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For the majority of the compounds listed in Appendix \(C_{r}\), the value of \(\Delta G_{f}^{\circ}\) is more positive (or less negative) than the value of \(\Delta H_{f}^{\circ}\) (a) Explain this observation, using \(\mathrm{NH}_{3}(\mathrm{~g}), \mathrm{CCl}_{4}(l)\), and \(\mathrm{KNO}_{3}(s)\) as examples. (b) An exception to this observation is \(\mathrm{CO}(g)\). Explain the trend in the \(\Delta H_{f}^{\circ}\) and \(\Delta G_{f}^{\circ}\) values for this molecule.

Short Answer

Expert verified
In most compounds, ΔG⁰f is more positive (or less negative) than ΔH⁰f due to a significant and positive entropy change (ΔS⁰f) during formation, as in the examples of NH₃(g), CCl₄(l), and KNO₃(s). This is attributed to these compounds forming from simpler components or having more degrees of freedom. However, in the case of CO(g), ΔG⁰f is not more positive than ΔH⁰f due to a smaller entropy change, possibly resulting from its linear structure and unusual chemical bonding that restrict its degrees of freedom.

Step by step solution

01

Understanding Gibbs free energy

The Gibbs free energy change (ΔG⁰f) is a thermodynamic quantity that determines the spontaneity of a process. It is related to the enthalpy change (ΔH⁰f) and the entropy change (ΔS⁰f) by the equation: ΔG⁰ = ΔH⁰ - TΔS⁰ where T is the temperature in Kelvin. Step 2: Explanation for NH₃(g), CCl₄(l), and KNO₃(s)
02

Explanation for NH₃(g), CCl₄(l), and KNO₃(s)

According to the observation stated in the exercise, for most compounds, ΔG⁰f is more positive (or less negative) than ΔH⁰f. To understand this, let's focus on the relationship between Gibbs free energy, enthalpy, and entropy in the formula: ΔG⁰ = ΔH⁰ - TΔS⁰ If ΔG⁰ is more positive than ΔH⁰, it means that the term TΔS⁰ is positive and significant. This means that for compounds like NH₃(g), CCl₄(l), and KNO₃(s), the entropy change (ΔS⁰f) during formation is positive, which is expected for substances forming from simpler components or substances with more degrees of freedom, such as gases and liquids. Step 3: Analysis of CO(g)
03

Analysis of CO(g)

The exercise mentions that CO(g) is an exception to the observed trend, and we need to explain why this is the case. Let's analyze the ΔH⁰f and ΔG⁰f values for CO(g): For CO(g), ΔG⁰f is not more positive than ΔH⁰f. To understand this, let's again focus on the relationship between Gibbs free energy, enthalpy, and entropy in the formula: ΔG⁰ = ΔH⁰ - TΔS⁰ Step 4: Explanation for the trend in CO(g)
04

Explanation for the trend in CO(g)

For CO(g), if ΔG⁰f is not more positive than ΔH⁰f, this implies that the term TΔS⁰ is either very small, close to zero, or even negative. This could occur if the entropy change (ΔS⁰f) during the formation of CO(g) was relatively smaller compared to the other examples. The smaller or negative value of TΔS⁰ could be attributed to the fact that CO(g) has a linear molecule structure and unusual chemical bonding that restrict its degrees of freedom, leading to low entropy change during its formation. In conclusion, the observed trend in ΔG⁰f and ΔH⁰f values can be explained by the entropy changes during the formation of the compounds. For most compounds like NH₃(g), CCl₄(l), and KNO₃(s), a positive and significant entropy change leads to the observed trend, while for CO(g) has a relatively smaller or negative entropy change, resulting in the exception.

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Key Concepts

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

Enthalpy Change
Enthalpy change, denoted as \( \Delta H \), is a central concept in thermodynamics that represents the heat absorbed or released by a system at constant pressure. It's a measure of the total energy change within a system during a chemical reaction or physical process. Imagine a reaction taking place where you mix chemicals and heat is either given off or absorbed.
Enthalpy change is the difference between the total energy of products and reactants.
  • If heat is released, the process is exothermic, resulting in a negative \( \Delta H \).
  • If heat is absorbed, the process is endothermic, resulting in a positive \( \Delta H \).
Take ammonia (NH₃) as an example. When formed, it releases energy into the surroundings, which is why \( \Delta H \) for NH₃'s formation is negative. This reinforces the fact that the energy of the products is less than that of the reactants. By examining enthalpy changes, we gain insight into the energetic characteristics of chemical reactions.
Entropy Change
Entropy, represented by \( \Delta S \), measures a system's level of disorder or randomness. It's an indicator of spontaneity in a process, akin to the unpredictability of a room getting messier over time. Generally, as systems evolve, they move toward higher entropy.
During a reaction, if a substance transforms from a more organized state to a less organized state, the entropy will increase. Entropy is associated with the distribution of energy within the system and freedom of particle movement.
  • If \( \Delta S \) is positive, the process favors more disorder, often leading to a spontaneous change.
  • If \( \Delta S \) is negative, the process results in less disorder.
In the case of carbon tetrachloride (CCl₄), as a liquid turning into a gas, the entropy increases given the gaseous particles have greater motion freedom than those in a liquid. This high increase in entropy results in a significant contribution to gibbs free energy change, affecting the system's overall spontaneity.
Thermodynamics
Thermodynamics is the overarching framework within which we study energy transformations in chemical systems. It guides us to understand how energy moves and changes form, affecting states of matter, temperatures, and energies within reactions.
The core principles include:
  • First Law, which asserts the conservation of energy - energy can neither be created nor destroyed.
  • Second Law, which states that the entropy of an isolated system always increases over time.
  • Third Law, establishing that at absolute zero, the entropy of a perfect crystal is zero.
In our exercise, thermodynamics helps interpret the changes in enthalpy and entropy during the formation of compounds and how these contribute to Gibbs free energy ($$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$).
For compounds like KNO₃ and CO, these transformations in energy and disorder define the stability and favorability of formation. Understanding thermodynamics, therefore, offers a pathway to predicting how reactions occur naturally and their potential applications or limitations in various scientific and industrial settings.

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Most popular questions from this chapter

Acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), is used in welding. (a) Write a balanced equation for the combustion of acetylene gas to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) How much heat is produced in burning \(1 \mathrm{~mol}\) of \(\mathrm{C}_{2} \mathrm{H}_{2}\) under standard conditions if both reactants and products are brought to \(298 \mathrm{~K} ?\) (c) What is the maximum amount of useful work that can be accomplished under standard conditions by this reaction?

Consider the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) .\) (a) Using data from Appendix \(C\), calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are \(0.40 \mathrm{~atm}\) and \(1.60 \mathrm{~atm}\), respectively.

Trouton's rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about \(88 \mathrm{~J} / \mathrm{mol}-\mathrm{K} .\) (a) Estimate the normal boiling point of bromine, \(\mathrm{Br}_{2}\), by determining \(\Delta H_{\text {vap }}^{\circ}\) for \(\mathrm{Br}_{2}\) using data from Appendix \(C\). Assume that \(\Delta H_{\text {vap }}^{\circ}\) remains constant with temperature and that Trouton's rule holds. (b) Look up the normal boiling point of \(\mathrm{Br}_{2}\) in a chemistry handbook or at the WebElements web site (www.webelements.com).

Which of the following processes are spontaneous, and \(\quad\) and \(1 \mathrm{~atm}\), (d) lightning, (e) formation of \(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\) which are nonspontaneous: (a) the ripening of a banana, molecules from \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) at room temperature and (b) dissolution of sugar in a cup of hot coffee, (c) the re- 1 atm of pressure? action of nitrogen atoms to form \(\mathrm{N}_{2}\) molecules at \(25^{\circ} \mathrm{C}\)

One way to derive Equation \(19.3\) depends on the observation that at constant \(T\) the number of ways, \(W\), of arranging \(m\) ideal-gas particles in a volume \(V\) is proportional to the volume raised to the \(m\) power: \(W \propto V^{m}\) Use this relationship and Boltzmann's relationship between entropy and number of arrangements (Equation 19.5) to derive the equation for the entropy change for the isothermal expansion or compression of \(n\) moles of an ideal gas.

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