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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 \(\mathrm{H}_{2}\) (at \(100^{\circ} \mathrm{C}, 0.5\) atm) b. 1 mole of \(\mathrm{N}_{2}\) (at \(\mathrm{STP}\) ) or 1 mole of \(\mathrm{N}_{2}\) (at \(100 \mathrm{K}, 2.0\) atm) c. 1 mole of \(\mathrm{H}_{2} \mathrm{O}(s)\) (at \(0^{\circ} \mathrm{C}\) ) or 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) (at \(20^{\circ} \mathrm{C}\) )

Short Answer

Expert verified
a. H2 at 100°C and 0.5 atm has a larger positional probability. b. N2 at STP has a larger positional probability. c. H2O(l) at 20°C has a larger positional probability.

Step by step solution

01

Find the volume at each condition

Using the ideal gas law: PV = nRT We can find the volume for each condition, where n=1 (number of moles), R is the ideal gas constant (0.0821 atm L/mol K). For 1 mole of H2 at STP: Temperature (T) = 0°C + 273.15 K = 273.15 K Pressure (P) = 1 atm \(V_1 = \frac{nRT_1}{P_1} = \frac{1 \times 0.0821 \times 273.15}{1} = 22.4 L\) For 1 mole of H2 at 100°C, 0.5 atm: Temperature (T) = 100°C + 273.15 K = 373.15 K Pressure (P) = 0.5 atm \(V_2 = \frac{nRT_2}{P_2} = \frac{1 \times 0.0821 \times 373.15}{0.5} = 61.3 L\)
02

Compare the volume occupied

As calculate above, H2 at STP occupies 22.4 L and H2 at 100°C and 0.5 atm occupies 61.3 L. Since the H2 at 100°C and 0.5 atm occupies a greater volume, it has a larger positional probability. #Case b#: Comparing Positonal Probability of N2 Given conditions: 1 mole of N2 at STP (Standard Temperature and Pressure) - 0°C, 1 atm 1 mole of N2 at 100 K, 2.0 atm
03

Find the volume at each condition

Using the ideal gas law (PV = nRT): For 1 mole of N2 at STP: Temperature (T) = 0°C + 273.15 K = 273.15 K Pressure (P) = 1 atm \(V_1 = \frac{nRT_1}{P_1} = \frac{1 \times 0.0821 \times 273.15}{1} = 22.4 L\) For 1 mole of N2 at 100 K, 2.0 atm: Temperature (T) = 100 K Pressure (P) = 2.0 atm \(V_2 = \frac{nRT_2}{P_2} = \frac{1 \times 0.0821 \times 100}{2} = 4.11 L\)
04

Compare the volume occupied

As calculated above, N2 at STP occupies 22.4 L and N2 at 100 K and 2.0 atm occupies 4.11 L. Since the N2 at STP occupies a greater volume, it has a larger positional probability. #Case c#: Comparing Positonal Probability of H2O Given conditions: 1 mole of H2O(s) - solid at 0°C 1 mole of H2O(l) - liquid at 20°C
05

Compare the available positions

In the case of H2O, we have it in solid and liquid forms. Generally, in a solid, particles are closely packed; in a liquid, they have relatively more space to move around.
06

Determine the substance with the larger positional probability

Since H2O in the liquid form has relatively more positions available for particles to move compared to H2O in the solid form, H2O(l) at 20°C has a larger positional probability. To summarize: a. H2 at 100°C and 0.5 atm has a larger positional probability. b. N2 at STP has a larger positional probability. c. H2O(l) at 20°C has a larger positional probability.

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Key Concepts

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

Ideal Gas Law
The ideal gas law, represented by the equation PV = nRT, is a cornerstone principle in chemistry and physics, describing the behavior of an ideal gas under varying conditions of pressure (P), volume (V), number of moles (n), and temperature (T). R is the ideal gas constant, which has a value of 0.0821 atm L/mol K.

This equation allows us to predict and calculate how a gas will behave when subjected to different environmental parameters. In educational exercises, students often use the ideal gas law to find the volume occupied by a gas at specific conditions, as seen in the given problem.

As an improvement, we could focus on explaining the significance of each variable in the formula to offer a more profound understanding. For instance, the pressure of a gas is directly proportional to the temperature and volume; when pressure increases, the volume tends to decrease at a constant temperature, and vice versa.

Impact of Variables on Gas Behavior

  • Number of moles (n): Indicates the amount of substance present.
  • Temperature (T): Higher temperatures increase the pressure and volume of a gas.
  • Pressure (P): Reflects the force exerted by the gas molecules against the walls of their container.
  • Volume (V): Describes the space the gas occupies and is inversely related to pressure.
State of Matter
Matter exists in different states, primarily as solids, liquids, and gases. This concept is pivotal when discussing positional probability within these states.

In solids, particles are tightly packed in a fixed structure with very limited movement, which translates to lower positional probability. Liquids have more freedom of movement due to a bit more space between particles, leading to a moderate positional probability. Gases, on the other hand, have particles that are far apart and move rapidly in all directions, resulting in the highest positional probability among the states of matter.

Understanding Positional Probability

Positional probability relates to the number of positions or configurations that particles can adopt within a given volume. In contexts where particles can spread out and arrange themselves in many possible ways, such as in gases, positional probability is high. The concept is essential for comprehending molecular behavior and predicting physical properties.

By delineating how positional probability varies across states of matter, students can better interpret problems involving state changes and apply this knowledge to real-world scenarios such as phase transitions.
Standard Temperature and Pressure (STP)
Standard Temperature and Pressure (STP) is a condition for measuring and comparing the behavior of gases where the temperature is set at 0°C (273.15 K) and the pressure at 1 atm. It serves as a reference point in scientific calculations to ensure consistency and accuracy when reporting on the properties of gases.

With STP conditions specified, one mole of an ideal gas occupies a volume of 22.4 liters, a fact that can be derived using the ideal gas law. This pre-established volume helps in making quick comparisons and inferring which substance has a larger positional probability under differing conditions.

Highlighting how standard conditions allow for predictable and comparable results, can enhance students' understanding of gas behaviors across different exercises. Additionally, explaining the relevance of STP in real-world applications, such as in industrial processes and experimental measurements, would provide practical insights into why mastering these concepts is so important.

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Most popular questions from this chapter

For the equilibrium $$\mathrm{A}(g)+2 \mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)$$ the initial concentrations are \([\mathrm{A}]=[\mathrm{B}]=[\mathrm{C}]=0.100 \mathrm{atm}\) Once equilibrium has been established, it is found that \([\mathrm{C}]=\) 0.040 atm. What is \(\Delta G^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C} ?\)

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 \(100.0^{\circ} \mathrm{C}\).

Consider the reaction $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ $$2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{S}_{\text {thombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$For each of the following mixtures of reactants and products at \(25^{\circ} \mathrm{C},\) predict the direction in which the reaction will shift to reach equilibrium. a. \(P_{\mathrm{NO}_{2}}=P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.0 \mathrm{atm}\) b. \(P_{\mathrm{NO}_{2}}=0.21 \mathrm{atm}, P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.50 \mathrm{atm}\) c. \(P_{\mathrm{NO}_{2}}=0.29 \mathrm{atm}, P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.6 \mathrm{atm}\)

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\) using the following data: $$\begin{aligned} &\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6} \text { at } 600 . \mathrm{K}\\\ &2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37} \mathrm{at} 600 . \mathrm{K} \end{aligned}$$

Using data from Appendix \(4,\) calculate \(\Delta H^{\circ}, \Delta G^{\circ},\) and \(K\) (at 298 K) for the production of ozone from oxygen: $$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$ At \(30 \mathrm{km}\) above the surface of the earth, the temperature is about \(230 . . \mathrm{K}\) and the partial pressure of oxygen is about \(1.0 \times 10^{-3}\) atm. Estimate the partial pressure of ozone in equilibrium with oxygen at \(30 \mathrm{km}\) above the earth's surface. Is it reasonable to assume that the equilibrium between oxygen and ozone is maintained under these conditions? Explain.

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