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Describe how the following changes affect the positional probability of a substance. a. increase in volume of a gas at constant T b. increase in temperature of a gas at constant V c. increase in pressure of a gas at constant T

Short Answer

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In summary: a. Increasing the volume of a gas at constant temperature results in a higher positional probability due to an increased number of positions for the particles to occupy. b. Increasing the temperature of a gas at constant volume leads to a higher positional probability as the particles move faster and have a higher probability of being found at different positions within the container. c. Increasing the pressure of a gas at constant temperature decreases the positional probability since the particles are forced into a smaller space, reducing the number of positions they can occupy.

Step by step solution

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1. Increase in volume of a gas at constant T

To analyze this situation, we can use the ideal gas law. If the temperature (T) is held constant, then any change in volume (V) must be accompanied by a proportional change in pressure (P). When the volume of a gas increases, the particles of the gas get more space to occupy, thereby increasing the positional probability of the substance.
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2. Increase in temperature of a gas at constant V

When the temperature of a gas is increased at a constant volume, the pressure of the gas (P) will also increase according to the ideal gas law. With a rise in temperature, the kinetic energy of the particles increases. This means that the particles will move faster and have a higher probability of being found at different positions within the container, hence increasing the positional probability of the substance.
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3. Increase in pressure of a gas at constant T

If the pressure of a gas is increased while the temperature remains constant, the volume must change according to the ideal gas law. In this case, the volume (V) will decrease in response to the increased pressure (P). When the volume of a gas decreases, the particles are forced into a smaller space, decreasing the number of positions the particles can occupy, which in turn reduces the positional probability of the substance.

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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 is a fundamental principle in chemistry and physics, describing the behavior of ideal gases. It is represented by the equation \( PV = nRT \), where \( P \) stands for pressure, \( V \) for volume, \( n \) for the amount of substance (in moles), \( R \) for the ideal gas constant, and \( T \) for temperature. This law shows the relationship between the pressure, volume, and temperature of a gas.

It helps us understand how gases react to changes in their environment. For instance, if the volume of a gas is increased while maintaining a constant temperature, the pressure will decrease to keep the equation balanced. Conversely, if the temperature of gas rises, the pressure or volume (or both) must adjust accordingly to maintain the equality.
Volume and Temperature Relationship
The relationship between volume and temperature of a gas is described by Charles's Law, which states that the volume of a gas is directly proportional to its temperature, provided the pressure remains constant (\( V \propto T \)).

This implies that as the temperature of a gas increases, its volume increases as well, assuming pressure doesn't change. When thinking about gases, picture the particles moving more quickly as the temperature rises; they need more space to roam.
- If temperature increases and volume remains constant, more collisions with the container walls happen, increasing pressure. - With increasing volume at constant temperature, particles have more space, which impacts how often they collide with each other and the container walls.
Kinetic Energy of Gas Particles
The kinetic energy of gas particles plays a crucial role in understanding the behavior of gases. The equation \( KE = \frac{3}{2}kT \) relates a particle's kinetic energy to temperature, where \( k \) is Boltzmann's constant and \( T \) is the absolute temperature.

When temperature rises, the kinetic energy of gas particles also increases. This translates to faster-moving particles, causing more frequent and forceful collisions with the container walls, thereby impacting pressure.
- Higher temperatures mean particles zip around faster, potentially influencing both volume and pressure.- Increased kinetic energy also means a higher probability of particles being found in new places within a container, enhancing positional probability.
Pressure and Volume Relationship
The inverse relationship between pressure and volume of a gas is dictated by Boyle's Law, represented as \( PV = k \), where \( k \) is a constant for a given amount of gas at constant temperature.

According to Boyle’s Law, if the volume of a gas decreases, the pressure increases, provided temperature remains unchanged. Consider a balloon being squeezed; the space decreases, forcing molecules into closer proximity, hence increasing pressure.
- A decrease in volume reduces the space available for particle movement, lowering positional probability.- Conversely, increasing the volume at constant pressure allows particles to disperse more, raising positional probability.

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Most popular questions from this chapter

Consider the reaction: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightarrow 2 \mathrm{NO}_{2}(g)$$ where \(P_{\mathrm{NO}_{2}}=0.29\) atm and \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.6 \mathrm{atm} .\) For this reaction at these conditions, \(\Delta G=-1000 \mathrm{J}\) and \(\Delta G^{\circ}=6000 \mathrm{J}\) . Which of the following statements about this reaction is(are) true? a. The reverse reaction is spontaneous at these conditions. b. At equilibrium, \(P_{\mathrm{N}_{2} \mathrm{O}_{4}}\) will be greater than 1.6 \(\mathrm{atm}\) . c. The value of K for this reaction is greater than 1. d. The maximum amount of work this reaction can produce at these conditions is –6000 J. e. The reaction is endothermic.

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from 99.90% to 99.99% purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix \(4,\) predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature-dependent. Predict the sign of \(\Delta S_{\text { sum }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\) at 298 \(\mathrm{K}\) . Using these values and data in Appendix 4 calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C} .\) The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at \(50 .^{\circ} \mathrm{C}\) f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\) . The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\) . g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\) . Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\) . The boiling point for Nic CO) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is 29.0 \(\mathrm{kJ} / \mathrm{mol} .\) [Hint: The phase change reaction and the corresponding equilibrium expression are \(\mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{NiCO} 4}\) greater than the \(K\) value. \(]\)

Consider the reaction: $$\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ for which \(\Delta H\) is \(-233 \mathrm{kJ}\) and \(\Delta S\) is \(-424 \mathrm{J} / \mathrm{K}\) . a. Calculate the free energy change for the reaction \((\Delta G)\) at 393 \(\mathrm{K} .\) b. Assuming \(\Delta H\) and \(\Delta S\) do not depend on temperature, at what temperatures is this reaction spontaneous?

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}_{\text { rhombic}}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{SO}_{3}(g) \longrightarrow 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\) c. \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\)

Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{f} / k_{\mathrm{r}},\) where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{2} / R T}\right)\) b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction."

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