The normal freezing point of n-octane (C 8 H 18 ) is -57 °C. (b) In what temperature range is the freezing of n-octane a spontaneous process?

insert step 1\u003e Determine the concept of spontaneity in terms of Gibbs free energy ($$ \Delta G ). A $$process is spontaneous when $$ \Delta G Use the Gibbs free energy equation: \( \Delta G = \Delta H - T\Delta S $$, where $$ \Delta H is $$the enthalpy change, $$ T is $$the temperature in Kelvin, and $$ \Delta S is $$the entropy change.. insert step 3\u003e Recognize that at the freezing point, $$ \Delta G = 0 $$, so $$ \Delta H = T_f \Delta S $$, where $$ T_f is $$the freezing point in Kelvin.. insert step 4\u003e Convert the freezing point from Celsius to Kelvin: $$ T_f = -57 + 273.15 .. $$insert step 5\u003e Determine the temperature range for spontaneity: For freezing to be spontaneous, the temperature must be below the freezing point, so $$ T \u003c T_f .$$

The normal freezing point of n-octane (C8H18) is -57 °C. (b) In what temperature range is the freezing of n-octane a spontaneous process?

insert step 1\u003e Determine the concept of spontaneity in terms of Gibbs free energy ($$ \Delta G ). A $$process is spontaneous when $$ \Delta G Use the Gibbs free energy equation: \( \Delta G = \Delta H - T\Delta S $$, where $$ \Delta H is $$the enthalpy change, $$ T is $$the temperature in Kelvin, and $$ \Delta S is $$the entropy change. insert step 3\u003e Recognize that at the freezing point, $$ \Delta G = 0 $$, so $$ \Delta H = T_f \Delta S $$, where $$ T_f is $$the freezing point in Kelvin. insert step 4\u003e Convert the freezing point from Celsius to Kelvin: $$ T_f = -57 + 273.15 . $$insert step 5\u003e Determine the temperature range for spontaneity: For freezing to be spontaneous, the temperature must be below the freezing point, so $$ T \u003c T_f .$$

Ch.19 - Chemical Thermodynamics

Brown - Chemistry: The Central Science 15th Edition

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Ch.19 - Chemical Thermodynamics

Problema 16b

Capitolo 19, Problema 16b

The normal freezing point of n-octane (C₈H₁₈) is -57 °C. (b) In what temperature range is the freezing of n-octane a spontaneous process?

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insert step 1> Determine the concept of spontaneity in terms of Gibbs free energy ($ \Delta G $). A process is spontaneous when $ \Delta G < 0 $.

insert step 2> Use the Gibbs free energy equation: $ \Delta G = \Delta H - T\Delta S $, where $ \Delta H $ is the enthalpy change, $ T $ is the temperature in Kelvin, and $ \Delta S $ is the entropy change.

insert step 3> Recognize that at the freezing point, $ \Delta G = 0 $, so $ \Delta H = T_f \Delta S $, where $ T_f $ is the freezing point in Kelvin.

insert step 4> Convert the freezing point from Celsius to Kelvin: $ T_f = -57 + 273.15 $.

insert step 5> Determine the temperature range for spontaneity: For freezing to be spontaneous, the temperature must be below the freezing point, so $ T < T_f $.

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Freezing Point

The freezing point of a substance is the temperature at which it transitions from a liquid to a solid state. For n-octane, this point is -57 °C, meaning that at temperatures below this value, n-octane will solidify. Understanding the freezing point is crucial for determining the conditions under which a substance will freeze spontaneously.

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Freezing Point Depression

Spontaneity of Processes

A spontaneous process is one that occurs without needing to be driven by an external force. In thermodynamics, spontaneity is often determined by changes in enthalpy and entropy. For freezing to be spontaneous, the system must favor lower energy states, which typically occurs at temperatures below the freezing point of the substance.

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Spontaneity of Processes

Thermodynamic Stability

Thermodynamic stability refers to the tendency of a system to remain in a particular state under given conditions. For n-octane, below its freezing point, the solid state is more stable than the liquid state. This stability is influenced by factors such as temperature and pressure, which dictate whether the freezing process will occur spontaneously.

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First Law of Thermodynamics

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