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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 \(T^{2} / \mathrm{K}^{2}\)

Short Answer

Expert verified
The standard entropy change, \(\Delta S^\circ\), for the reaction \(3\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)\) at 725K is -198.40 J/mol路K.

Step by step solution

01

Find the standard entropy values for the reactants and products.

In order to solve this problem, first, we need to gather the standard entropy values (\(\mathrm{S}_{\text{molar}}^\circ\)) of all reactants and products involved in the reaction. In this case, we have H鈧(g), N鈧(g), and NH鈧(g). The values are as follows (given in textbook or reference tables): S鈧樷伆 (H鈧) = 130.60 J/mol路K S鈧樷伆 (N鈧) = 191.60 J/mol路K S鈧樷伆 (NH鈧) = 192.50 J/mol路K
02

Apply the equation and compute the entropy change.

Now, we can apply the following equation to determine the standard entropy change, \(\Delta S^\circ\): \[\Delta S^\circ = \sum \text{ (moles of products * S鈧樷伆 of products)} - \sum \text{ (moles of reactants * S鈧樷伆 of reactants)}\] For this reaction, the equation will look like this: \[\Delta S^\circ = [2(\text{moles of NH鈧儅) * S鈧樷伆 (\text{NH鈧儅)] - [ 3(\text{moles of H鈧倉) * S鈧樷伆 (\text{H鈧倉) + (\text{moles of N鈧倉) * S鈧樷伆 (\text{N鈧倉) ]\] Now plug in the values: \[\Delta S^\circ = [2(192.50 \ \mathrm{J/mol \cdot K})] - [ 3(130.60 \ \mathrm{J/mol \cdot K}) + (191.60 \ \mathrm{J/mol \cdot K}) ]\]
03

Calculate the value of 螖S鈦.

Now we can calculate the value of 螖S鈦 by performing the arithmetic operations. \[\Delta S^\circ = (385.00 \ \mathrm{J/mol \cdot K}) - (391.80 + 191.60 \ \mathrm{J/mol \cdot K})\] \[\Delta S^\circ = (385.00 - 583.40) \ \mathrm{J/mol \cdot K}\] \[\Delta S^\circ = -198.40 \ \mathrm{J/mol \cdot K}\] Therefore, the standard entropy change, \(\Delta S^\circ\), for the reaction at 725K is -198.40 J/mol路K.

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Key Concepts

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

Entropy
Entropy is a fundamental concept in thermodynamics, which describes the degree of disorder or randomness in a system. Entropy, represented by the symbol S, plays a vital role in determining the direction and feasibility of chemical reactions.
It provides a measure of the number of ways energy can be distributed among the molecules of a system, indicating how spread out or dispersed the energy is. For students, it鈥檚 helpful to think of entropy as how messy the system is becoming.
A key point to remember: an increase in entropy generally means more disorder.
  • If the total entropy of a system and its surroundings increases, the process can occur spontaneously.
  • Entropy is measured in units of joules per mole per kelvin (J/mol路K).
  • In chemical reactions, the standard molar entropy \(S_{\text{molar}}^\circ\) is often used, indicating entropy under standard conditions, typically 1 atm and 298 K.
Understanding entropy helps in determining why certain reactions happen and others do not. It shifts perspectives from just looking at energy (enthalpy) to also considering randomness and dispersal of particles.
Reaction Entropy
Reaction entropy, represented as \(\Delta S^\circ\), is the change in entropy that occurs in a chemical reaction when the reactants form the products. This value indicates whether a reaction leads to an increase or decrease in the disorder of the system.
Here's how to calculate reaction entropy:
  • The formula for calculating \(\Delta S^\circ\) in reactions is: \[\Delta S^\circ = \sum (\text{S}_{\text{molar}}^\circ \times \text{moles of products}) - \sum (\text{S}_{\text{molar}}^\circ \times \text{moles of reactants})\]
  • For each substance, multiply its standard molar entropy by the number of moles in the balanced equation.
  • Subtract the total entropy of reactants from the total entropy of products.
In this exercise, we calculated \(\Delta S^\circ\) for the reaction \(3\ \text{H}_{2}(g)+\text{N}_{2}(g) \rightarrow 2\ \text{NH}_{3}(g)\) and found that it results in a negative value of -198.40 J/mol路K. This implies a decrease in disorder, as the reaction favors the formation of a more ordered product, ammonia (NH鈧), from gaseous hydrogen and nitrogen. This negative change reflects a shift towards reduced randomness as the smaller, lighter gases are combined into a more complex molecule.
Temperature-dependent Heat Capacities
Temperature-dependent heat capacities refer to the ability of a substance to absorb heat as the temperature changes, impacting the entropy of systems. In thermodynamics, adjusting for temperature effects is critical because heat capacity can vary with temperature.
For reactions taking place under various temperatures, this characteristic dictates how the entropy and enthalpy values might alter due to the additional thermal energy involved.
  • When calculating changes in entropy over different temperatures, ignoring higher order terms like \(T^2/\text{K}^2\) simplifies calculations while providing reasonable accuracy under certain conditions.
  • While keeping calculations simpler, this assumption usually works when temperature changes are mild or within a limited range.
In the provided exercise, such simplifications were applied by omitting terms in temperature-dependent heat capacities higher than \(T^2/\text{K}^2\). This allows students to focus on core concepts of reaction entropy change without getting lost in deeper complexities that higher-order terms might introduce. Remember, though simplifications can make calculations easier, they also require careful judgment about when and how they are used.

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Most popular questions from this chapter

Under anaerobic conditions, glucose is broken down in muscle tissue to form lactic acid according to the reaction \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \rightarrow 2 \mathrm{CH}_{3} \mathrm{CHOHCOOH}\). Thermodynamic data at \(T=298 \mathrm{K}\) for glucose and lactic acid are given in the following table: $$\begin{array}{lccc} & \Delta \boldsymbol{H}_{f}^{\circ}\left(\mathbf{k J} \mathbf{~ m o l}^{-1}\right) & C_{P, m}\left(\mathbf{J} \mathbf{K}^{-1} \mathbf{m o l}^{-\mathbf{1}}\right) & S_{m}^{\circ}\left(\mathbf{J} \mathbf{K}^{-\mathbf{1}} \mathbf{m o l}^{-\mathbf{1}}\right) \\ \hline \text { Glucose } & -1273.1 & 219.2 & 209.2 \\ \text { Lactic } & -673.6 & 127.6 & 192.1 \end{array}$$ Calculate, \(\Delta S\) for the system, the surroundings, and the universe at \(T=325 \mathrm{K}\). Assume the heat capacities are constant \\[ \text { between } T=298 \mathrm{K} \text { and } T=330 . \mathrm{K} \\]

An ideal gas sample containing 1.75 moles for which \(C_{V, m}=5 R / 2\) undergoes the following reversible cyclical process from an initial state characterized by \(T=275 \mathrm{K}\) and \(P=1.00\) bar: a. It is expanded reversibly and adiabatically until the volume triples. b. It is reversibly heated at constant volume until \(T\) increases to \(275 \mathrm{K}\) c. The pressure is increased in an isothermal reversible compression until \(P=1.00\) bar. Calculate \(q, w, \Delta U, \Delta H,\) and \(\Delta S\) for each step in the cycle, and for the total cycle.

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 \(\mathrm{H}_{2} \mathrm{O}\left(s, 0^{\circ} \mathrm{C}\right) \rightarrow \mathrm{H}_{2} \mathrm{O}\left(g, 100 \cdot^{\circ} \mathrm{C}\right)\)

An air conditioner is a refrigerator with the inside of the house acting as the cold reservoir and the outside atmosphere acting as the hot reservoir. Assume that an air conditioner consumes \(1.70 \times 10^{3} \mathrm{W}\) of electrical power and that it can be idealized as a reversible Carnot refrigerator. If the coefficient of performance of this device is 3.30 how much heat can be extracted from the house in a day?

The maximum theoretical efficiency of an internal combustion engine is achieved in a reversible Carnot cycle. Assume that the engine is operating in the Otto cycle and that \(C_{V, m}=5 R / 2\) for the fuel-air mixture initially at \(273 \mathrm{K}\) (the temperature of the cold reservoir). The mixture is compressed by a factor of 6.9 in the adiabatic compression step. What is the maximum theoretical efficiency of this engine? How much would the efficiency increase if the compression ratio could be increased to \(15 ?\) Do you see a problem in doing so?

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