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Chapter 17: Problem 96

For rubidium \(\Delta H_{\mathrm{vap}}^{\circ}=69.0 \mathrm{~kJ} / \mathrm{mol}\) at \(686^{\circ} \mathrm{C}\), its boiling point. Calculate \(\Delta S^{\circ}, q, w\), and \(\Delta E\) for the vaporization of \(1.00 \mathrm{~mole}\) of rubidium at \(686^{\circ} \mathrm{C}\) and \(1.00\) atm pressure.

Short Answer

Expert verified
In this problem, we calculated the thermodynamic properties of the vaporization of 1.00 mole of rubidium at its boiling point and 1.00 atm pressure. We found the following values: 1. The standard entropy change, \(\Delta S^{\circ} = 71.95 \, \mathrm{J/(mol \cdot K)}\) 2. The heat, q = 69.0 × 10³ J/mol 3. The work, w = -8036 J/mol 4. The internal energy change, \(\Delta E = 61000 \, \mathrm{J/mol}\)

Step by step solution

01

Convert the boiling point to Kelvin

Before we proceed, let's convert the boiling point from Celsius to Kelvin. Temperature in Kelvin: T = T(°C) + 273.15 T = 686°C + 273.15 = 959.15 K
02

Calculate the entropy change using the Clausius-Clapeyron equation

The equation we need is: \( \Delta S^\circ = \frac{\Delta H_{\mathrm{vap}}^\circ}{T_\mathrm{boil}}\) Using the given values, the entropy change can be calculated as: \( \Delta S^\circ = \frac{69.0 \, \mathrm{kJ/mol}}{959.15 \, \mathrm{K}} \) First, we convert the enthalpy from kJ to J: \( \Delta S^\circ = \frac{69.0 \times 10^3 \, \mathrm{J/mol}}{959.15 \, \mathrm{K}}\) Now, calculate the entropy change: \( \Delta S^\circ = 71.95 \, \mathrm{J/(mol \cdot K)} \)
03

Calculate the heat q using the enthalpy of vaporization

The heat involved in the vaporization process is equal to the enthalpy change. Since 1 mole of rubidium vaporizes, the heat q is: q = ΔHvap = 69.0 kJ/mol = 69.0 × 10^3 J/mol
04

Calculate the work w using the formula w = -PΔV

First, we need to find the change in volume (ΔV) during the vaporization process. We can use the ideal gas law equation, PV = nRT. For the initial state (liquid rubidium): P = 1 atm = 101325 Pa V = V(liquid), we assume this to be very small compared to the gas phase n = 1 mol R = 8.314 J/(mol·K) T = 959.15 K For the final state (rubidium vapor): P = 1 atm = 101325 Pa V = V(gas) n = 1 mol R = 8.314 J/(mol·K) T = 959.15 K The change in volume during vaporization is (ΔV = V(gas) - V(liquid)). Since V(liquid) is very small compared to V(gas), ΔV ≈ V(gas). We can rewrite the ideal gas equation for V(gas): \( V_\mathrm{gas} = \frac{nRT}{P} \) Now, plug in the values: \( V_\mathrm{gas} = \frac{1 \cdot 8.314 \, \mathrm{J/(mol \cdot K)} \cdot 959.15 \, \mathrm{K}}{1 \cdot 101325 \, \mathrm{Pa}} \) Calculate V(gas): \( V_\mathrm{gas} = 0.07931 \, \mathrm{m^3/mol} \) Now, we calculate the work w using the formula w = -PΔV: w = -P × ΔV ≈ -P × V(gas) w = -101325 Pa × 0.07931 m³/mol = -8036 J/mol
05

Calculate the change in internal energy ΔE using the first law of thermodynamics

The first law of thermodynamics states that ΔE = q + w. We already have the values for q and w calculated in Steps 3 and 4. ΔE = q + w ΔE = (69.0 × 10^3 J/mol) + (-8036 J/mol) ΔE = 61000 J/mol Summary: ΔS° = 71.95 J/(mol·K) q = 69.0 × 10^3 J/mol w = -8036 J/mol ΔE = 61000 J/mol

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Key Concepts

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

Clausius-Clapeyron equation
When studying the phase transitions of substances, the Clausius-Clapeyron equation is essential. It provides a way to quantify the relationship between pressure and temperature during phase changes, like the vaporization of a substance. This equation is particularly useful when we're interested in the entropy change, \( \Delta S^\circ \), of a vaporization process.

Typically, the equation is used to calculate the slope of the line on a phase diagram that represents phase equilibrium between two phases, such as liquid and vapor. In the context of rubidium vaporization, the Clausius-Clapeyron equation connects the enthalpy of vaporization, \( \Delta H_{\mathrm{vap}}^\circ \), with \( \Delta S^\circ \) and the boiling point temperature. By dividing the enthalpy of vaporization by the absolute temperature at which the vaporization occurs, we obtain the entropy change for the process.
Entropy change
The concept of entropy in thermodynamics often refers to the level of disorder or randomness in a system. During the phase transition, especially vaporization, the entropy of a substance increases as it goes from a more ordered phase (liquid) to a less ordered phase (gas).

The entropy change, \( \Delta S^\circ \), can be understood as a measure of how much the disorder in the system has changed due to vaporization. In calculations involving rubidium vaporization, \( \Delta S^\circ \) plays a pivotal role and can be directly found using the Clausius-Clapeyron equation by dividing the enthalpy of vaporization \( (\Delta H_{\mathrm{vap}}^\circ) \) by the temperature in Kelvin.
Enthalpy of vaporization
Enthalpy of vaporization, \( \Delta H_{\mathrm{vap}}^\circ \), is a term used to describe the amount of heat needed to turn a substance from its liquid phase into gas without changing its temperature, which in this case, is for rubidium.

This energy change reflects the amount of work needed to overcome the attractive forces between particles in the liquid state. It is a crucial value that can be measured or provided in problems like the one for rubidium. The larger the enthalpy of vaporization, the stronger the intermolecular forces that must be overcome. This value not only allows us to calculate the entropy change but is also equal to the heat, \( q \), absorbed during the vaporization of a mole of substance at its boiling point.
Internal energy change
Internal energy change, \( \Delta E \), gives insight into the overall change in energy within a system after a process occurs, such as rubidium vaporization. The first law of thermodynamics, which asserts that energy cannot be created or destroyed, only transformed, is used to calculate \( \Delta E \).

During our rubidium vaporization process, internal energy change includes both the heat added to the system (from the enthalpy of vaporization) and the work done by the system as it expands (volume work). The internal energy change is particularly important because it represents the total change of energy, accounting for both heat and work, thus painting a full picture of the energetic event.
Ideal gas law
The ideal gas law is a fundamental equation in chemistry and physics, which describes the behavior of an ideal gas by relating its pressure (\( P \)), volume (\( V \)), number of moles (\( n \)), and temperature (\( T \)) with the gas constant (\( R \)). The equation, \( PV = nRT \), assumes that the gas particles are in constant motion and collide elastically with no intermolecular forces.

In the context of rubidium vaporization, the ideal gas law enables us to estimate the volume of one mole of rubidium gas at a given temperature and pressure. This information is vital when calculating work done by the system or the gas as it expands against atmospheric pressure during vaporization. The change in volume (\( \Delta V \)) is a key factor in finding the work (\( w \)) involved in the vaporization process.

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Most popular questions from this chapter

The equilibrium constant \(K\) for the reaction $$ 2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g) $$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus \(1 / T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{~K}\) and a \(y\) -intercept of \(-14.51\). Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise 79

Consider the reaction $$ 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) $$ a. Calculate \(\Delta G^{\circ}\) for this reaction. The \(\Delta G_{\mathrm{f}}^{\circ}\) values for \(\mathrm{POCl}_{3}(g)\) and \(\mathrm{PCl}_{3}(g)\) are \(-502 \mathrm{~kJ} / \mathrm{mol}\) and \(-270 . \mathrm{kJ} / \mathrm{mol}\), respectively. b. Is this reaction spontaneous under standard conditions at \(298 \mathrm{~K}\) ? c. The value of \(\Delta S^{\circ}\) for this reaction is \(179 \mathrm{~J} / \mathrm{K} \cdot\) mol. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

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

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

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
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