_{r}^{circ}=-906.4kj/mol$\n5. $ch_{4}(g)+2o_{2}(g)\rightarrow co_{2}(g)+2h_{2}o(g)$ $delta h_{r}^{circ}=-802.7kj/mol$\n6. complete combustion of ethyne in oxygen $delta h_{r}^{circ}=-802.7kj/mol$\n7. complete combustion of octane in oxygen $delta h_{r}^{circ}=-5074.1kj/mol$

14.4 - 1 standard enthalpies of formation and $delta h_{f}^{circ}$\nuse the table below to answer the following questions.\ntable 2 standard enthalpies of formation (kj/mol)\n| compound | $delta h_{f}^{circ}$ (kj/mol) | compound | $delta h_{f}^{circ}$ (kj/mol) |\n| ---- | ---- | ---- | ---- |\n| $ch_{3}oh(l)$ | - 239.1 | $fe_{2}o_{3}(s)$ | - 824.2 |\n| $ch_{4}(g)$ | - 74.4 | $h_{2}o(l)$ | - 285.8 |\n| $c_{2}h_{2}(g)$ | + 228.2 | $h_{2}o(g)$ | - 241.8 |\n| $c_{8}h_{18}(l)$ | - 250.1 | $nh_{3}(g)$ | - 45.9 |\n| $co(g)$ | - 110.5 | $no(g)$ | + 90.2 |\n| $co_{2}(g)$ | - 393.5 | $no_{2}(g)$ | + 33.2 |\ndetermine the enthalpy change of reaction from standard enthalpies of formation for each of these reactions. (for reactions 6 and 7, first write the chemical equation using a coefficient of 1 for the hydrocarbon reactant.)\n1. $2fe(s)+3o_{2}(g)\rightarrow fe_{2}o_{3}(s)$ $delta h_{r}^{circ}=-824.2kj/mol$\n2. $co(g)+2h_{2}(g)\rightarrow ch_{3}oh(l)$ $delta h_{r}^{circ}=-128.6kj/mol$\n3. $2h_{2}(g)+o_{2}(g)\rightarrow 2h_{2}o(l)$ $delta h_{r}^{circ}=-571.6kj/mol$\n4. $4nh_{3}(g)+5o_{2}(g)\rightarrow 4no(g)+6h_{2}o(g)$ $delta h_{r}^{circ}=-906.4kj/mol$\n5. $ch_{4}(g)+2o_{2}(g)\rightarrow co_{2}(g)+2h_{2}o(g)$ $delta h_{r}^{circ}=-802.7kj/mol$\n6. complete combustion of ethyne in oxygen $delta h_{r}^{circ}=-802.7kj/mol$\n7. complete combustion of octane in oxygen $delta h_{r}^{circ}=-5074.1kj/mol$

Answer

Explanation:

Step1: Identify $\Delta H_f^\circ$ values from the table.

The standard enthalpies of formation ($\Delta H_f^\circ$) for the reactants and products are: $\Delta H_f^\circ (\text{Fe}_2\text{O}_3\text{(s)}) = -824.2 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{Fe(s)}) = 0 \text{ kJ/mol}$ (element in standard state) $\Delta H_f^\circ (\text{O}_2\text{(g)}) = 0 \text{ kJ/mol}$ (element in standard state)

Step2: Apply Hess's Law formula.

The enthalpy change of reaction ($\Delta H_r^\circ$) is calculated using: $\Delta H_r^\circ = \sum (n \cdot \Delta H_f^\circ){\text{products}} - \sum (m \cdot \Delta H_f^\circ){\text{reactants}}$ For $2 \text{ Fe(s)} + 3 \text{ O}_2\text{(g)} \rightarrow \text{Fe}_2\text{O}_3\text{(s)}$: $\Delta H_r^\circ = [1 \cdot \Delta H_f^\circ (\text{Fe}_2\text{O}_3\text{(s)})] - [2 \cdot \Delta H_f^\circ (\text{Fe(s)}) + 3 \cdot \Delta H_f^\circ (\text{O}_2\text{(g)})]$

Step3: Substitute values and calculate.

$\Delta H_r^\circ = [1 \cdot (-824.2 \text{ kJ/mol})] - [2 \cdot (0 \text{ kJ/mol}) + 3 \cdot (0 \text{ kJ/mol})]$ $\Delta H_r^\circ = -824.2 \text{ kJ/mol} - 0 \text{ kJ/mol}$ $\Delta H_r^\circ = -824.2 \text{ kJ/mol}$

Answer:

  1. $\Delta H_r^\circ = -824.2 \text{ kJ/mol}$

Explanation:

Step1: Identify $\Delta H_f^\circ$ values from the table.

$\Delta H_f^\circ (\text{CH}_3\text{OH(l)}) = -239.1 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{CO(g)}) = -110.5 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{H}_2\text{(g)}) = 0 \text{ kJ/mol}$ (element in standard state)

Step2: Apply Hess's Law formula.

For $\text{CO(g)} + 2 \text{ H}_2\text{(g)} \rightarrow \text{CH}_3\text{OH(l)}$: $\Delta H_r^\circ = [1 \cdot \Delta H_f^\circ (\text{CH}_3\text{OH(l)})] - [1 \cdot \Delta H_f^\circ (\text{CO(g)}) + 2 \cdot \Delta H_f^\circ (\text{H}_2\text{(g)})]$

Step3: Substitute values and calculate.

$\Delta H_r^\circ = [1 \cdot (-239.1 \text{ kJ/mol})] - [1 \cdot (-110.5 \text{ kJ/mol}) + 2 \cdot (0 \text{ kJ/mol})]$ $\Delta H_r^\circ = -239.1 \text{ kJ/mol} - (-110.5 \text{ kJ/mol})$ $\Delta H_r^\circ = -239.1 \text{ kJ/mol} + 110.5 \text{ kJ/mol}$ $\Delta H_r^\circ = -128.6 \text{ kJ/mol}$

Answer:

  1. $\Delta H_r^\circ = -128.6 \text{ kJ/mol}$

Explanation:

Step1: Identify $\Delta H_f^\circ$ values from the table.

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

Step2: Apply Hess's Law formula.

For $2 \text{ H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2 \text{ H}_2\text{O(l)}$: $\Delta H_r^\circ = [2 \cdot \Delta H_f^\circ (\text{H}_2\text{O(l)})] - [2 \cdot \Delta H_f^\circ (\text{H}_2\text{(g)}) + 1 \cdot \Delta H_f^\circ (\text{O}_2\text{(g)})]$

Step3: Substitute values and calculate.

$\Delta H_r^\circ = [2 \cdot (-285.8 \text{ kJ/mol})] - [2 \cdot (0 \text{ kJ/mol}) + 1 \cdot (0 \text{ kJ/mol})]$ $\Delta H_r^\circ = -571.6 \text{ kJ/mol} - 0 \text{ kJ/mol}$ $\Delta H_r^\circ = -571.6 \text{ kJ/mol}$

Answer:

  1. $\Delta H_r^\circ = -571.6 \text{ kJ/mol}$

Explanation:

Step1: Identify $\Delta H_f^\circ$ values from the table.

$\Delta H_f^\circ (\text{NO(g)}) = +90.2 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{H}_2\text{O(g)}) = -241.8 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{NH}_3\text{(g)}) = -45.9 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{O}_2\text{(g)}) = 0 \text{ kJ/mol}$ (element in standard state)

Step2: Apply Hess's Law formula.

For $4 \text{ NH}_3\text{(g)} + 5 \text{ O}_2\text{(g)} \rightarrow 4 \text{ NO(g)} + 6 \text{ H}_2\text{O(g)}$: $\Delta H_r^\circ = [4 \cdot \Delta H_f^\circ (\text{NO(g)}) + 6 \cdot \Delta H_f^\circ (\text{H}_2\text{O(g)})] - [4 \cdot \Delta H_f^\circ (\text{NH}_3\text{(g)}) + 5 \cdot \Delta H_f^\circ (\text{O}_2\text{(g)})]$

Step3: Substitute values and calculate.

$\Delta H_r^\circ = [4 \cdot (90.2 \text{ kJ/mol}) + 6 \cdot (-241.8 \text{ kJ/mol})] - [4 \cdot (-45.9 \text{ kJ/mol}) + 5 \cdot (0 \text{ kJ/mol})]$ $\Delta H_r^\circ = [360.8 \text{ kJ/mol} - 1450.8 \text{ kJ/mol}] - [-183.6 \text{ kJ/mol} + 0 \text{ kJ/mol}]$ $\Delta H_r^\circ = -1090.0 \text{ kJ/mol} - (-183.6 \text{ kJ/mol})$ $\Delta H_r^\circ = -1090.0 \text{ kJ/mol} + 183.6 \text{ kJ/mol}$ $\Delta H_r^\circ = -906.4 \text{ kJ/mol}$

Answer:

  1. $\Delta H_r^\circ = -906.4 \text{ kJ/mol}$

Explanation:

Step1: Identify $\Delta H_f^\circ$ values from the table.

$\Delta H_f^\circ (\text{CO}_2\text{(g)}) = -393.5 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{H}_2\text{O(g)}) = -241.8 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{CH}_4\text{(g)}) = -74.4 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{O}_2\text{(g)}) = 0 \text{ kJ/mol}$ (element in standard state)

Step2: Apply Hess's Law formula.

For $\text{CH}_4\text{(g)} + 2 \text{ O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2 \text{ H}_2\text{O(g)}$: $\Delta H_r^\circ = [1 \cdot \Delta H_f^\circ (\text{CO}_2\text{(g)}) + 2 \cdot \Delta H_f^\circ (\text{H}_2\text{O(g)})] - [1 \cdot \Delta H_f^\circ (\text{CH}_4\text{(g)}) + 2 \cdot \Delta H_f^\circ (\text{O}_2\text{(g)})]$

Step3: Substitute values and calculate.

$\Delta H_r^\circ = [1 \cdot (-393.5 \text{ kJ/mol}) + 2 \cdot (-241.8 \text{ kJ/mol})] - [1 \cdot (-74.4 \text{ kJ/mol}) + 2 \cdot (0 \text{ kJ/mol})]$ $\Delta H_r^\circ = [-393.5 \text{ kJ/mol} - 483.6 \text{ kJ/mol}] - [-74.4 \text{ kJ/mol} + 0 \text{ kJ/mol}]$ $\Delta H_r^\circ = -877.1 \text{ kJ/mol} - (-74.4 \text{ kJ/mol})$ $\Delta H_r^\circ = -877.1 \text{ kJ/mol} + 74.4 \text{ kJ/mol}$ $\Delta H_r^\circ = -802.7 \text{ kJ/mol}$

Answer:

  1. $\Delta H_r^\circ = -802.7 \text{ kJ/mol}$

Explanation:

Step1: Write balanced combustion equation.

The complete combustion of ethyne ($\text{C}_2\text{H}_2\text{(g)}$) with a coefficient of 1 for ethyne: $\text{C}_2\text{H}_2\text{(g)} + \frac{5}{2} \text{ O}_2\text{(g)} \rightarrow 2 \text{ CO}_2\text{(g)} + \text{H}_2\text{O(g)}$

Step2: Identify $\Delta H_f^\circ$ values from the table.

$\Delta H_f^\circ (\text{CO}_2\text{(g)}) = -393.5 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{H}_2\text{O(g)}) = -241.8 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{C}_2\text{H}_2\text{(g)}) = +228.2 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{O}_2\text{(g)}) = 0 \text{ kJ/mol}$ (element in standard state)

Step3: Apply Hess's Law formula.

$\Delta H_r^\circ = [2 \cdot \Delta H_f^\circ (\text{CO}_2\text{(g)}) + 1 \cdot \Delta H_f^\circ (\text{H}_2\text{O(g)})] - [1 \cdot \Delta H_f^\circ (\text{C}_2\text{H}_2\text{(g)}) + \frac{5}{2} \cdot \Delta H_f^\circ (\text{O}_2\text{(g)})]$

Step4: Substitute values and calculate.

$\Delta H_r^\circ = [2 \cdot (-393.5 \text{ kJ/mol}) + 1 \cdot (-241.8 \text{ kJ/mol})] - [1 \cdot (228.2 \text{ kJ/mol}) + \frac{5}{2} \cdot (0 \text{ kJ/mol})]$ $\Delta H_r^\circ = [-787.0 \text{ kJ/mol} - 241.8 \text{ kJ/mol}] - [228.2 \text{ kJ/mol}]$ $\Delta H_r^\circ = -1028.8 \text{ kJ/mol} - 228.2 \text{ kJ/mol}$ $\Delta H_r^\circ = -1257.0 \text{ kJ/mol}$

Answer:

  1. $\text{C}_2\text{H}_2\text{(g)} + \frac{5}{2} \text{ O}_2\text{(g)} \rightarrow 2 \text{ CO}_2\text{(g)} + \text{H}_2\text{O(g)}$ $\Delta H_r^\circ = -1257.0 \text{ kJ/mol}$

Explanation:

Step1: Write balanced combustion equation.

The complete combustion of octane ($\text{C}8\text{H}{18}\text{(l)}$) with a coefficient of 1 for octane: $\text{C}8\text{H}{18}\text{(l)} + \frac{25}{2} \text{ O}_2\text{(g)} \rightarrow 8 \text{ CO}_2\text{(g)} + 9 \text{ H}_2\text{O(g)}$

Step2: Identify $\Delta H_f^\circ$ values from the table.

$\Delta H_f^\circ (\text{CO}_2\text{(g)}) = -393.5 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{H}_2\text{O(g)}) = -241.8 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{C}8\text{H}{18}\text{(l)}) = -250.1 \text{ kJ/mol}$ $\Delta H_f^\circ (\text{O}_2\text{(g)}) = 0 \text{ kJ/mol}$ (element in standard state)

Step3: Apply Hess's Law formula.

$\Delta H_r^\circ = [8 \cdot \Delta H_f^\circ (\text{CO}_2\text{(g)}) + 9 \cdot \Delta H_f^\circ (\text{H}_2\text{O(g)})] - [1 \cdot \Delta H_f^\circ (\text{C}8\text{H}{18}\text{(l)}) + \frac{25}{2} \cdot \Delta H_f^\circ (\text{O}_2\text{(g)})]$

Step4: Substitute values and calculate.

$\Delta H_r^\circ = [8 \cdot (-393.5 \text{ kJ/mol}) + 9 \cdot (-241.8 \text{ kJ/mol})] - [1 \cdot (-250.1 \text{ kJ/mol}) + \frac{25}{2} \cdot (0 \text{ kJ/mol})]$ $\Delta H_r^\circ = [-3148.0 \text{ kJ/mol} - 2176.2 \text{ kJ/mol}] - [-250.1 \text{ kJ/mol}]$ $\Delta H_r^\circ = -5324.2 \text{ kJ/mol} - (-250.1 \text{ kJ/mol})$ $\Delta H_r^\circ = -5324.2 \text{ kJ/mol} + 250.1 \text{ kJ/mol}$ $\Delta H_r^\circ = -5074.1 \text{ kJ/mol}$

Answer:

  1. $\text{C}8\text{H}{18}\text{(l)} + \frac{25}{2} \text{ O}_2\text{(g)} \rightarrow 8 \text{ CO}_2\text{(g)} + 9 \text{ H}_2\text{O(g)}$ $\Delta H_r^\circ = -5074.1 \text{ kJ/mol}$