Chloroform has ΔHvaporization = 29.2 kJ>mol and boils at 61.2 °C. What is the value of ΔSvaporization for chloroform?

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Identify the given values: ΔHvaporization = 29.2 kJ/mol, boiling point = 61.2 °C.

Convert the boiling point from Celsius to Kelvin by adding 273.15 to the Celsius temperature. T(K) = 61.2 °C + 273.15.

Use the formula for entropy change during vaporization, ΔSvaporization = ΔHvaporization / T, where T is the temperature in Kelvin.

Substitute the values of ΔHvaporization and T into the formula to find ΔSvaporization.

Ensure the units of ΔSvaporization are in J/(mol·K) by converting kJ to J if necessary.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Enthalpy of Vaporization (ΔHvaporization)

The enthalpy of vaporization is the amount of energy required to convert a unit mass of a liquid into vapor without a change in temperature. It is typically expressed in kJ/mol and is a crucial factor in understanding phase changes, particularly when calculating the energy changes associated with boiling or evaporating substances.

Entropy of Vaporization (ΔSvaporization)

The entropy of vaporization is a measure of the disorder or randomness that occurs when a liquid transitions to a gas. It is calculated using the relationship ΔSvaporization = ΔHvaporization / T, where T is the temperature in Kelvin. This concept is essential for understanding the thermodynamic favorability of phase changes.

Temperature Conversion

In thermodynamics, it is important to use the absolute temperature scale (Kelvin) when performing calculations involving energy changes. To convert Celsius to Kelvin, add 273.15 to the Celsius temperature. This conversion is necessary for accurately applying formulas that involve temperature, such as those relating to enthalpy and entropy.

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