Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

For each of the following pairs of substances, which substance has the greater value of \(S^{\circ} ?\) a. \(\mathrm{C}_{\text {graphite }}(s)\) or \(\mathrm{C}_{\text {diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)

Short Answer

Expert verified
For each pair of substances, the substance with the greater value of \(S^\circ\) is as follows: a. Graphite carbon: \(\mathrm{C}_{\text{graphite}}(s)\) b. Gaseous ethanol: \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}(g)\) c. Gaseous carbon dioxide: \(\mathrm{CO}_{2}(g)\)

Step by step solution

01

a. Graphite and diamond carbon

Graphite and diamond both have the same chemical formula (C, carbon), but they have different crystalline structures, which influence their entropy. In graphite, carbon atoms are arranged in a layered, planar structure with relatively weak intermolecular forces between those layers, allowing them to slide past each other. In contrast, diamond has a covalently bonded tetrahedral structure, which is much more rigid. As a result, graphite has higher entropy since its structure allows for greater freedom of motion for its carbon atoms. So, graphite has a greater value of S° than diamond.
02

b. Liquid and gaseous ethanol

Liquid and gaseous ethanol (C2H5OH) have the same molecular formula, but each exists in a different phase. Entropy increases when moving from solid to liquid to gas due to increased molecular freedom. Therefore, gaseous ethanol (C2H5OH(g)) has a larger value of S° than liquid ethanol (C2H5OH(l)) due to the increased degree of freedom of motion for molecules within the gas phase.
03

c. Solid and gaseous carbon dioxide

Carbon dioxide (CO2) exists as a solid (dry ice) and as a gas (familiar to us as the product of combustion and respiration). Entropy increases when moving from solid to liquid to gas. Solid CO₂ has a very ordered crystal lattice, and its molecules have limited freedom to move. In contrast, gaseous CO₂ molecules have a much higher degree of freedom and energy dispersal, as they are not held together in any definite structure. Therefore, gaseous carbon dioxide (CO₂(g)) has a greater value of S° than solid carbon dioxide (CO₂(s)).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Graphite vs Diamond
Graphite and diamond are both allotropes of carbon, which means they are made entirely of carbon atoms. However, these two substances differ significantly in their chemical structure, impacting their entropy or the measure of disorder. In graphite, carbon atoms are organized in stacked, planar layers, resembling a honeycomb lattice. This arrangement results in weak interactions between the layers, allowing them to slide past one another easily. Hence, graphite exhibits higher entropy due to the flexibility and movement permitted within its structure.
Diamond, on the other hand, forms a three-dimensional tetrahedral lattice, where each carbon atom is strongly bonded to four other carbon atoms. This rigid, strong bonding network leads to a very structured and orderly arrangement, limiting motion and increasing predictability. As a result, diamond has a lower entropy compared to graphite.
  • Graphite structure: layered, less order, higher entropy
  • Diamond structure: rigid tetrahedral lattice, lower entropy
Phase Changes
When a substance undergoes a phase change from solid to liquid to gas, there is generally an increase in entropy. This is because the molecules in a gas have more freedom of motion and are spread further apart compared to when they are in a solid or liquid state. Take ethanol (\(C_2H_5OH\)), for instance. As a liquid, ethanol molecules are close together, but they have the freedom to move past one another, something they cannot do in a solid phase where such movement is restricted.
As ethanol transitions to a gaseous state, its molecules gain additional freedom and energy, dispersing and moving independently of one another. This increased molecular freedom is why the entropy of gaseous ethanol is greater than that of its liquid form.
  • Solid to liquid to gas: increasing entropy
  • More freedom in gas phase = higher entropy
Chemical Structure
The chemical structure of a molecule plays a crucial role in determining its entropy. Let's consider carbon dioxide (\(CO_2\)) as it exists in two distinct phases: solid (dry ice) and gas. As a solid, CO₂ is tightly packed in a rigid crystal lattice structure, maintaining a high degree of order. This order reduces the entropy of the molecule, as there is limited freedom for molecular movement.
In its gaseous form, the CO₂ molecules are not bound in any fixed arrangement, allowing them to move randomly and freely. This randomness and freedom increase the entropy of the gaseous phase when compared to the solid phase. Generally, the more rigid and organized a chemical structure is, the lower its entropy, while flexible, dispersed structures lead to higher entropy levels.
  • Order in structure = lower entropy
  • Dispersed, flexible structures = higher entropy

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The following reaction occurs in pure water: $$\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ which is often abbreviated as $$\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ For this reaction, \(\Delta G^{\circ}=79.9 \mathrm{kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C} .\) Calculate the value of \(\Delta G\) for this reaction at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{OH}^{-}\right]=0.15 M\) and \(\left[\mathrm{H}^{+}\right]=0.71 M\)

Consider the system $$\mathrm{A}(g) \longrightarrow \mathrm{B}(g)$$ at $$25^{\circ} \mathrm{C}$$ a. Assuming that \(G_{A}^{\circ}=8996 \mathrm{J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{J} / \mathrm{mol}\) calculate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if 1.00 mole of \(\mathrm{A}(g)\) at 1.00 atm and 1.00 mole of \(\mathrm{B}(g)\) at 1.00 atm are mixed at \(25^{\circ} \mathrm{C}.\) c. Show by calculations that \(\Delta G=0\) at equilibrium.

Consider two perfectly insulated vessels. Vessel 1 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and water at \(0^{\circ} \mathrm{C}\). Vessel 2 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and a saltwater solution at \(0^{\circ} \mathrm{C}\). Consider the process \(\mathrm{H}_{2} \mathrm{O}(s) \rightarrow \mathrm{H}_{2} \mathrm{O}(l).\) a. Determine the sign of \(\Delta S, \Delta S_{\text {sur, }}\) and \(\Delta S_{\text {univ }}\) for the process in vessel 1. b. Determine the sign of \(\Delta S, \Delta S_{\text {sur, }}\) and \(\Delta S_{\text {univ }}\) for the process in vessel 2. (Hint: Think about the effect that a salt has on the freezing point of a solvent.)

At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\). For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text {surr }}\) ? \(\Delta S ?\) \(\Delta S_{\text {univ }}\) ? Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined

Consider the following reaction at \(25.0^{\circ} \mathrm{C}:\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{kJ} / \mathrm{mol}\) and -176.6 \(\text{J/K}\) \cdot mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\). Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free