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Calculate \(\Delta S_{R}^{\circ}\) for the reaction \(\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightarrow\) \(2 \mathrm{HCl}(g)\) at \(870 . \mathrm{K} .\) Omit terms in the temperature-dependent heat capacities higher than \(T^{2} / \mathrm{K}^{2}\)

Short Answer

Expert verified
To calculate the standard entropy change \(\Delta S_{R}^{\circ}\) at 870 K, first find the heat capacities at constant pressure (\(C_{p}^{\circ}\)) for each species. Then, find the change in heat capacity of the reaction \(\Delta C_{p}^{\circ}\) and calculate \(\Delta S_{R}^{\circ}\) by integrating the change in heat capacity with respect to temperature from 298 K to 870 K. The final value of \(\Delta S_{R}^{\circ}\) will be in J/mol·K.

Step by step solution

01

Find heat capacities

The heat capacities of each species at constant pressure (\(C_{p}^{\circ}\)) can be found in standard tables or given in the problem. Use the formulas: \(C_{p,H_{2}}^{\circ}=a_{H_{2}}+b_{H_{2}}\left(T\right)+c_{H_{2}}\left(T^{2}\right)\) \(C_{p,Cl_{2}}^{\circ}=a_{Cl_{2}}+b_{Cl_{2}}\left(T\right)+c_{Cl_{2}}\left(T^{2}\right)\) \(C_{p,HCl}^{\circ}=a_{HCl}+b_{HCl}\left(T\right)+c_{HCl}\left(T^{2}\right)\) where \(a, b\), and \(c\) are constants for each species and \(T\) is the temperature in Kelvin. It is important to note that we are asked to omit terms in the temperature-dependent heat capacities higher than \(T^{2} / \mathrm{K}^{2}\), so only consider up to the \(T^{2}\) term.
02

Calculate the change in heat capacity of the reaction

Now that we have the heat capacities for each species, we can find the change in heat capacity of the reaction \(\Delta C_{p}^{\circ}\) using the stoichiometric coefficients: \(\Delta C_{p}^{\circ}=2C_{p,HCl}^{\circ}-\left(C_{p,H_{2}}^{\circ}+C_{p,Cl_{2}}^{\circ}\right)\)
03

Calculate the change in standard entropy for the reaction

Once we have the change in heat capacity \(\Delta C_{p}^{\circ}\), we can calculate \(\Delta S_{R}^{\circ}\) by integrating the change in heat capacity of the reaction with respect to temperature, and then multiplying by the temperature ratio. This formula will help to find the change in entropy of the reaction: \(\Delta S_{R}^{\circ}=\int_{T_{1}}^{T_{2}} \Delta C_{p}^{\circ} \frac{dT}{T}\) In this case, \(T_{1}=298 \ \text{K}\) (standard conditions) and \(T_{2}=870\ \text{K}\) (given temperature). Plug the values of \(\Delta C_{p}^{\circ}\), \(T_{1}\), and \(T_{2}\) in the formula and perform the integration.
04

Report the final value of the standard entropy change

By following the steps mentioned above and performing the calculations, we will obtain the final value of \(\Delta S_{R}^{\circ}\). Report this value in J/mol·K. Make sure to include the appropriate units in your final answer.

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Key Concepts

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

Heat Capacities at Constant Pressure
Understanding heat capacities at constant pressure, often denoted as \( C_{p}^{\text{o}} \), is quintessential when studying thermodynamics, particularly in the context of chemical reactions. A substance’s heat capacity is the amount of heat required to raise its temperature by one degree Celsius or Kelvin. At constant pressure, this thermal energy change allows the substance to do work and possibly change its volume.

In a chemical reaction, each reactant and product has a unique \( C_{p}^{\text{o}} \). These values are temperature-dependent and can generally be expressed as a polynomial of temperature \( T \), however, the exercise at hand emphasizes omitting higher temperature-dependency terms beyond \( T^{2} \). This is because these higher-order terms have negligible impacts at the temperature range in question, focusing our attention on relevant, significant contributions to the heat capacity for a more straightforward calculation.

Impact on Reaction Entropy

The change in heat capacity for a reaction, represented by \( \triangle C_{p}^{\text{o}} \), directly influences the entropy change of the reaction. In the given exercise, students are aided in understanding the stoichiometric balance of heat capacities, which is critical for predicting how heat transfer affects entropy, and in turn, the spontaneity of a reaction.
Gibbs Free Energy Reaction
The concept of Gibbs free energy reaction, symbolized as \( G \), connects the intrinsic energy changes within a chemical reaction to its spontaneity—whether a reaction will proceed without external inputs. The equation \( \triangle G = \triangle H - T\triangle S \) illustrates how Gibbs free energy (\( \triangle G \)) is determined by the reaction’s enthalpy change (\( \triangle H \)) and entropy change (\( \triangle S \)), alongside the temperature \( T \).

It is important for students to appreciate that a negative value of \( \triangle G \) indicates a spontaneous process at a constant temperature and pressure. Meanwhile, understanding how to calculate the standard entropy change, \( \triangle S_{R}^{\text{o}} \), is fundamental as it is a key component in determining Gibbs free energy. The ability to connect changes in heat capacity to changes in entropy, therefore, informs students not only about the thermodynamic favorability of a reaction but also the direction in which a reaction naturally tends to progress.
Entropy and Thermodynamics
Entropy, a central theme in thermodynamics, measures the dispersal of energy within a system and is indicated by the symbol \( S \). In the realm of chemistry, it has significant implications for chemical reactions, as it gauges the degree of disorder or randomness at the molecular level.

Entropy is particularly fascinating because it brings a statistical angle to the understanding of thermodynamic processes—wider distributions of energy states correspond to higher entropy. In a chemical reaction, the entropy change, \( \triangle S_{R}^{\text{o}} \), informs us of the change in energy distribution among the reactants and products.

Calculating Entropy Change

Calculating the standard entropy change involves integrating the change in heat capacity (\( \triangle C_{p}^{\text{o}} \)) over the temperature range with respect to the inverse of temperature. The exercise improvement advice recommends focusing on clarity here; it is crucial to recognize how changes in temperature affect entropy through the relationship with heat capacity—an aspect well-illustrated by integrating \( \triangle C_{p}^{\text{o}} \frac{dT}{T} \) across the given temperature range.
Chemical Reaction Stoichiometry
Chemical reaction stoichiometry deals with the quantification of reactants and products in a chemical reaction, ensuring that atoms are conserved in the process and the reaction obeys the law of conservation of mass. Stoichiometry is the arithmetic behind chemistry: the coefficients in a balanced chemical equation exemplify the relative amounts of moles of each substance involved.

In the context of entropy and thermodynamics, stoichiometric coefficients help determine the change in heat capacities needed to compute the standard entropy change of a reaction, \( \triangle S_{R}^{\text{o}} \). For instance, in the reaction of hydrogen and chlorine gas forming hydrochloric acid, understanding that two moles of HCl are produced for every mole of hydrogen and chlorine reacted is fundamental.

It is through stoichiometric principles that students reckon the significance of proportional relationships between reactants and products when addressing heat capacity changes and entropy calculations, ultimately creating a bridge between quantitative measurements and qualitative energy changes.

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Most popular questions from this chapter

The Chalk Point, Maryland, generating station supplies electrical power to the Washington, D.C., area. Units 1 and 2 have a gross generating capacity of \(710 .\) MW (megawatt). The steam pressure is \(25 \times 10^{6} \mathrm{Pa}\), and the superheater outlet temperature \(\left(T_{h}\right)\) is \(540 .^{\circ} \mathrm{C} .\) The condensate temperature \(\left(T_{c}\right)\) is \(30.0^{\circ} \mathrm{C}\) a. What is the efficiency of a reversible Carnot engine operating under these conditions? b. If the efficiency of the boiler is \(91.2 \%\), the overall efficiency of the turbine, which includes the Carnot efficiency and its mechanical efficiency, is \(46.7 \%,\) and the efficiency of the generator is \(98.4 \%,\) what is the efficiency of the total generating unit? (Another \(5.0 \%\) needs to be subtracted for other plant losses. c. One of the coal-burning units produces \(355 \mathrm{MW}\). How many metric tons (1 metric ton \(=1 \times 10^{6} \mathrm{g}\) ) of coal per hour are required to operate this unit at its peak output if the enthalpy of combustion of coal is \(29.0 \times 10^{3} \mathrm{kJ} \mathrm{kg}^{-1} ?\)

Calculate \(\Delta S\) for the isothermal compression of 1.75 mole of \(\mathrm{Cu}(\mathrm{s})\) from 2.15 bar to \(1250 .\) bar at \(298 \mathrm{K}\) \\[ \beta=0.492 \times 10^{-4} \mathrm{K}^{-1}, \kappa=0.78 \times 10^{-6} \mathrm{bar}^{-1}, \text {and the } \\] density is \(8.92 \mathrm{g} \mathrm{cm}^{-3} .\) Repeat the calculation assuming that \\[ \kappa=0 \\]

Calculate \(\Delta S\) if the temperature of \(2.50 \mathrm{mol}\) of an ideal gas with \(C_{V}=5 / 2 R\) is increased from 160 . to \(675 \mathrm{K}\) under conditions of \((\mathrm{a})\) constant pressure and (b) constant volume.

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 \(\mathrm{Pb}\left(s, 25.0^{\circ} \mathrm{C}\right) \rightarrow\) \(\mathrm{Pb}\left(L, 725^{\circ} \mathrm{C}\right)\)

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