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Tetrachloromethane (carbon tetrachloride), \(\mathrm{CCl}_{4}\), has a normal boiling point of \(76.7^{\circ} \mathrm{C}\) and an enthalpy of vaporization, \(\Delta H_{v a p}\), of \(29.82 \mathrm{~kJ} / \mathrm{mol}\). Estimate the entropy of vaporization, \(\Delta S_{v a p} \circ\). Estimate the free energy of vaporization, \(\Delta G_{\text {van }}^{\circ}\), at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
\(\Delta S_{vap}^\circ = 85.23 \text{ J/mol⋅K}\); \(\Delta G_{vap}^\circ = 426.67 \text{ J/mol}\) at 25°C.

Step by step solution

01

Understanding the Given Information

We are given the normal boiling point of CCl4 as 76.7°C and its enthalpy of vaporization as 29.82 kJ/mol. We need to find the entropy of vaporization, \(\Delta S_{vap}^\circ\), and the free energy of vaporization, \(\Delta G_{vap}^\circ\), at 25°C.
02

Convert the Boiling Point to Kelvin

The normal boiling point is given as 76.7°C. To work with thermodynamic equations, convert this to Kelvin: \(T_{b} = 76.7 + 273.15 = 349.85\, \text{K}\).
03

Calculate the Entropy of Vaporization \(\Delta S_{vap}^\circ\)

Use the formula \(\Delta S_{vap}^\circ = \frac{\Delta H_{vap}}{T_{b}}\). Substitute \(\Delta H_{vap} = 29.82\, \text{kJ/mol} = 29820\, \text{J/mol}\) and \(T_{b} = 349.85\, \text{K}\): \[\Delta S_{vap}^\circ = \frac{29820}{349.85} = 85.23\, \text{J/mol}\cdot\text{K}\].
04

Calculate the Free Energy of Vaporization \(\Delta G_{vap}^\circ\)

Use the Gibbs free energy relation \(\Delta G_{vap}^\circ = \Delta H_{vap} - T\Delta S_{vap}^\circ\). At 25°C (298.15 K), substitute \(\Delta H_{vap} = 29820\, \text{J/mol}\), \(T = 298.15\, \text{K}\), and \(\Delta S_{vap}^\circ = 85.23\, \text{J/mol}\cdot\text{K}\): \[\Delta G_{vap}^\circ = 29820 - 298.15 \times 85.23 = 426.67\, \text{J/mol}\].

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Key Concepts

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

Entropy of Vaporization
Entropy of vaporization, denoted as \( \Delta S_{vap} \), measures the change in entropy when a substance transitions from a liquid to a gas. This concept is pivotal in thermodynamics as it describes the degree of disorder or randomness introduced in the system as molecules move from being closely packed in a liquid state to being spread out in a gaseous state.
For example, when carbon tetrachloride, \( \text{CCl}_4 \), vaporizes, the entropy increases due to the greater spacing between molecules in the vapor phase. To compute the entropy of vaporization, use the formula:
\[\Delta S_{vap} = \frac{\Delta H_{vap}}{T_b}\]where \(\Delta H_{vap}\) is the enthalpy of vaporization, and \(T_b\) is the boiling point temperature in Kelvin.
This equation highlights the relationship between enthalpy and entropy, providing insight into how much energy disperses during vaporization.
Enthalpy of Vaporization
The enthalpy of vaporization, \( \Delta H_{vap} \), is the amount of energy required to convert one mole of a liquid into its gaseous state at constant pressure. It essentially represents the energy needed to overcome intermolecular forces in the liquid.
For carbon tetrachloride, this value is \( 29.82 \, \text{kJ/mol} \).
This value is significant because it indicates the strength of intermolecular attractions: stronger forces require more energy to break, resulting in a higher enthalpy.
  • Enthalpy is dependent on the nature of the liquid.
  • Temperature can also affect the value slightly.
For computations in thermodynamics, it's common to convert enthalpy from kilojoules per mole to joules per mole for consistency in units when calculating entropy or free energy.
Gibbs Free Energy
Gibbs free energy, denoted as \( \Delta G \), is a thermodynamic quantity serving as an indicator of spontaneity for a process. It combines enthalpy, entropy, and temperature, expressed by:
\[\Delta G = \Delta H - T\Delta S\]If \( \Delta G \) is negative, the process occurs spontaneously.
During the vaporization of carbon tetrachloride at \( 25^{\circ} \mathrm{C} \), we calculate \( \Delta G_{vap} \) using the given enthalpy and the previously calculated entropy.
  • \( \Delta G \) gives insight into the feasibility of vaporization under non-boiling conditions.
  • The enthalpy and entropy values guide the direction and magnitude of the free energy.
The calculations reveal \( \Delta G_{vap}^\circ = 426.67 \, \text{J/mol} \), showing that vaporization at this temperature is not entirely spontaneous, as the process would not result in a large energy decrease.

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Most popular questions from this chapter

Consider a reaction in which \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are positive. Suppose the reaction is nonspontaneous at room temperature. How would you estimate the temperature at which the reaction becomes spontaneous?

The following equation shows how nitrogen dioxide reacts with water to produce nitric acid: $$ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(l)+\mathrm{NO}(g) $$ Predict the sign of \(\Delta S^{\circ}\) for this reaction.

For each of the following reactions, decide whether there is an increase or a decrease in entropy. Why do you think so? (No calculations are needed.) a.\(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) b.\(\mathrm{NH}_{4} \mathrm{Cl}(s) \stackrel{\mathrm{NH}_{3}}(g)+\mathrm{HCl}(g)\) c.\(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\) d.\(\mathrm{Li}_{3} \mathrm{~N}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{LiOH}(a q)+\mathrm{NH}_{3}(g)\)

Adenosine triphosphate, ATP, is used as a freeenergy source by biological cells. (See the essay on page \(624 .)\) ATP hydrolyzes in the presence of enzymes to give ADP: $$ \begin{aligned} \mathrm{ATP}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow & \mathrm{ADP}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \\ & \Delta G^{\circ}=-30.5 \mathrm{~kJ} / \mathrm{mol} \text { at } 25^{\circ} \mathrm{C} \end{aligned} $$ Consider a hypothetical biochemical reaction of molecule A to give molecule \(\mathrm{B}\) : $$ \mathrm{A}(a q) \longrightarrow \mathrm{B}(a q) ; \Delta G^{\circ}=+15.0 \mathrm{~kJ} / \mathrm{mol} \text { at } 25^{\circ} \mathrm{C} $$ a.Calculate the ratio \([\mathrm{B}] /[\mathrm{A}]\) at \(25^{\circ} \mathrm{C}\) at equilibrium. b.Now consider this reaction "coupled" to the reaction for the hydrolysis of ATP: $$ \begin{array}{r} \mathrm{A}(a q)+\mathrm{ATP}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{B}(a q)+\operatorname{ADP}(a q)+ \\ \mathrm{H}_{2} \mathrm{PO}_{4}(a q) \end{array} $$ If a cell maintains a high ratio of ATP to ADP and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) by continuously making ATP, the conversion of \(\mathrm{A}\) to \(\mathrm{B}\) can be made highly spontaneous. A characteristic value of this ratio is $$ \frac{[\mathrm{ATP}]}{[\mathrm{ADP}]\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]}=500 $$ Calculate the ratio [B][A] in this case and compare it with the uncoupled reaction. Compared with the uncoupled reaction, how much larger is this ratio when coupled to the hydrolysis of ATP?

Hypothetical elements \(\mathrm{A}(g)\) and \(\mathrm{B}(g)\) are introduced into a container and allowed to react according to the reaction \(\mathrm{A}(g)+2 \mathrm{~B}(g) \longrightarrow \mathrm{AB}_{2}(g)\). The container depicts the reaction mixture after equilibrium has been attained. a.Is the value of \(\Delta S\) for the reaction positive, negative, or zero? b.Is the value of \(\Delta H\) for the reaction positive, negative, or zero? c.Prior to equilibrium, is the value of \(\Delta G\) for the reaction positive, negative, or zero? d.At equilibrium, is the value of \(\Delta G\) for the reaction positive, negative, or zero?

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