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

Question

The freezing point of isobutane is \(-160^{\circ} \mathrm{C} . \Delta \mathrm{H}_{\text {(solid } \rightarrow \text { liquid })}\) 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 Given Information

We need to calculate the value of \( y \) given that the entropy change \( \Delta S \) of the fusion process is expressed as \( 10y \) and the enthalpy change \( \Delta H \) is \( +4520 \, \mathrm{J} \, \mathrm{mol}^{-1} \). The freezing point of isobutane is \(-160^{\circ} \mathrm{C} \), which is equivalent to \(113.15 \mathrm{~K} \) in Kelvin.

Step 2: Use the Gibbs Free Energy Equation at Equilibrium

At the freezing point, the system is in equilibrium between solid and liquid. Hence, \( \Delta G = 0 \). The Gibbs free energy equation is \( \Delta G = \Delta H - T \Delta S \). At equilibrium, this becomes \( \Delta H = T \Delta S \).

Step 3: Solve for Entropy Change \( \Delta S \)

Substitute the known values into the equation: \( 4520 = 113.15 \times (10y) \). This can be expressed as \( 4520 = 113.15 \times 10y \).

Step 4: Solve for \( y \)

Rearrange the equation to solve for \( y \): \[ y = \frac{4520}{1131.5} \]. Simplifying this gives \( y \approx 4 \).

Key Concepts

Entropy Change

Entropy change, symbolized as \( \Delta S \), measures the degree of disorder or randomness in a system. During phase changes, like the fusion of isobutane from solid to liquid, entropy change is significant. When isobutane melts at its freezing point of \(-160^{\circ} \mathrm{C}\), which is 113.15 K in Kelvin, molecules transition from a more ordered solid state to a less ordered liquid state. Disorder increases because molecules are free to move about more in the liquid than in the solid. Entropy change is expressed in \( \mathrm{J} \mathrm{mol}^{-1} \kern{0.1em} \mathrm{K}^{-1} \). In the given problem, the entropy change is expressed as \( 10y \). At equilibrium, the entropy change can be calculated using the equation relating entropy \( \Delta H = T \Delta S \). Thus,

• A positive \( \Delta S \) indicates an increase in entropy, as seen in melting.
• Understanding how to relate \( \Delta S \) to temperature and enthalpy changes is crucial.

Enthalpy Change

The enthalpy change, represented as \( \Delta H \), is the measure of heat change at constant pressure during a reaction or phase transition. It provides vital insights into the energy required in processes like melting. For isobutane, the given \( \Delta H \) is \(+4520 \mathrm{~J} \mathrm{~mol}^{-1}\), depicting how much heat is absorbed for each mole of isobutane during melting. This positive \( \Delta H \) tells us the process is endothermic, requiring energy input to overcome the attractive forces in the solid state, facilitating the transition to liquid. Here are some key takeaways:

• A positive \( \Delta H \) symbolizes an endothermic process, crucial for phase changes.
• Understanding \( \Delta H \) helps in calculating energy requirements for transitions, like melting, which is essential for comprehension of thermodynamic properties.

Gibbs Free Energy

Gibbs free energy (\( \Delta G \)) is a thermodynamic quantity that combines enthalpy and entropy to predict reaction spontaneity. At equilibrium, such as at the freezing point of isobutane, \( \Delta G = 0 \). This indicates no net change in the phase distribution of the substances involved, signifying equilibrium.The formula for Gibbs free energy is \( \Delta G = \Delta H - T \Delta S \). When \( \Delta G = 0 \), it simplifies to \( \Delta H = T \Delta S \), which is used to compute the entropy change at equilibrium conditions, critical in determining phases shifts.Key aspects to understand:

• Gibbs free energy helps determine if a reaction will occur spontaneously.
• At \( \Delta G = 0 \), the process has no net direction, indicating equilibrium.
• This understanding aids in managing and manipulating conditions to control phase changes effectively.