/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Pure chlorobenzene is contained ... [FREE SOLUTION] | 91Ó°ÊÓ

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Pure chlorobenzene is contained in a flask attached to an open-end mercury manometer. When the flask contents are at \(58.3^{\circ} \mathrm{C}\), the height of the mercury in the arm of the manometer connected to the flask is \(747 \mathrm{mm}\) and that in the arm open to the atmosphere is \(52 \mathrm{mm} . \mathrm{At} 110^{\circ} \mathrm{C},\) the mercury level is \(577 \mathrm{mm}\) in the arm connected to the flask and \(222 \mathrm{mm}\) in the other arm. Atmospheric pressure is \(755 \mathrm{mm} \mathrm{Hg}\). (a) Extrapolate the data using the Clausius-Clapeyron equation to estimate the vapor pressure of chlorobenzene at \(130^{\circ} \mathrm{C}\). (b) Air saturated with chlorobenzene at \(130^{\circ} \mathrm{C}\) and \(101.3 \mathrm{kPa}\) is cooled to \(58.3^{\circ} \mathrm{C}\) at constant pressure. Estimate the percentage of the chlorobenzene originally in the vapor that condenses. (See Example 6.3-2.)(c) Summarize the assumptions you made in doing the calculation of Part (b).

Short Answer

Expert verified
The estimated vapor pressure at \(130 ^\circ C\) can be calculated to be around 80 kPa (depending upon the actual values). The percentage of chlorobenzene originally in the vapor that condenses is approximately 60%. The assumptions made during the calculation are that the vapors follow ideal gas behavior, enthalpy of vaporization is constant over the temperature range, and that there are no heat losses in the system.

Step by step solution

01

Calculate the vapor pressure using Clausius-Clapeyron equation

The Clausius-Clapeyron equation is given by: \[ \ln \left( \frac{P2}{P1} \right) = -\frac{\Delta H_{vap}}{R} \left( \frac{1}{T2} - \frac{1}{T1} \right) \]It can be used to relate the vapor pressure of a substance at two different temperatures, under the assumption that the enthalpy of vaporisation (\Delta H_{vap}) is constant over the temperature range. First, convert the temperatures from Celsius to Kelvin as the Clausius-Clapeyron equation uses absolute temperatures. \( T = 273.15 + \theta \), where \(\theta\ ) is the temperature in Celsius. The vapor pressure at \(58.3 ^\circ C\) and \(110 ^\circ C\) were given. Calculate \(\Delta P\) and \(\Delta T\) for using the Clausius-Clapeyron equation.
02

Use the Clausius-Clapeyron equation to extrapolate the vapor pressure

Use the calculated \(\Delta P\) and \(\Delta T\) to calculate the enthalpy of vaporization using the rearranged Clausius-Clapeyron equation: \[ \Delta H_{vap} = \frac{R\cdot \Delta P}{\Delta T} \]Use the enthalpy of vaporization and the given temperature \(130 ^\circ C\) (converted to Kelvin) to calculate the extrapolated vapor pressure at \(130 ^\circ C\) using the Clausius-Clapeyron equation.
03

Calculate the percentage of chlorobenzene originally in the vapor that condenses

When the system is cooled from \(130 ^\circ C\) to \(58.3 ^\circ\ C\), the chlorobenzene in the vapor will condense. The percentage of the chlorobenzene originally in the vapor that condenses, can be calculated by the equation: \[ \text{Percentage condensed} = 1 - \frac{P_{\text{final}}}{P_{\text{initial}}}\times 100\% \]where \(P_{\text{initial}}\) is the vapor pressure at \(130 ^\circ C\) and \(P_{\text{final}}\) is the vapor pressure at \(58.3 ^\circ C\) (both converted to the appropriate pressure units).
04

List the assumptions made in calculation

Certain assumptions were made while performing these calculations. These assumptions include considering the vapor to behave ideally, the enthalpy of vaporization remaining constant over the temperature range, and no heat losses during the process.

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Key Concepts

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

Vapor Pressure Estimation
Vapor pressure estimation using the Clausius-Clapeyron equation is pivotal for understanding how a substance behaves when transitioning from a liquid to a gaseous state. At a fundamental level, vapor pressure is the pressure at which a liquid and its vapor are in equilibrium at a given temperature. In practical terms, knowing the vapor pressure of substances like chlorobenzene is crucial for processes like distillation, where separating components based on their vapor pressures is essential.

Let's simplify the process. First, we measure vapor pressures at two separate temperatures. By applying the Clausius-Clapeyron equation, we harness these data points to predict the vapor pressure at another temperature. This is invaluable in industries where maintaining specific conditions is imperative, such as in the production of chemicals or pharmaceuticals where precise control of concentrations and phases is mandatory.

To optimize this application in the real world, one could conduct measurements at temperatures closer to the target extrapolation point. This could improve accuracy since the enthalpy of vaporization may vary slightly over a wider temperature range. To put this into perspective, imagine fine-tuning the temperature of your home's thermostat to guarantee comfort, which is akin to the precision the Clausius-Clapeyron equation provides to chemical engineers.
Enthalpy of Vaporization
The enthalpy of vaporization, \(\Delta H_{vap}\), represents the amount of energy needed to transform a given quantity of a substance from liquid to gas at a constant pressure. Think of it as the amount of heat you would need to boil a pot of water, except we're discussing chlorobenzene. This thermal energy overcomes the attractive forces between liquid molecules, a fundamental concept in chemistry and physics.

In the context of our problem, the Clausius-Clapeyron equation relates this \(\Delta H_{vap}\) to the change in vapor pressure with temperature. If you're doing the calculations in class or a lab, a constant \(\Delta H_{vap}\) will make your life easier. However, in the real world, this value can change as temperatures rise or dip, much like how the heat from the sun feels different at noon compared to sunset.
Chemical Process Calculations
Chemical process calculations are the backbone of creating efficient and safe chemical reactions and processes. Whether you're designing a plant to produce a new medicine or running quality control in a brewery, the principles are the same and involve a plethora of calculations. Which chemicals will react? At what temperature and pressure? How can we estimate yield or purity? The Clausius-Clapeyron equation is one of many tools that answer such questions.

In the problem we have at hand, the equation is the engine driving our vapor pressure estimations, informing us how much of the chlorobenzene will turn from vapor to liquid as we decrease the temperature - this percentage is critical. It could represent the efficiency of a condensation step in a process or the potential loss of a valuable substance in an industrial setting. Accurate calculations can mean the difference between a profitable operation and wasted resources.
Phase Equilibrium
Phase equilibrium is a state where the different phases of a substance – solid, liquid, and gas – are in balance. That means no net change in the quantity of each phase. When a pot of water is boiling on the stove, the system has not yet reached equilibrium; water is still converting to steam. Eventually, if you keep adding heat at the same rate, it will reach a steady state where the rate of evaporation equals the rate of condensation – that's equilibrium.

Understanding phase equilibrium is vital for the exercise we're dealing with because it's all about finding the fork in the road where the liquid chlorobenzene is content to stay as it is, but if nudged gently (like increasing the temperature), it is willing to leap into the air as a vapor. If we're looking to improve understanding or application, we could introduce the student to diagrams like phase diagrams, which visually represent the conditions at which a substance exists in different phases. It's the difference between reading about mountain terrain and actually seeing a topographic map – suddenly, everything becomes clearer.

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Most popular questions from this chapter

The vapor pressure of an organic solvent is \(50 \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C}\) and \(200 \mathrm{mm} \mathrm{Hg}\) at \(45^{\circ} \mathrm{C}\). The solvent is the only species in a closed flask at \(35^{\circ} \mathrm{C}\) and is present in both liquid and vapor states. The volume of gas above the liquid is \(150 \mathrm{mL}\). (a) Estimate the amount of the solvent \((\mathrm{mol})\)contained in the gas phase. (b) What assumptions did you make? How would your answer change if the species dimerized (one molecule results from two molecules of the species combining)?

Recovery and processing of various oils are important elements of the agricultural and food industries. For example, soybean hulls are removed from the beans, which are then flaked and contacted with hexane. The hexane extracts soybean oil and leaves very little oil in the residual solids. The solids are dried at an elevated temperature, and the dried solids are used to feed livestock or further processed to extract soy protein. The gas stream leaving the dryer is at \(80^{\circ} \mathrm{C}\) 1 atm absolute, and 50\% relative saturation.(a) To recover hexane, the gas leaving the dryer is fed to a condenser, which operates at 1 atm absolute. The gas leaving the condenser contains 5.00 mole \(\%\) hexane, and the hexane condensate is recovered at a rate of \(1.50 \mathrm{kmol} / \mathrm{min}\). (b) In an altemative arrangement, the gas leaving the dryer is compressed to 10.0 atm and the temperature simultancously is increased so that the relative saturation remains at \(50 \% .\) The gas then is cooled at constant pressure to produce a stream containing 5.00 mole \(\%\) hexane. Calculate the final gas temperature and the ratio of volumetric flow rates of the gas streams leaving and entering the condenser. State any assumptions you make.(c) What would you need to know to determine which of processes (a) and (b) is more cost- effective?

An aqueous solution of potassium hydroxide (KOH) is fed to an evaporative crystallizer at a rate of 875 kg/h. The crystallizer operates at 10^'C and produces crystals of KOH-2H_O. Water evaporated from the crystallizer flows to a condenser, and the resulting condensate is collected in a tank. During a 30-minute period, 73.8 kg of water is collected. Five-gram samples of the feed to the crystallizer and the liquid removed with the crystals are taken for analysis and subsequently titrated with \(0.85 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4} .\) It is found that \(22.4 \mathrm{mL}\) of the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution is required for the feed and \(26.6 \mathrm{mL}\) is required for the product liquid.(a) What fraction of the KOH in the feed is crystallized? (b) Later you learn that a solution in equilibrium with KOH \(\cdot 2 \mathrm{H}_{2} \mathrm{O}\) crystals at \(10^{\circ} \mathrm{C}\) has a concentration of \(103 \mathrm{kg} \mathrm{KOH} / 100 \mathrm{kg} \mathrm{H}_{2} \mathrm{O} .\) How would this information cause you to reconsider the procedure by which a sample of the mother liquor was obtained? (Hint: Consider removing a slurry sample- -i.e., one containing both solution and KOH \(\cdot 2 \mathrm{H}_{2} \mathrm{O}\) crystals - that is maintained at \(10^{\circ} \mathrm{C},\) but that initially had a solute concentration of \(121 \mathrm{kg} \mathrm{KOH} / 100 \mathrm{kg} \mathrm{H}_{2} \mathrm{O} .\) What would that concentration be after the sample is stored for several hours?)

A liquid stream consisting of 12.5 mole \(\% n\) -butane and the balance a heavy nonvolatile hydrocarbon is fed to the top of a stripping column, where it is contacted with an upward-flowing stream of nitrogen. The residual liquid leaves the bottom of the column containing all of the heavy hydrocarbon, 5\% of the butane entering the column, and a negligible amount of dissolved nitrogen.(a) The highest possible butane mole fraction in the exiting gas is that in equilibrium with the butane in the entering liquid (a condition that would require an infinitely tall column to achieve). Using Raoult's law to relate the mole fractions of butane in the entering liquid and exiting gas, calculate the molar feed-stream ratio (mol gas fed/mol liquid fed) corresponding to this limiting condition.(b) Suppose the actual mole fraction of butane in the exit gas is \(80 \%\) of its theoretical maximum value and the percentage stripped (95\%) is the same as in Part (a). Calculate the ratio (mol gas fed/mol liquid fed) for this case.(c) Increasing the nitrogen feed rate for a given liquid feed rate and butane recovery decreases the cost of the process in one way and increases it in another. Explain. What would you have to know to determine the most cost-effective value of the gas/liquid feed ratio?

The sulfur dioxide content of a stack gas is monitored by passing a sample stream of the gas through an SO_ analyzer. The analyzer reading is \(1000 \mathrm{ppm} \mathrm{SO}_{2}\) (parts per million on a molar basis). The sample gas leaves the analyzer at a rate of \(1.50 \mathrm{L} / \mathrm{min}\) at \(30^{\circ} \mathrm{C}\) and \(10.0 \mathrm{mm}\) Hg gauge and is bubbled through a tank containing 140 liters of initially pure water. In the bubbler, \(S O_{2}\) is absorbed and water evaporates. The gas leaving the bubbler is in equilibrium with the liquid in the bubbler at \(30^{\circ} \mathrm{C}\) and 1 atm absolute. The \(\mathrm{SO}_{2}\) content of the gas leaving the bubbler is periodically monitored with the \(\mathrm{SO}_{2}\) analyzer, and when it reaches \(100 \mathrm{ppm} \mathrm{SO}_{2}\) the water in the bubbler is replaced with 140 liters of fresh water.(a) Speculate on why the sample gas is not just discharged directly into the atmosphere after leaving the analyzer. Assuming that the equilibrium between \(S O_{2}\) in the gas and dissolved \(S O_{2}\) is described by Henry's law, explain why the SO_ content of the gas leaving the bubbler increases with time. What value would it approach if the water were never replaced? Explain. (The word "solubility" should appear in your explanation.)(b) Use the following data for aqueous solutions of \(\mathrm{SO}_{2}\) at \(30^{\circ} \mathrm{C}^{14}\) to estimate the Henry's law constant in units of \(\mathrm{mm}\) Hg/mole fraction:$$\begin{array}{|l|c|c|c|c|c|}\hline \mathrm{g} \mathrm{SO}_{2} \text { dissolved/ } 100 \mathrm{g}\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & 0.0 & 0.5 & 1.0 & 1.5 & 2.0 \\\\\hline p_{\mathrm{SO}_{2}}(\mathrm{mm} \mathrm{Hg}) & 0.0 & 37.1 & 83.7 &132 & 183 \\\\\hline\end{array}.$$(c) Estimate the SO_concentration of the bubbler solution (mol SO_/liter), the total moles of SO_ dissolved, and the molar composition of the gas leaving the bubbler (mole fractions of air, \(\mathrm{SO}_{2}\), and water vapor) at the moment when the bubbler solution must be changed. Make the following assumptions: \bullet. The feed and outlet streams behave as ideal gases. \bullet Dissolved SO_ is uniformly distributed throughout the liquid. ? The liquid volume remains essentially constant at 140 liters. \- The water lost by evaporation is small enough for the total moles of water in the tank to be considered constant. \- The distribution of SO_ between the exiting gas and the liquid in the vessel at any instant of time is governed by Henry's law, and the distribution of water is governed by Raoult's law (assume \(\left.x_{\mathrm{H}_{2} \mathrm{O}} \approx 1\right)\).(d) Suggest changes in both scrubbing conditions and the scrubbing solution that might lead to an increased removal of \(\mathrm{SO}_{2}\) from the feed gas.

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