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Choose the substance with the larger positional probability in each case. a. 1 mole of \(\mathrm{H}_{2}\) (at \(\mathrm{STP} )\) or 1 mole of \(\mathrm{H}_{2}\left(\text { at } 100^{\circ} \mathrm{C}, 0.5 \mathrm{atm}\right)\) b. 1 mole of \(\mathrm{N}_{2}(\text { at } \mathrm{STP})\) or 1 mole of \(\mathrm{N}_{2}(\text { at } 100 \mathrm{K}, 2.0 \mathrm{atm})\) c. 1 mole of \(\mathrm{H}_{2} \mathrm{O}(s)\) (at \(0^{\circ} \mathrm{C} )\) or 1 \(\mathrm{mole}\) of \(\mathrm{H}_{2} \mathrm{O}(l)\left(\mathrm{at} 20^{\circ} \mathrm{C}\right)\)

Short Answer

Expert verified
a. 1 mole of \(\mathrm{H}_{2}\) at 100°C and 0.5 atm has a larger positional probability. b. 1 mole of \(\mathrm{N}_{2}\) at STP has a larger positional probability. c. 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) at 20°C has a larger positional probability.

Step by step solution

01

Case a: Comparing \(\mathrm{H}_{2}\) at STP and at 100°C, 0.5 atm

In this case, we need to compare 1 mole of \(\mathrm{H}_{2}\) at STP to 1 mole of \(\mathrm{H}_{2}\) at 100°C and 0.5 atm. Since higher temperatures and lower pressures lead to higher positional probabilities, let's compare both the conditions: 1. Temperature: At STP, the temperature is 0°C, while in the second case, it is 100°C. The second case has a higher temperature, which leads to higher positional probability. 2. Pressure: At STP, the pressure is 1 atm, while in the second case, it is 0.5 atm. The second case has lower pressure, which leads to higher positional probability. As both the temperature and pressure lead to higher positional probability in the second case, we can conclude that 1 mole of \(\mathrm{H}_{2}\) at 100°C and 0.5 atm has a larger positional probability than at STP.
02

Case b: Comparing \(\mathrm{N}_{2}\) at STP and at 100 K, 2.0 atm

Now, we need to compare 1 mole of \(\mathrm{N}_{2}\) at STP to 1 mole of \(\mathrm{N}_{2}\) at 100 K and 2.0 atm: 1. Temperature: At STP, the temperature is 273.15 K, while in the second case, it is 100 K. The first case has a higher temperature, which leads to higher positional probability. 2. Pressure: At STP, the pressure is 1 atm, while in the second case, it is 2.0 atm. The first case has lower pressure, which leads to higher positional probability. Since both the temperature and pressure lead to higher positional probability in the first case, we can conclude that 1 mole of \(\mathrm{N}_{2}\) at STP has a larger positional probability than at 100 K and 2.0 atm.
03

Case c: Comparing \(\mathrm{H}_{2} \mathrm{O}(s)\) at 0°C and \(\mathrm{H}_{2} \mathrm{O}(l)\) at 20°C

Finally, we need to compare 1 mole of \(\mathrm{H}_{2} \mathrm{O}(s)\) at 0°C to 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) at 20°C. In this case, we are comparing solid water (ice) to liquid water. It is important to note that the positional probability of particles in a liquid is higher than that of particles in a solid due to the increased freedom of movement in the liquid state. Since liquid water has a higher positional probability than solid water, we can conclude that 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) at 20°C has a larger positional probability than 1 mole of \(\mathrm{H}_{2} \mathrm{O}(s)\) at 0°C.

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Key Concepts

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

STP (Standard Temperature and Pressure)
Understanding Standard Temperature and Pressure (STP) is essential in chemistry because it establishes common conditions to compare how substances behave. STP is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm. These conditions serve as a reference point for comparing gases because they offer a benchmark to assess and predict gas behavior under different circumstances. When a gas is at STP, its properties such as density, volume, and positional probability—an indicator of how dispersed its particles are—can be consistently measured and compared. Higher temperature and lower pressure generally lead to greater positional probability, meaning the particles are spread out more, as seen when we compare substances at different conditions. For example, hydrogen gas at higher temperatures and lower pressures than at STP might have a higher positional probability, meaning more potential positions for its particles.
mole concept
The mole is a fundamental concept in chemistry, representing a specific number of particles, usually atoms or molecules. One mole is defined as the quantity containing exactly 6.022 × 10^23 particles, known as Avogadro's number. This allows chemists to translate between the microscopic scale of atoms and the macroscopic scale of grams and liters that we see in the laboratory. Understanding moles is vital for calculating positional probability because it helps define "1 mole of substance" whether in gaseous, liquid, or solid forms. It allows comparison of equivalent amounts in terms of sheer particle count. For instance, when comparing 1 mole of oxygen gas at STP to 1 mole at other conditions, mole concept enables a direct comparison of particle behavior under different conditions.
phase comparison (solid vs liquid)
Phases of matter—solid, liquid, and gas—display different characteristics that affect positional probability. In solids, particles are tightly packed in a fixed arrangement, leading to lower positional probability. Conversely, liquids have more freedom and mobility. Particle movement in liquids allows for higher positional probability due to increased space and energy levels. For example, water as ice (solid) at 0°C has very structured, immobile molecules, whereas liquid water at 20°C has molecules that move freely and occupy more positions in space. This difference demonstrates why liquid water possesses higher positional probability compared to ice since its particles can move more freely, thereby accessing more positions in space.

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Most popular questions from this chapter

At what temperatures will the following processes be spontaneous? a. \(\Delta H=-18 \mathrm{kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) b. \(\Delta H=+18 \mathrm{kJ}\) and \(\Delta S=+60 . \mathrm{JK}\) c. \(\Delta H=+18 \mathrm{kJ}\) and \(\Delta S=-60 . \mathrm{J} / \mathrm{K}\) d. \(\Delta H=-18 \mathrm{kJ}\) and \(\Delta S=+60 . \mathrm{J} / \mathrm{K}\)

Calculate \(\Delta S_{\text { surr }}\) for the following reactions at \(25^{\circ} \mathrm{C}\) and 1 \(\mathrm{atm}\) . a. \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\Delta H^{\circ}=-2221 \mathrm{kJ}\) b. \(2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \qquad \Delta H^{\rho}=112 \mathrm{kJ}\)

Which of the following processes are spontaneous? a. Salt dissolves in \(\mathrm{H}_{2} \mathrm{O}\) . b. A clear solution becomes a uniform color after a few drops of dye are added. c. Iron rusts. d. You clean your bedroom.

For the reaction $$\mathrm{CS}_{2}(g)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{SO}_{2}(g)$$ \(\Delta S^{\circ}\) is equal to \(-143 \mathrm{JK}\) . Use this value and data from Appendix 4 to calculate the value of \(S^{\circ}\) for \(\mathrm{CS}_{2}(g) .\)

Which of the following reactions (or processes) are expected to have a negative value for \(\Delta S^{\circ} ?\) a. \(\mathrm{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)\) b. \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\) c. \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\) d. \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) e. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)

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