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Chapter 18: Problem 32

The molar heats of fusion and vaporization of ethanol are 7.61 and \(26.0 \mathrm{~kJ} / \mathrm{mol}\), respectively. Calculate the molar entropy changes for the solid-liquid and liquidvapor transitions for ethanol. At 1 atm pressure, ethanol melts at \(-117.3^{\circ} \mathrm{C}\) and boils at \(78.3^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The molar entropy change is 48.8 J/mol K for fusion and 74.0 J/mol K for vaporization.

Step by step solution

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01

Understanding the Formula

Entropy change for a phase transition is given by the formula: \( \Delta S = \frac{\Delta H}{T} \). Here, \( \Delta H \) is the molar enthalpy change (either for fusion or vaporization), and \( T \) is the absolute temperature in Kelvin.
02

Calculate the Melting Temperature in Kelvin

Convert the melting point of ethanol from Celsius to Kelvin using the formula: \( T(K) = T(^{\circ}C) + 273.15 \). For ethanol melting: \( T = -117.3 + 273.15 = 155.85 \) K.
03

Calculate the Boiling Temperature in Kelvin

Convert the boiling point of ethanol from Celsius to Kelvin using the formula: \( T(K) = T(^{\circ}C) + 273.15 \). For ethanol boiling: \( T = 78.3 + 273.15 = 351.45 \) K.
04

Compute Molar Entropy Change for Fusion

Use the formula \( \Delta S = \frac{\Delta H}{T} \) to calculate the entropy change for fusion. Here, \( \Delta H = 7.61 \) kJ/mol and \( T = 155.85 \) K. \[ \Delta S_{fusion} = \frac{7.61 \times 10^3}{155.85} = 48.8 \text{ J/mol K} \].
05

Compute Molar Entropy Change for Vaporization

Use the formula \( \Delta S = \frac{\Delta H}{T} \) to calculate the entropy change for vaporization. Here, \( \Delta H = 26.0 \) kJ/mol and \( T = 351.45 \) K. \[ \Delta S_{vaporization} = \frac{26.0 \times 10^3}{351.45} = 74.0 \text{ J/mol K} \].

Key Concepts

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

Molar Heat of Fusion
When a substance transitions from a solid to a liquid, it absorbs a specific amount of energy without changing its temperature. This energy is known as the molar heat of fusion. The molar heat of fusion is crucial because it represents the energy required to overcome intermolecular forces maintaining the solid state. For ethanol, this is measured as 7.61 kJ/mol.To find out how much chaos or disorder—or entropy—increases during this transition, we can utilize the formula:- \[\Delta S = \frac{\Delta H}{T}\] Here, \(\Delta H\) represents the molar heat of fusion, and \(T\) is the absolute temperature in Kelvin. In calculating the molar entropy change for the melting of ethanol, this energy gets divided by the melting temperature (in Kelvin) to find the increase in entropy.
Molar Heat of Vaporization
The molar heat of vaporization refers to the amount of energy required to transform a substance from a liquid into a gas. This process occurs at a constant temperature and pressure. For ethanol, this value is 26.0 kJ/mol. This energy not only breaks intermolecular forces but also allows molecules to freely move in a gaseous state.Just like in fusion, the change in entropy, \(\Delta S\), for vaporization can be calculated using: - \[\Delta S = \frac{\Delta H}{T}\] Where \(\Delta H\) is the molar heat of vaporization, and \(T\) is the boiling temperature of ethanol in Kelvin. The increase in entropy for vaporization is generally larger than for fusion, as the transition from liquid to gas involves more freedom of movement.
Phase Transition
Phase transition involves changes in state of matter, such as solid to liquid or liquid to gas. During these transitions, substances absorb or release energy, but the temperature remains constant throughout the change.The key phase transitions include: - **Fusion**: Solid to liquid, which involves the molar heat of fusion. This transition for ethanol happens at \(-117.3^{\circ} \mathrm{C}\), when ethanol melts.- **Vaporization**: Liquid to gas, involving the molar heat of vaporization. Ethanol boils at \(78.3^{\circ} \mathrm{C}\). During these transitions, the substance's entropy changes, reflecting how disordered or chaotic the system becomes. Calculating molar entropy change involves dividing the heat absorbed or released by the transition's temperature in Kelvin.
Temperature Conversion
Temperature conversion between Celsius and Kelvin is essential when calculating changes in molar entropy. This is because entropy calculations rely on absolute temperature (Kelvin scale).The conversion formula is straightforward:- \[T(\mathrm{K}) = T(^{\circ} \mathrm{C}) + 273.15\]For ethanol:- **Melting point conversion**: From \(-117.3^{\circ} \mathrm{C}\) to \(155.85 \mathrm{K}\).- **Boiling point conversion**: From \(78.3^{\circ} \mathrm{C}\) to \(351.45 \mathrm{K}\). Accurately converting temperatures is crucial as even a minor error could lead to significant discrepancies in entropy calculations.

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Most popular questions from this chapter

A certain reaction is known to have a \(\Delta G^{\circ}\) value of \(-122 \mathrm{~kJ} / \mathrm{mol}\). Will the reaction necessarily occur if the reactants are mixed together?

The standard enthalpy of formation and the standard entropy of gaseous benzene are \(82.93 \mathrm{~kJ} / \mathrm{mol}\) and \(269.2 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\), respectively. Calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) for the given process at \(25^{\circ} \mathrm{C}\). Comment on your answers: $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g) $$

For the autoionization of water at \(25^{\circ} \mathrm{C}:\) $$ \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ \(K_{\mathrm{w}}\) is \(1.0 \times 10^{-14}\). What is \(\Delta G^{\circ}\) for the process?

When a native protein in solution is heated to a high enough temperature, its polypeptide chain will unfold to become the denatured protein. The temperature at which a large portion of the protein unfolds is called the melting temperature. The melting temperature of a certain protein is found to be \(46^{\circ} \mathrm{C},\) and the enthalpy of denaturation is \(382 \mathrm{~kJ} / \mathrm{mol}\). Estimate the entropy of denaturation, assuming that the denaturation is a twostate process; that is, native protein \(\longrightarrow\) denatured protein. The single polypeptide protein chain has 122 amino acids. Calculate the entropy of denaturation per amino acid. Comment on your result.

Calculate the pressure of \(\mathrm{O}_{2}\) (in \(\mathrm{atm}\) ) over a sample of \(\mathrm{NiO}\) at \(25^{\circ} \mathrm{C}\) if \(\Delta G^{\circ}=212 \mathrm{~kJ} / \mathrm{mol}\) for the reaction: $$ \mathrm{NiO}(s) \rightleftharpoons \mathrm{Ni}(s)+\frac{1}{2} \mathrm{O}_{2}(g) $$
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