Problem 140 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The value of \( y \) is 4.

Step 1: Understanding the Problem

We are given the freezing point of isobutane, the enthalpy change for the fusion process, and the entropy change in the form of \(10y\) J/mol. We need to determine the value of \(y\).

Step 2: Formula for Entropy Change

The entropy change \( \Delta S \) for a phase transition can be calculated using the formula \( \Delta S = \frac{\Delta H}{T} \), where \( \Delta H \) is the enthalpy change and \( T \) is the temperature in Kelvin.

Step 3: Convert Temperature to Kelvin

The freezing point of isobutane is given as \(-160^{\circ} \mathrm{C}\). Convert this to Kelvin using the formula \( T(K) = T(^{\circ}C) + 273.15 \). Therefore, \( T = -160 + 273.15 = 113.15 \text{ K} \).

Step 4: Calculate the Entropy Change

Using \( \Delta S = \frac{\Delta H}{T} \), substitute \( \Delta H = 4520 \text{ J/mol} \) and \( T = 113.15 \text{ K} \).\[\Delta S = \frac{4520}{113.15} \approx 39.95 \text{ J/mol} \]

Step 5: Determine the Value of y

We know the entropy change is given as \(10y\). Set this equal to the calculated \( \Delta S \).\[10y = 39.95\]Solve for \(y\):\[ y = \frac{39.95}{10} = 3.995 \approx 4 \]

Key Concepts

Freezing Point Depression

Freezing point depression is a fascinating phenomenon involving the lowering of a liquid's freezing point when a solute is added. This concept is crucial in understanding how different substances can alter the overall temperature reactions during phase changes. In simpler terms, when a solute is mixed with a solvent, like adding salt to water, the freezing temperature decreases compared to the pure solvent on its own.

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.

Recognizing these effects provides insights into thermodynamics and solutions' behaviors in various conditions.

Enthalpy of Fusion

The enthalpy of fusion is the amount of energy needed to change a substance from solid to liquid without changing its temperature. It's a key player in understanding phase transitions and how energy flows during these processes. The energy is required to overcome the forces holding the molecules in the solid phase to allow for the more fluid liquid phase.

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.

Such calculations are vital for industries reliant on chemical processes involving phase changes.

Kelvin Temperature Conversion

Temperature conversion to Kelvin is a fundamental skill required for thermodynamics, a key area in physics and chemistry. Kelvin is the SI unit for temperature and is used extensively in scientific calculations because it starts at absolute zero, the theoretical point where molecular motion ceases. Converting Celsius to Kelvin is straightforward with the formula \(T(K) = T(^{\circ}C) + 273.15\).

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.

The correct application of Kelvin conversion allows for accurate scientific calculations and deeper insights into material properties and reactions.