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Determine the entropy difference between solid and liquid states for a substance melting at \(100^{\circ} \mathrm{C}\) and having a heat of fusion of \(10,000 \mathrm{~J} / \mathrm{mol}\)

Short Answer

Expert verified
The entropy difference between the solid and liquid states for the substance melting at \(100^{\circ} \mathrm{C}\) and having a heat of fusion of \(10,000 \mathrm{~J} / \mathrm{mol}\) is 26.81 J/(mol*K).

Step by step solution

01

Convert the temperature from Celsius to Kelvin

The temperature is given in Celsius, so we'll need to convert it to Kelvin before performing our calculation. To do this, add 273.15 to the Celsius temperature: \(T = 100^{\circ} \mathrm{C} + 273.15\) \(T = 373.15 \mathrm{K}\)
02

Compute the entropy difference

Now that we have the temperature in Kelvin, we can calculate the entropy difference. The formula to determine the entropy difference is: ∆S = ∆H/T where ∆H represents the heat of fusion, given as 10,000 J/mol and T is the temperature in Kelvin, which we calculated as 373.15 K. We can now substitute the values into the formula: ∆S = (10,000 J/mol)/(373.15 K) ∆S = 26.81 J/(mol*K) The entropy difference between the solid and liquid states for the substance is 26.81 J/(mol*K).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat of Fusion
When a substance changes from a solid to a liquid, it absorbs a certain amount of energy without a change in temperature. This energy is called the heat of fusion. The heat of fusion is the energy required to change 1 mole of a solid substance to a liquid at its melting point. It is typically expressed in joules per mole (J/mol). In the given exercise, the heat of fusion is specified as 10,000 J/mol, which means this amount of energy is needed for each mole of the substance to transition from solid to liquid.
  • Important because it is indicative of a material's phase change properties.
  • Varies for different substances.
  • Used in calculating entropy changes during melting.
Understanding heat of fusion is crucial when studying thermodynamic processes, as it helps us quantify the energy changes occurring during phase transitions.
Temperature Conversion
In thermodynamics, temperatures are often needed in Kelvin rather than Celsius for calculations. The Kelvin scale is an absolute temperature scale that starts from absolute zero, the coldest possible temperature. To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. As shown in the exercise solution, the conversion of 100°C to Kelvin results in 373.15 K. This conversion is vital in many calculations:
  • Ensures accuracy when using thermodynamic formulas.
  • Enables consistent use of energy units like Joules.
  • Prevents negative temperatures in computations, which aren't meaningful in energy work.
Understanding how to convert temperatures to Kelvin ensures that thermodynamic calculations are performed correctly.
Thermodynamics
Thermodynamics is the study of energy, heat, work, and their transformations. It focuses on the movement and flow of energy which occurs in a variety of processes. The entropy change calculation in the exercise is a common thermodynamic task. Entropy measures the disorder or randomness in a system and dictates how energy flows and is distributed:
  • Key in calculating how energy is transferred during phase changes.
  • Helps determine the efficiency of thermal systems.
  • Provides insight into the spontaneity of processes.
Understanding the principles of thermodynamics is essential for analyzing processes that involve energy changes, such as melting.
Phase Change
A phase change refers to the transition of a substance from one state of matter (solid, liquid, gas) to another. During phase changes, like melting or freezing, the substance will either absorb or release heat energy. In this exercise, the phase change from solid to liquid occurs at a specific temperature, which is the melting point. Key points about phase changes are:
  • Temperature remains constant during the transition until the phase change completes.
  • Involves a change in the internal energy without changing temperature.
  • Important in calculating entropy, which measures energy distribution.
By understanding phase changes, we can better predict and calculate the energy and entropy changes that accompany these transitions.

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Most popular questions from this chapter

If the standard free energy of formation of HI from \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) at \(490^{\circ} \mathrm{C}\) is \(-12.1 \mathrm{~kJ} /\) mole of \(\mathrm{HI}\), what is the equilibrium constant for this reaction? Assume \(\mathrm{R}=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{deg}^{-1}\)

For sublimation of iodine crystals, $$ \mathrm{I}_{2}(\mathrm{~s}) \rightleftarrows \mathrm{I}_{2}(\mathrm{~g}) $$ at \(25^{\circ} \mathrm{C}\) and atmospheric pressure, it is found that the change in enthalpy, \(\Delta \mathrm{H}=9.41 \mathrm{Kcal} /\) mole and the change in entropy, \(\Delta \mathrm{S}=20.6 \mathrm{cal} /\) deg \(-\) mole. At what temperature will solid iodine be in equilibrium with gaseous iodine?

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), can be synthesized in two ways. The first method involves reduction of oxygen by hydrogen, $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}(\ell) $$ The second method involves oxidation of water: $$ 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}_{2}(\ell) $$ Find the free energy of formation, \(\Delta \mathrm{G}^{\circ}\), for both processes and predict which process is more efficient for the commercial preparation of hydrogen peroxide.

Determine \(\Delta \mathrm{G}^{\circ}\) for the reaction \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\ell)\) \(\Delta \mathrm{G}^{\circ} \mathrm{f}\) of \(\mathrm{NH}_{3}(\mathrm{~g})=-4.0 \mathrm{Kcal} / \mathrm{mole}\) \(\Delta \mathrm{G}^{\circ} \mathrm{f}\) of \(\mathrm{NO}(\mathrm{g})=20.7 \mathrm{Kcal} / \mathrm{mole}\) \(\Delta \mathrm{G}^{\circ} \mathrm{f}\) of \(\mathrm{H}_{2} \mathrm{O}(\ell)=-56.7 \mathrm{Kcal} / \mathrm{mole}\)

Given, for acetic acid that \(\Delta \mathrm{H}_{\text {fus }}=2592\) cal/mole at its melting point, \(16.6^{\circ} \mathrm{C}\) and \(\Delta \mathrm{H}_{\mathrm{VAP}}=5808 \mathrm{cal} / \mathrm{mole}\) at its boiling point, \(118.3^{\circ} \mathrm{C}\), calculate the change in entropy that takes place when 1 mole of the vapor is condensed at its boiling point and changed to a solid at its melting point, all under constant pressure, taken as 1 atm. Assume that the molar heat capacity of acetic acid is \(27.6 \mathrm{cal} /\) deg \(-\) mole.

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