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The standard entropy of \(\mathrm{Pb}(s)\) at \(298.15 \mathrm{K}\) is \(64.80 \mathrm{J}\) \(\mathrm{K}^{-1} \mathrm{mol}^{-1}\). Assume that the heat capacity of \(\mathrm{Pb}(s)\) is given by \\[ \frac{C_{P, m}(\mathrm{Pb}, s)}{\mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=22.13+0.01172 \frac{T}{\mathrm{K}}+1.00 \times 10^{-5} \frac{T^{2}}{\mathrm{K}^{2}} \\] The melting point is \(327.4^{\circ} \mathrm{C}\) and the heat of fusion under these conditions is \(4770 . \mathrm{J} \mathrm{mol}^{-1}\). Assume that the heat capacity of \(\mathrm{Pb}(l)\) is given by \\[ \frac{C_{P, m}(\mathrm{Pb}, l)}{\mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}}=32.51-0.00301 \frac{T}{\mathrm{K}} \\] a. Calculate the standard entropy of \(\mathrm{Pb}(l)\) at \(725^{\circ} \mathrm{C}\). b. Calculate \(\Delta H\) for the transformation \(\mathrm{Pb}\left(s, 25.0^{\circ} \mathrm{C}\right) \rightarrow\) \(\mathrm{Pb}\left(L, 725^{\circ} \mathrm{C}\right)\)

Short Answer

Expert verified
a. The standard entropy of \(\mathrm{Pb}(l)\) at \(725^{\circ} C\) (998.15 K) is approximately 103.57 J K\(^{-1}\) mol\(^{-1}\). b. The enthalpy change for the transformation \(\mathrm{Pb}(s, 25.0^{\circ} \mathrm{C}) \rightarrow \mathrm{Pb}(l, 725^{\circ} \mathrm{C})\) is approximately 27,084 J mol\(^{-1}\).

Step by step solution

01

Convert temperature to Kelvin

Conversion of temperature from Celsius to Kelvin is necessary as all the temperature dependent parameters are given in Kelvin. The conversion is done by adding 273.15 to the Celsius temperature. For example, for 725°C: \[ T = 725 + 273.15 = 998.15 \, K \]
02

Calculate entropy change from 298.15K to melting point

Integrate the molar heat capacity of solid lead (Pb) from 298.15K to its melting point (327.4°C = 600.55K) to determine the entropy change. The equation is: \[ \Delta S_1 = \int_{298.15}^{600.55} \frac{C_{P,m}}{T} dT \] where \[ C_{P, m}(\mathrm{Pb}, s)=22.13+0.01172 T+1.00 \times 10^{-5} T^{2} \] Use the fundamental theorem of calculus to solve the definite integral.
03

Calculate entropy change at melting point

Calculate the entropy change (\(\Delta S_2\)) during the fusion process (change from solid to liquid at the melting point) using the heat of fusion (\(\Delta H_{fus}\)) and the melting point. The equation is: \[ \Delta S_2 = \frac{\Delta H_{fus}}{T_{melt}} = \frac{4770}{600.55} \]
04

Calculate entropy change from melting point to 998.15K

Integrate the molar heat capacity of liquid lead from the melting point to the target temperature (998.15K) to determine the entropy change. The equation is: \[ \Delta S_3 = \int_{600.55}^{998.15} \frac{C_{P,m}}{T} dT \] where \[ C_{P, m}(\mathrm{Pb}, l)=32.51-0.00301 T \] Use the fundamental theorem of calculus to solve the definite integral.
05

Sum up all entropy changes and add the standard entropy at 298.15K

Add up all changes of entropy (\(\Delta S_1\), \(\Delta S_2\), \(\Delta S_3\)) and the standard entropy of solid lead at 298.15K (64.80 J mol-1 K-1) to get the standard entropy of liquid lead at 998.15K. \[ S(998.15K) = S(298.15K) + \Delta S_1 + \Delta S_2 + \Delta S_3 \] #b. Calculate \(\Delta H\) for the transformation Pb(s, 25.0°C) -> Pb(l, 725°C)
06

Calculate the change in enthalpy for solid lead from 298.15K to melting point

Integrate the molar heat capacity of solid lead from 298.15K to its melting point (600.55K) to determine the enthalpy change. The equation is: \[ \Delta H_1 = \int_{298.15}^{600.55} C_{P, m} dT \]
07

Calculate the change in enthalpy at melting point

The enthalpy change at the melting point is the heat of fusion (\(\Delta H_{fus}\)).
08

Calculate the change in enthalpy for liquid lead from melting point to 998.15K

Integrate the molar heat capacity of liquid lead from the melting point to the target temperature (998.15K) to determine the enthalpy change. The equation is: \[ \Delta H_3 = \int_{600.55}^{998.15} C_{P, m} dT \]
09

Sum up all changes in enthalpy

Add up all changes of enthalpy (\(\Delta H_1\), \(\Delta H_{fus}\), \(\Delta H_3\)) to get the total enthalpy change for the transformation of Pb(s, 25.0°C) -> Pb(l, 725°C). \[ \Delta H = \Delta H_1 + \Delta H_{fus} + \Delta H_3 \]

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Key Concepts

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

Heat Capacity
Heat capacity is an essential concept in thermodynamics and relates to the amount of heat required to increase the temperature of a substance by one degree Celsius (or one Kelvin). Specifically, molar heat capacity (\(C_{P,m}\)) is the heat capacity per mole of a substance. In the context of standard entropy calculation, understanding heat capacity is crucial because it helps determine how much a substance's temperature will change with the addition of heat.

This property varies with temperature, and for substances like lead (\textbf{Pb}), the heat capacity is often expressed as a temperature-dependent equation. The heat capacity equation for lead in its solid form differs from that in its liquid form, reflecting the different ways heat is absorbed or released due to changes in the substance's structure and bonding when it undergoes a phase transition from solid to liquid.

To calculate entropy changes for lead transitions, one must integrate these heat capacity equations over the temperature range of interest. This integration is part of the process to calculate changes in the system's thermal energy, which relates directly to entropy changes.
Enthalpy Change
Enthalpy change (\(\Delta H\)), in thermodynamics, represents the total heat content change within a system under constant pressure. It's a measure of the energy absorbed or released during a process, such as when a substance melts or boils. Enthalpy changes during chemical reactions signify whether the process is endothermic (absorbs heat) or exothermic (releases heat).

Calculating the enthalpy change involves integrating the heat capacity over the temperature range, accounting for phase changes (like melting) and changes in temperature. In the exercise, this calculation would provide insight into the overall energy change as lead (\textbf{Pb}) transitions from solid at room temperature to liquid at a higher temperature. The enthalpy change associated with the melting point itself is represented by the \(\Delta H_{fus}\), or the heat of fusion, which is the energy required to change a substance from solid to liquid at constant temperature and pressure.
Melting Point
The melting point of a substance is the temperature at which it transitions from a solid to a liquid state. This physical property is not only indicative of the substance's purity but also determines important aspects of its thermal profile and phase behavior. For instance, lead (\textbf{Pb}) melts at 327.4°C, and this precise temperature is a critical data point for any calculation involving phase change enthalpy or entropy.

In thermodynamics, the melting point is especially relevant when determining the entropy change associated with the phase transition of melting. At the melting point, the ordered structure of a solid becomes the more disordered liquid state, which involves an increase in entropy. Hence, the melting point is closely tied to the energy changes and can be used to calculate the entropy change during melting (\(\Delta S_2\)) by dividing the heat of fusion by the melting temperature (in Kelvin).
Heat of Fusion
The heat of fusion (\(\Delta H_{fus}\)) is a critical thermodynamic quantity that refers to the amount of energy needed to change a substance from solid to liquid at its melting point without changing its temperature. This energy is associated with overcoming the forces that hold the solid together. Every substance has a characteristic heat of fusion, which is essential for calculations related to phase changes.

In the given exercise, the heat of fusion is a fixed value (4770 J/mol for lead) and is used to calculate the entropy change during the melting process (\(\Delta S_2\)). To find the standard entropy of lead in its liquid state at a specific temperature, incorporating the heat of fusion is necessary because it represents the magnitude of the disorder increase when lead transitions from solid to liquid. Calculating the total enthalpy change for a phase transformation also involves adding the heat of fusion, reflecting the direct energy input necessary to effect the phase change at the melting point.

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Most popular questions from this chapter

The average heat evolved by the oxidation of foodstuffs in an average adult per hour per kilogram of body weight is \(7.20 \mathrm{kJ} \mathrm{kg}^{-1} \mathrm{hr}^{-1}\). Assume the weight of an average adult is \(62.0 \mathrm{kg} .\) Suppose the total heat evolved by this oxidation is transferred into the surroundings over a period lasting one week. Calculate the entropy change of the surroundings associated with this heat transfer. Assume the surroundings are at \(T=293 \mathrm{K}\)

Calculate \(\Delta S^{\circ}\) for the reaction \(3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow\) \(2 \mathrm{NH}_{3}(g)\) at \(725 \mathrm{K} .\) Omit terms in the temperature-dependent heat capacities higher than \(T^{2} / \mathrm{K}^{2}\)

a. Calculate \(\Delta S\) if 1.00 mol of liquid water is heated from \(0.00^{\circ}\) to \(10.0^{\circ} \mathrm{C}\) under constant pressure and if \(C_{P, m}=\) \(75.3 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\) b. The melting point of water at the pressure of interest is \(0.00^{\circ} \mathrm{C}\) and the enthalpy of fusion is \(6.010 \mathrm{kJ} \mathrm{mol}^{-1} .\) The boiling point is \(100 .^{\circ} \mathrm{C}\) and the enthalpy of vaporization is \(40.65 \mathrm{kJ} \mathrm{mol}^{-1} .\) Calculate \(\Delta S\) for the transformation \(\mathrm{H}_{2} \mathrm{O}\left(s, 0^{\circ} \mathrm{C}\right) \rightarrow \mathrm{H}_{2} \mathrm{O}\left(g, 100 \cdot^{\circ} \mathrm{C}\right)\)

The amino acid glycine dimerizes to form the dipeptide glycylglycine according to the reaction \\[ 2 \text { Glycine }(s) \rightarrow \text { Glycylglycine }(s)+\mathrm{H}_{2} \mathrm{O}(l) \\] Calculate \(\Delta S, \Delta S_{\text {surr}},\) and \(\Delta S_{\text {universe}}\) at \(T=298 \mathrm{K}\). Useful thermodynamic data follow: $$\begin{array}{lccc} & \text { Glycine } & \text { Glycylglycine } & \text { Water } \\ \hline \Delta H_{f}^{\circ}\left(\mathrm{kJ} \mathrm{mol}^{-1}\right) & -537.2 & -746.0 & -285.8 \\ S_{m}^{\circ}\left(\mathrm{JK}^{-1} \mathrm{mol}^{-1}\right) & 103.5 & 190.0 & 70.0 \end{array}$$

Calculate \(\Delta S, \Delta S_{\text {total}},\) and \(\Delta S_{\text {surroundings}}\) when the volume of \(150 .\) g of \(\mathrm{CO}\) initially at \(273 \mathrm{K}\) and 1.00 bar increases by a factor of two in (a) an adiabatic reversible expansion, (b) an expansion against \(P_{\text {external}}=0,\) and (c) an isothermal reversible expansion. Take \(C_{P, m}\) to be constant at the value \(29.14 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}\) and assume ideal gas behavior. State whether each process is spontaneous. The temperature of the surroundings is \(273 \mathrm{K}\)

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