Chapter 6: Problem 140
The freezing point of isobutane is \(-160^{\circ} \mathrm{C} \cdot \Delta \mathrm{H}_{\text {(solid } \rightarrow \text { liquid) }}\) is \(+4520 \mathrm{~J} \mathrm{~mol}^{-1}\). For this fusion process, entropy change in \(\mathrm{J} \mathrm{mol}^{-1}\) is \(10 \mathrm{y}\). The value of \(\mathrm{y}\) is
Short Answer
Step by step solution
Understanding the Problem
Formula for Entropy Change
Convert Temperature to Kelvin
Calculate the Entropy Change
Determine the Value of y
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Freezing Point Depression
This occurs due to the interference of solute particles with the formation of a solid structure in the solvent. By disrupting the uniformity of ice formation, the freezing point of the solution is lowered.
- In isobutane's case, we see its freezing point is extraordinarily low at \(-160^{\circ}C\) due to its molecular structure.
- Understanding the interaction between solute and solvent helps in practical applications, such as de-icing roads or making ice cream smoother.
Enthalpy of Fusion
In the example of isobutane, it's given that the enthalpy of fusion is \(\Delta H = 4520 \, \text{J/mol}\). This means that each mole of isobutane requires 4520 joules to transition between these states.
- Phase transitions, such as melting or freezing, involve breaking or forming molecular bonds requiring or releasing energy.
- Enthalpy changes in these processes provide insights into molecular interactions and stability.
- By knowing the enthalpy of fusion, we can calculate other thermodynamic properties, like the change in entropy.
Kelvin Temperature Conversion
In the context of the given problem, the freezing point of isobutane at \(-160^{\circ}C\) was converted to Kelvin:
\[T = -160 + 273.15 = 113.15 \, \text{K}\]
- The conversion ensures accuracy when applying thermodynamic formulas, such as calculating entropy changes.
- Working in Kelvin provides consistency across scientific disciplines and allows for comparisons between different materials and their phase transitions.
- Understanding and applying Kelvin conversions is essential for solving problems involving energy changes and temperature-dependent processes.