Answer :
Let's evaluate the heat released by each of the substances diborane [tex]\(\left( \text{B}_2\text{H}_6 \right)\)[/tex], methane [tex]\(\left( \text{CH}_4 \right)\)[/tex], and hydrazine [tex]\(\left( \text{N}_2\text{H}_4 \right)\)[/tex] when reacting with oxygen, using the enthalpy changes in kilojoules per mole and the total heats in megajoules for a given mass (3.90 kg) of each fuel.
To begin, we'll use the enthalpy of reaction ([tex]\(\Delta H\)[/tex]) for each substance, calculate the number of moles in 3.90 kg, and then compute the total heat released.
### Diborane (B[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex])
1. Molar Mass Calculation:
- Boron (B) = 10.81 g/mol
- Hydrogen (H) = 1.01 g/mol
- Molar mass of [tex]\(\text{B}_2\text{H}_6\)[/tex] = [tex]\(2 \times 10.81\)[/tex] + [tex]\(6 \times 1.01\)[/tex] = 27.64 g/mol
2. Moles in 3.90 kg:
[tex]\[\frac{3900 \text{ g}}{27.64 \text{ g/mol}} = 141.11 \text{ mol}\][/tex]
3. Enthalpy of Reaction:
[tex]\(\Delta H_r^\circ = -2034 \text{ kJ/mol}\)[/tex] (assumed based on standard data)
4. Total Heat ([tex]\(\text{q}_{\text{surr}}\)[/tex]) Released:
[tex]\[\text{q}_{\text{surr}} = 141.11 \text{ mol} \times (-2034 \text{ kJ/mol})\][/tex]
[tex]\[\text{q}_{\text{surr}} = -287,026 \text{ kJ} = -287.03 \text{ MJ}\][/tex]
### Methane (CH[tex]\(_4\)[/tex])
1. Molar Mass Calculation:
- Carbon (C) = 12.01 g/mol
- Hydrogen (H) = 1.01 g/mol
- Molar mass of [tex]\(\text{CH}_4\)[/tex] = [tex]\(12.01 + 4 \times 1.01 = 16.05 \text{ g/mol}\)[/tex]
2. Moles in 3.90 kg:
[tex]\[\frac{3900 \text{ g}}{16.05 \text{ g/mol}} = 242.36 \text{ mol}\][/tex]
3. Enthalpy of Reaction:
[tex]\(\Delta H_f^\circ = -890 \text{ kJ/mol}\)[/tex] (assumed based on standard data)
4. Total Heat ([tex]\(\text{q}_{\text{surr}}\)[/tex]) Released:
[tex]\[\text{q}_{\text{surr}} = 242.36 \text{ mol} \times (-890 \text{ kJ/mol})\][/tex]
[tex]\[\text{q}_{\text{surr}} = -215,700 \text{ kJ} = -215.70 \text{ MJ}\][/tex]
### Hydrazine (N[tex]\(_2\)[/tex]H[tex]\(_4\)[/tex])
1. Molar Mass Calculation:
- Nitrogen (N) = 14.01 g/mol
- Hydrogen (H) = 1.01 g/mol
- Molar mass of [tex]\(\text{N}_2\text{H}_4\)[/tex] = [tex]\(2 \times 14.01 + 4 \times 1.01 = 32.05 \text{ g/mol}\)[/tex]
2. Moles in 3.90 kg:
[tex]\[\frac{3900 \text{ g}}{32.05 \text{ g/mol}} = 121.66 \text{ mol}\][/tex]
3. Enthalpy of Reaction:
[tex]\(\Delta H_r^\circ = -622 \text{ kJ/mol}\)[/tex] (assumed based on standard data)
4. Total Heat ([tex]\(\text{q}_{\text{surr}}\)[/tex]) Released:
[tex]\[\text{q}_{\text{surr}} = 121.66 \text{ mol} \times (-622 \text{ kJ/mol})\][/tex]
[tex]\[\text{q}_{\text{surr}} = -75,640 \text{ kJ} = -75.64 \text{ MJ}\][/tex]
### Evaluation for Rocket Fuel Application
Given the heat released from each substance:
- Diborane: [tex]\( -287 \text{ MJ}\)[/tex]
- Methane: [tex]\( -216 \text{ MJ}\)[/tex]
- Hydrazine: [tex]\( -76 \text{ MJ}\)[/tex]
In firepower terms for the same 3.90 kg mass, diborane releases the most energy, followed by methane, and finally hydrazine. Based on the amount of heat released, diborane would be the most potent rocket fuel, as it provides the most thermal energy, which is desirable in rocket design.
To begin, we'll use the enthalpy of reaction ([tex]\(\Delta H\)[/tex]) for each substance, calculate the number of moles in 3.90 kg, and then compute the total heat released.
### Diborane (B[tex]\(_2\)[/tex]H[tex]\(_6\)[/tex])
1. Molar Mass Calculation:
- Boron (B) = 10.81 g/mol
- Hydrogen (H) = 1.01 g/mol
- Molar mass of [tex]\(\text{B}_2\text{H}_6\)[/tex] = [tex]\(2 \times 10.81\)[/tex] + [tex]\(6 \times 1.01\)[/tex] = 27.64 g/mol
2. Moles in 3.90 kg:
[tex]\[\frac{3900 \text{ g}}{27.64 \text{ g/mol}} = 141.11 \text{ mol}\][/tex]
3. Enthalpy of Reaction:
[tex]\(\Delta H_r^\circ = -2034 \text{ kJ/mol}\)[/tex] (assumed based on standard data)
4. Total Heat ([tex]\(\text{q}_{\text{surr}}\)[/tex]) Released:
[tex]\[\text{q}_{\text{surr}} = 141.11 \text{ mol} \times (-2034 \text{ kJ/mol})\][/tex]
[tex]\[\text{q}_{\text{surr}} = -287,026 \text{ kJ} = -287.03 \text{ MJ}\][/tex]
### Methane (CH[tex]\(_4\)[/tex])
1. Molar Mass Calculation:
- Carbon (C) = 12.01 g/mol
- Hydrogen (H) = 1.01 g/mol
- Molar mass of [tex]\(\text{CH}_4\)[/tex] = [tex]\(12.01 + 4 \times 1.01 = 16.05 \text{ g/mol}\)[/tex]
2. Moles in 3.90 kg:
[tex]\[\frac{3900 \text{ g}}{16.05 \text{ g/mol}} = 242.36 \text{ mol}\][/tex]
3. Enthalpy of Reaction:
[tex]\(\Delta H_f^\circ = -890 \text{ kJ/mol}\)[/tex] (assumed based on standard data)
4. Total Heat ([tex]\(\text{q}_{\text{surr}}\)[/tex]) Released:
[tex]\[\text{q}_{\text{surr}} = 242.36 \text{ mol} \times (-890 \text{ kJ/mol})\][/tex]
[tex]\[\text{q}_{\text{surr}} = -215,700 \text{ kJ} = -215.70 \text{ MJ}\][/tex]
### Hydrazine (N[tex]\(_2\)[/tex]H[tex]\(_4\)[/tex])
1. Molar Mass Calculation:
- Nitrogen (N) = 14.01 g/mol
- Hydrogen (H) = 1.01 g/mol
- Molar mass of [tex]\(\text{N}_2\text{H}_4\)[/tex] = [tex]\(2 \times 14.01 + 4 \times 1.01 = 32.05 \text{ g/mol}\)[/tex]
2. Moles in 3.90 kg:
[tex]\[\frac{3900 \text{ g}}{32.05 \text{ g/mol}} = 121.66 \text{ mol}\][/tex]
3. Enthalpy of Reaction:
[tex]\(\Delta H_r^\circ = -622 \text{ kJ/mol}\)[/tex] (assumed based on standard data)
4. Total Heat ([tex]\(\text{q}_{\text{surr}}\)[/tex]) Released:
[tex]\[\text{q}_{\text{surr}} = 121.66 \text{ mol} \times (-622 \text{ kJ/mol})\][/tex]
[tex]\[\text{q}_{\text{surr}} = -75,640 \text{ kJ} = -75.64 \text{ MJ}\][/tex]
### Evaluation for Rocket Fuel Application
Given the heat released from each substance:
- Diborane: [tex]\( -287 \text{ MJ}\)[/tex]
- Methane: [tex]\( -216 \text{ MJ}\)[/tex]
- Hydrazine: [tex]\( -76 \text{ MJ}\)[/tex]
In firepower terms for the same 3.90 kg mass, diborane releases the most energy, followed by methane, and finally hydrazine. Based on the amount of heat released, diborane would be the most potent rocket fuel, as it provides the most thermal energy, which is desirable in rocket design.
