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Calculate the melting point of benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right)\), given the following data: \(\Delta H_{\text {fusion }}=18.02 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta S_{\text {fusion }}=\) \(45.56 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})\)

Short Answer

Expert verified
The melting point of benzoic acid is 395.67 K.

Step by step solution

01

Understand the Relationship

To find the melting point, we use the equation that relates enthalpy and entropy changes during fusion: \( T = \frac{\Delta H_{\text{fusion}}}{\Delta S_{\text{fusion}}} \). This equation comes from the equality established at phase change equilibrium: \( \Delta G = 0 = \Delta H - T\Delta S \).
02

Insert Given Values

Use the values provided: \( \Delta H_{\text{fusion}} = 18.02 \text{ kJ/mol} = 18020 \text{ J/mol} \) and \( \Delta S_{\text{fusion}} = 45.56 \text{ J/(K} \cdot \text{mol)} \).
03

Calculate the Melting Point

Substitute the values into the equation: \[ T = \frac{18020 \text{ J/mol}}{45.56 \text{ J/(K} \cdot \text{mol)}} = 395.67 \text{ K} \]. Evaluate the fraction to get the melting point in Kelvin.

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Key Concepts

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

Phase Change Equilibrium
During a phase change, such as melting, there is a special state where both the solid and liquid phases of a substance coexist. This state is called phase change equilibrium. At this point, the temperature remains constant despite the energy exchange taking place.
When a substance, like benzoic acid, melts, its internal structure is disrupted, allowing a transition from the ordered solid state to a more disordered liquid state. The energy required to achieve this transition is exactly balanced by the heat being absorbed or released, a condition described by the equation \[ \Delta G = 0 = \Delta H - T\Delta S \]This implies no net change in Gibbs free energy (\(\Delta G = 0\)), signifying equilibrium between phases. By solving this equation for temperature, \[ T = \frac{\Delta H}{\Delta S} \]we can find the melting point where this balance occurs. This is the condition used in the original exercise to determine the melting point of benzoic acid.
Enthalpy of Fusion
Enthalpy of fusion, denoted as \(\Delta H_{\text{fusion}}\), is the amount of energy needed to change a substance from a solid to a liquid at its melting point. This value represents the heat absorbed by one mole of the substance during melting without any temperature change.
Benzoic acid requires \(18.02 \text{ kJ/mol}\) of energy for fusion. Understanding this value is essential because it indicates the strength of interactions within a solid. Greater enthalpy values suggest stronger forces holding the molecules together in a solid structure. In the case of benzoic acid, this gives us an insight into its relatively stable solid form compared to when it transitions to a liquid.
This concept not only helps in calculating the melting point but also provides a deeper understanding of the material's thermal properties and stability.
Entropy of Fusion
Entropy of fusion, represented as \(\Delta S_{\text{fusion}}\), measures the change in molecular randomness or disorder as a substance transitions from a solid to a liquid.
In the context of benzoic acid, the entropy of fusion is \(45.56 \text{ J/(K} \cdot \text{mol)}\). This value helps explain how molecules in the structured, ordered solid state rearrange into a less ordered liquid state. Higher entropy values suggest more significant changes in molecular freedom and disorder during melting.
In calculating melting points, entropy gives insight into the degree of disorder associated with phase changes. In our exercise, combining entropy with enthalpy helps us compute the exact temperature at which benzoic acid melts, showcasing the interplay of energy and disorder in phase transitions.
Benzoic Acid
Benzoic acid is a simple aromatic carboxylic acid with the formula \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{CO}_{2}\mathrm{H}\). It is widely used as a food preservative and has various chemical applications.
As it undergoes a phase change from solid to liquid, understanding its melting point is crucial for its use in industry and research. The stability in solid form and controlled melting conditions are essential for its storage and handling.
By applying calculated thermodynamic data, such as enthalpy and entropy of fusion, we can precisely determine conditions under which benzoic acid can transition between states. This allows users to better predict behaviors and optimize its various practical applications, from preserving food to synthesizing polymers.

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Most popular questions from this chapter

Sorbitol \(\left(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}_{6}\right)\), a substance used as a sweetener in foods, is prepared by the reaction of glucose with hydrogen in the presence of a catalyst: $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}(a q)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}_{6}(a q)$$ Which of the following quantities are affected by the catalyst? (a) Rate of the forward reaction (b) Rate of the reverse reaction (c) Spontaneity of the reaction (d) \(\Delta H^{\circ}\) (e) \(\Delta S^{\circ}\) (f) \(\Delta G^{\circ}\) (g) The equilibrium constant (h) Time required to reach equilibrium

What is meant by the standard state of a substance?

Which state has higher entropy? Explain in terms of probability. (a) \(1 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) gas at STP or \(1 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) gas at \(273 \mathrm{~K}\) and \(0.10\) atm (b) A crystal of \(\mathrm{NaCl}\) at \(0{ }^{\circ} \mathrm{C}\) or the same crystal at \(50^{\circ} \mathrm{C}\)

Consider the following endothermic decomposition of \(\mathrm{AB}_{2}\) molecules: (a) What is the sign \((+,-\), or 0\()\) of \(\Delta S^{\circ}\) for the reaction? (b) Is the reaction more likely to be spontaneous at high temperatures or at low temperatures? Explain.

Predict the sign of \(\Delta S\) in the system for each of the following reactions: (a) \(\mathrm{PCl}_{5}(s) \longrightarrow \mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g)\) (b) \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (c) \(2 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \longrightarrow \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{Mg}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{MgCl}_{2}(s)\)

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