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The melting point of a solid is \(27^{\circ} \mathrm{C}\) and it's latent heat of fusion is 600 calories per mole. The entropy change for the fusion of one mole of the solid (in \(\mathrm{cal} \mathrm{K}^{-1}\) ) is a. 180 b. \(22.2\) c. 2 d. \(0.5\)

Short Answer

Expert verified
The correct answer is c. 2 cal K鈦宦.

Step by step solution

01

Understand the concept

Entropy change \((\Delta S)\) during fusion (melting) can be calculated using the formula:\[\Delta S = \frac{\text{latent heat of fusion}}{\text{temperature in Kelvin}}\]. This formula relates the entropy change to the amount of heat absorbed during the phase change at a constant temperature.
02

Convert melting point to Kelvin

The melting point is given in Celsius. Convert it to Kelvin using the formula: \[T(\text{K}) = T(^{\circ}C) + 273.15\]. For this problem, \[T(\text{K}) = 27 + 273.15 = 300.15\text{ K}\].
03

Apply the formula for entropy change

Now use the formula for entropy change: \[\Delta S = \frac{600 \text{ cal/mole}}{300.15 \text{ K}}\].
04

Calculate the entropy change

Perform the division to find the entropy change: \[\Delta S = \frac{600}{300.15} \approx 2 \text{ cal K}^{-1}\].
05

Choose the correct option

The calculated entropy change is approximately 2 \(\text{cal K}^{-1}\), therefore the correct answer is option c. 2.

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Key Concepts

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

Melting Point
The melting point is a key property of a substance. It is the temperature at which a solid turns into a liquid. It's important to know that at the melting point, the solid and liquid phases exist in equilibrium. This means that the amount of solid and liquid remains constant as long as the temperature stays the same. When we talk about the melting point in scientific terms, it's usually given in Celsius or Kelvin. Remember, the melting point is the same regardless of the amount of substance. - **In practical terms:** Ice melts at 0掳C or 273.15 K. - When a solid reaches its melting point, it needs energy, called heat of fusion, to proceed with the phase change. Understanding the melting point helps us predict how materials will behave in different environments. It also gives us valuable information about the purity of the substance. The more pure the substance, the sharper and more precise its melting point.
Latent Heat of Fusion
Latent heat of fusion is the energy required to change a substance from solid to liquid at its melting point. The term "latent" signifies that this energy does not raise the temperature but instead changes the state. This energy breaks the bonds holding the molecules together in a solid. - **Significance:** Knowing the latent heat helps us calculate how much energy is needed for phase transitions. - It's usually measured in calories per mole or Joules per mole. For example, water has a latent heat of fusion of about 80 calories per gram. This means you need 80 calories to turn one gram of ice at 0掳C into water without changing its temperature. In our exercise, the latent heat of fusion is given as 600 calories per mole. This essential data point is used to compute the entropy change during phase transitions. Understanding latent heat of fusion is crucial in fields like meteorology and material science, where temperature and phase changes play a vital role.
Kelvin Temperature Conversion
Temperature is a measure of the average kinetic energy of the particles in a substance. When dealing with thermodynamic equations, using the Kelvin scale is essential. The Kelvin scale is absolute, meaning it starts at absolute zero, the point where no more thermal energy can be removed. - **Converting Celsius to Kelvin:** This involves a simple equation: \(T(\text{K}) = T(^{\circ}C) + 273.15\)This is critical for calculations related to entropy and other thermodynamic properties, where the use of Kelvin ensures consistency and accuracy. - To convert 27掳C to Kelvin, you add 273.15, resulting in 300.15 K.Remember, Kelvin is widely used in science and engineering because it avoids negative numbers in equations, simplifying calculations. Understanding Kelvin temperature conversion makes applying formulas like the one for entropy change straightforward and error-free.

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Most popular questions from this chapter

Standard molar enthalpy of formation of \(\mathrm{CO}_{2}\) is equal to a. zero b. the standard molar enthalpy of combustion of gaseous carbon. c. the sum of standard molar enthalpies of formation of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) d. the standard molar enthalpy of combustion of carbon (graphite) [IIT 1997]

For which of the following reaction is the standard entropy of reaction \(\Delta \mathrm{S}^{\circ}\) positive? a. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) c. \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) d. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Calculate \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}\) for chloride ion from the following data: \(1 / 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{L} / 2 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{HCl}(\mathrm{g})\) \(\Delta \mathrm{H}_{\mathrm{f}}^{\circ}=-92.4 \mathrm{~kJ}\) \(\mathrm{HCl}(\mathrm{g})+\mathrm{nH}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\) \(\Delta \mathrm{H}_{298}=-74.8 \mathrm{~kJ}\) \(\Delta \mathrm{H}_{\mathrm{f}}^{\mathrm{o}}\left[\mathrm{H}^{+}\right]=0.0 \mathrm{~kJ}\) a. \(-189 \mathrm{~kJ}\) b. \(-167 \mathrm{~kJ}\) c. \(+167 \mathrm{~kJ}\) d. \(-191 \mathrm{~kJ}\)

(A): Endothermic reaction is spontaneous at all temperatures. \((\mathbf{R}): \Delta \mathrm{G}\) is negative when \(\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H}\)

Write the correct sequence of entropy change when a solid melt, a gas liquefies, liquid vapourizes and a solid dissolves in water. a. Increases, increases, increases, increases respectively b. Increases, decreases, increases, decreases respectively c. Increases, decreases, increases, increases respectively d. Decreases, increases, decreases, increases respectively

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