Octane (C8H18), an important component of gasoline, is a flammable compound with a boiling point of 125.67 C. The �Hvap of octane is 34.41 kJ/mol. Its freezing point is -56.82°C and its �Hfus is 20.73 kJ/mol. The specific heats of octane liquid and vapor are 2.15 J/gK and 1.71 J/gK, respectively:  Although it is nonpolar, octane has a higher boiling point than water. Explain why this is reasonable. Although octane is a liquid at room temperature, it is dangerous not only to light the liquid itself, but also to have an open flame in its vicinity. Why is this the case? What could be done to mitigate this hazard? Calculate the amount of heat loss that would be just sufficient to freeze 3.5 g of octane vapor beginning at 227°C.
Octane (C8H18), an important component of gasoline, is a flammable compound with a boiling point of 125.67°C. The �Hvap of octane is 34.41 kJ/mol. Its freezing point is -56.82°C and its �Hfus is 20.73 kJ/mol. The specific heats of octane liquid and vapor are 2.15 J/gK and 1.71 J/gK, respectively:
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Although it is nonpolar, octane has a higher boiling point than water. Explain why this is reasonable.
Although octane is a liquid at room temperature, it is dangerous not only to light the liquid itself, but also to have an open flame in its vicinity. Why is this the case? What could be done to mitigate this hazard?
Calculate the amount of heat loss that would be just sufficient to freeze 3.5 g of octane vapor beginning at 227°C.
The Correct Answer and Explanation is :
Explanation of Boiling Point of Octane vs Water
The boiling point of a substance depends on the strength of the intermolecular forces between its molecules. While octane (C8H18) is nonpolar and primarily experiences London dispersion forces, these forces are still quite significant due to the large number of electrons in the molecule. Octane is a long, relatively heavy molecule with a large surface area for these forces to act over, which leads to a higher boiling point than smaller nonpolar molecules. Water, on the other hand, has hydrogen bonds, which are much stronger intermolecular forces than London dispersion forces. Despite this, the hydrogen bonds in water are more sensitive to temperature fluctuations because water molecules are smaller and more easily vaporized. Thus, octane’s higher molecular weight leads to a higher boiling point, despite the lack of stronger intermolecular forces like hydrogen bonding.
Danger of Open Flame Near Octane
Octane is dangerous not only because of its flammability, but also because it vaporizes easily at room temperature. When octane vapor mixes with air and is exposed to an ignition source such as an open flame, it can cause a fire or explosion. Even if the liquid is not directly ignited, the vaporized octane can ignite at temperatures well below its boiling point (125.67°C). The liquid’s low flash point of around -43°C makes it volatile and able to form combustible mixtures with air easily. To mitigate this hazard, the handling of octane should be done in well-ventilated areas, and precautions like using closed containers, flame arresters, and maintaining a safe distance from ignition sources are essential.
Calculation of Heat Loss to Freeze 3.5 g of Octane Vapor at 227°C
To calculate the heat required to freeze 3.5 g of octane vapor at 227°C, we need to consider both the cooling of the vapor to its freezing point and the phase change from vapor to solid.
- Heat required to cool vapor from 227°C to the freezing point of octane (-56.82°C):
The heat change can be calculated using the formula:
[
q = m \cdot C \cdot \Delta T
]
where:
- (m = 3.5 \, \text{g}) (mass of octane)
- (C = 1.71 \, \text{J/g°C}) (specific heat of octane vapor)
- (\Delta T = 227°C – (-56.82°C) = 283.82°C) (temperature change)
[
q = 3.5 \, \text{g} \cdot 1.71 \, \text{J/g°C} \cdot 283.82°C = 1700.74 \, \text{J}
]
- Heat required for phase change from vapor to solid (freezing):
Next, we calculate the heat required for the phase change using the heat of fusion:
[
q = n \cdot \Delta H_{\text{fus}}
]
where:
- (n = \frac{3.5 \, \text{g}}{114.23 \, \text{g/mol}}) (moles of octane, given the molar mass of octane is 114.23 g/mol)
- (\Delta H_{\text{fus}} = 20.73 \, \text{kJ/mol}) (heat of fusion for octane)
First, calculate the number of moles of octane:
[
n = \frac{3.5 \, \text{g}}{114.23 \, \text{g/mol}} = 0.0306 \, \text{mol}
]
Now, calculate the heat required for the phase change:
[
q = 0.0306 \, \text{mol} \cdot 20.73 \, \text{kJ/mol} = 0.635 \, \text{kJ} = 635 \, \text{J}
]
- Total heat loss:
Finally, sum the heat required for cooling and phase change:
[
q_{\text{total}} = 1700.74 \, \text{J} + 635 \, \text{J} = 2335.74 \, \text{J}
]
Thus, the total heat loss required to freeze 3.5 g of octane vapor starting at 227°C is approximately 2336 J.
Conclusion
This answer illustrates how the heat energy required to cool and freeze a substance involves both sensible heat (changing temperature) and latent heat (phase change). In the case of octane, this process is relatively energy-intensive due to its high heat of vaporization and fusion, making octane a highly energetic compound in various chemical processes.