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Chapter 6: Problem 8

Liquid methyl ethyl ketone \((\mathrm{MEK})\) is introduced into a vessel containing air. The system temperature is increased to \(55^{\circ} \mathrm{C},\) and the vessel contents reach equilibrium with some MEK remaining in the liquid state. The equilibrium pressure is \(1200 \mathrm{mm} \mathrm{Hg}\).(a) Use the Gibbs phase rule to determine how many degrees of freedom exist for the system at equilibrium. State the meaning of your result in your own words.(b) Mixtures of MEK vapor and air that contain between 1.8 mole\% MEK and 11.5 mole\% MEK can ignite and burn explosively if exposed to a flame or spark. Determine whether or not the given vessel constitutes an explosion hazard.

Short Answer

Expert verified
The degrees of freedom for the system in equilibrium is 2, which means there are two independent variables that can be changed without affecting the state of the system. The mole fraction of MEK in the vapor phase is approximately 21.96%, so although the vapor is flammable, the present mixture does not constitute an explosion hazard.

Step by step solution

01

Gibbs phase rule to determine degrees of freedom

According to the Gibbs phase rule, the number of degrees of freedom (F) for a system can be determined by the formula \( F = C - P + 2 \), where \( C \) is the number of components and \( P \) is the number of phases. Given that in this system, the liquid and vapor forms of MEK and air are present, we have 2 components (MEK and air) and 2 phases (liquid and gas). Substituting these values into the given formula, we find that \( F = 2 - 2 + 2 = 2 \). This means that there are two independent variables that can be changed without changing the state of the system. In other words, once the temperature and pressure are defined, the composition of both phases are set.
02

Calculating mole percentage of MEK in vapor phase

We'll use the concept of partial pressure here, which states that the total pressure exerted by a mixture of gases is the sum of the individual pressures they would exert by themselves. We're given that the total pressure inside the vessel is 1200 mm Hg. To find the mole fraction of MEK in the air, we need to know the partial pressure of MEK, which can be looked up and is 263.5 mm Hg at 55 掳C. Thus the mole fraction of MEK in the air is then the partial pressure of the MEK divided by the total pressure, or \(\frac{263.5 \, \mathrm{mm} \, \mathrm{Hg}}{1200 \, \mathrm{mm} \, \mathrm{Hg}} \times 100 = 21.96 \% \)
03

Assessing explosion hazard

Given that mixtures of MEK and air that range between 1.8% and 11.5% can ignite and burn explosively, we see that 21.96% is outside of this range. This indicates that the given vessel does not pose an explosion hazard at this temperature and pressure.

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Key Concepts

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

Degrees of Freedom
In the context of the Gibbs phase rule, "degrees of freedom" refer to the number of independent variables that can be adjusted in a thermodynamic system without altering the system's equilibrium state. It's a helpful concept for understanding how much flexibility you have in changing conditions like pressure, temperature, or composition, while keeping the system stable.

The Gibbs phase rule formula is: \[ F = C - P + 2 \]where:
  • \( F \) is the number of degrees of freedom
  • \( C \) is the number of components (here, MEK and air)
  • \( P \) is the number of phases (liquid and gas)
For the MEK and air mixture, we have 2 components and 2 phases. Plugging in these values gives:\[ F = 2 - 2 + 2 = 2 \]This tells us that two variables, such as temperature and pressure, can independently be changed. Therefore, once temperature and pressure are set, the composition in each phase will be determined.
Explosion Hazard
An explosion hazard occurs when the concentration of a volatile substance in air falls within its explosive limits, making it ignitable by a spark or flame. Understanding these limits is crucial to avoid unsafe conditions in chemical processes involving flammable substances.

In the case of methyl ethyl ketone (MEK), the explosive range is between 1.8% and 11.5% by mole in air. If the concentration of MEK vapor is within this range, the mixture can ignite explosively.

Based on the calculations, the mole percentage of MEK vapor in the vessel is 21.96%, which is above the explosive limit. This indicates that under the given conditions, the vessel is not an explosion hazard, because the concentration is higher than the maximum explosive limit. This is helpful for engineers and safety officers to make informed decisions and ensure the safe operation of chemical processes.
Partial Pressure
Partial pressure is a crucial concept for understanding gas mixtures. It describes the pressure exerted individually by each component in a mixture as if it occupied the entire space by itself. This is especially important in systems like the one in the exercise, where we determine the proportion of a substance in a gas mixture.

The total pressure measured in a gas mixture is the sum of the partial pressures of all individual gases present. Mathematically, it can be expressed as:\[ P_{ ext{total}} = P_1 + P_2 + \ldots + P_n \]where each \( P \) represents the partial pressure of a component gas.

For the MEK and air system, knowing the total pressure (1200 mm Hg) and the partial pressure of MEK (263.5 mm Hg at 55 掳C), we can calculate the mole fraction of MEK:\[ \text{Mole fraction of MEK} = \frac{P_{ ext{MEK}}}{P_{ ext{total}}} \times 100 = \frac{263.5}{1200} \times 100 = 21.96\% \]This tells us how much MEK vapor is in the air, making it possible to gauge the potential for explosion and to control safety conditions.

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Most popular questions from this chapter

Nitric acid is a chemical intermediate primarily used in the synthesis of ammonium nitrate, which is used in the manufacture of fertilizers. The acid also is important in the production of other nitrates and in the separation of metals from ores. Nitric acid may be produced by oxidizing ammonia to nitric oxide over a platinum-rhodium catalyst, then oxidizing the nitric oxide to nitrogen dioxide in a separate unit where it is absorbed in water to form an aqueous solution of nitric acid.The reaction sequence is as follows:$$\begin{aligned} 4 \mathrm{NH}_{3}+5 \mathrm{O}_{2} & \rightarrow 4 \mathrm{NO}+6 \mathrm{H}_{2} \mathrm{O} \\\4 \mathrm{NO}+2 \mathrm{O}_{2} & \rightarrow 4 \mathrm{NO}_{2} \\\4 \mathrm{NO}_{2}+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{O}_{2} & \rightarrow 4 \mathrm{HNO}_{3}(\mathrm{aq}) \end{aligned}$$.Ammonia vapor produced by vaporizing pure liquid ammonia at 820 kPa absolute is mixed with air, and the combined stream enters the ammonia oxidation unit. Air at \(30^{\circ} \mathrm{C}, 1\) atm absolute, and \(50 \%\) relative humidity is compressed and fed to the process. A fraction of the air is sent to the cooling and hydration units, while the remainder is passed through a heat exchanger and mixed with the ammonia. The total oxygen fed to the process is the amount stoichiometrically required to convert all of the ammonia to HNO \(_{3},\) while the fraction sent to the ammonia oxidizer corresponds to the stoichiometric amount required to convert ammonia to NO.The ammonia reacts completely in the oxidizer, with \(97 \%\) forming NO and the rest forming \(\mathrm{N}_{2}\). Only a negligible amount of \(\mathrm{NO}_{2}\) is formed in the oxidizer. However, the gas leaving the oxidizer is subjected to a series of cooling and hydration steps in which the NO is completely oxidized to \(\mathrm{NO}_{2}\) which in turn combines with water (some of which is present in the gas from the oxidizer and the rest is added) to form a 55 wt\% aqueous solution of nitric acid. The product gas from the process may be taken to contain only \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\). (a) Taking a basis of \(100 \mathrm{kmol}\) of ammonia fed to the process, calculate (i) the volumes \(\left(\mathrm{m}^{3}\right)\) of the ammonia vapor and air fed to the process using the compressibility-factor equation of state; (ii) the amount (kmol) and composition (in mole fractions) of the gas leaving the oxidation unit; (iii) the required volume of liquid water \(\left(\mathrm{m}^{3}\right)\) that must be fed to the cooling and hydration units; and (iv) the fraction of the air fed to the ammonia oxidizer. (b) Scale the results from Part (a) to a new basis of 100 metric tons per hour of 55\% nitric acid solution.(c) Nitrogen oxides (collectively referred to as \(\mathrm{NO}_{x}\) ) are a category of pollutants that are formed in many ways, including processes like that described in this problem. List the annual emission rates of the three largest sources of \(\mathrm{NO}_{x}\) emissions in your home region. What are the effects of exposure to excessive concentrations of \(\mathrm{NO}_{x} ?\) (d) A platinum-rhodium catalyst is used in ammonia oxidation. Fxplain the function of the catalyst, describe its structure, and explain the relationship of the structure to the function.

A stream of 5.00 wt\% oleic acid in cottonseed oil enters an extraction unit at a rate of \(100.0 \mathrm{kg} / \mathrm{h}\). The unit operates as a single equilibrium stage (the streams leaving the unit are in equilibrium) at \(85^{\circ} \mathrm{C} . \mathrm{At}\) this temperature, propane and cottonseed oil are essentially immiscible, the vapor pressure of propane is 34 atm, and the distribution coefficient (oleic acid mass fraction in propane/oleic acid mass fraction in cottonseed oil) is \(0.15 .\)(a) Calculate the rate at which liquid propane must be fed to the unit to extract \(90 \%\) of the oleic acid. (b) Estimate the minimum operating pressure of the extraction unit. Explain your answer.(c) High-pressure operation is costly and introduces potential safety hazards. Suggest two possible reasons for using propane as the solvent when other less volatile hydrocarbons are equally good solvents for oleic acid.

A quantity of methyl acetate is placed in an open, transparent, three-liter flask and boiled long enough to purge all air from the vapor space. The flask is then sealed and allowed to equilibrate at \(30^{\circ} \mathrm{C},\) at which temperature methyl acetate has a vapor pressure of \(269 \mathrm{mm}\) Hg. Visual inspection shows \(10 \mathrm{mL}\) of liquid methyl acetate present.(a) What is the pressure in the flask at equilibrium? Explain your reasoning.(b) What is the total mass (grams) of methyl acetate in the flask? What fraction is in the vapor phase at equilibrium?(c) The above answers would be different if the species in the vessel were ethyl acetate because methyl acetate and ethyl acetate have different vapor pressures. Give a rationale for that difference.

\- A solution is prepared by dissolving \(0.5150 \mathrm{g}\) of a solute \((\mathrm{MW}=110.1)\) in \(100.0 \mathrm{g}\) of an organic solvent \((\mathrm{M} \mathrm{W}=94.10) .\) The solution is observed to have a freezing point \(0.41^{\circ} \mathrm{C}\) below that of the pure solvent. A second solution is prepared by dissolving \(0.4460 \mathrm{g}\) of a solute having an unknown molecular weight in 95.60 g of the original solvent. A freezing point depression of \(0.49^{\circ} \mathrm{C}\) is observed. Determine the molecular weight of the second solute and the heat of fusion ( \(\mathrm{kJ} / \mathrm{mol}\) ) of the solvent. The melting point of the pure solvent is \(-5.000^{\circ} \mathrm{C}\).

An important parameter in the design of gas absorbers is the ratio of the flow rate of the feed liquid to that of the feed gas. The lower the value of this ratio, the lower the cost of the solvent required to process a given quantity of gas but the taller the absorber must be to achieve a specified separation.Propane is recovered from a 7 mole \(\%\) propane \(-93 \%\) nitrogen mixture by contacting the mixture with liquid \(n\) -decane. An insignificant amount of decane is vaporized in the process, and \(98.5 \%\) of the propane entering the unit is absorbed.(a) The highest possible propane mole fraction in the exiting liquid is that in equilibrium with the propane mole fraction in the feed gas (a condition requiring an infinitely tall column). Using Raoult's law to relate the mole fractions of propane in the feed gas and liquid, calculate the ratio \(\left(\dot{n}_{L_{1}} / \dot{n}_{G_{2}}\right)\) corresponding to this limiting condition.(b) Suppose the actual feed ratio \(\left(\dot{n}_{L_{1}} / \dot{n}_{G_{2}}\right)\) is 1.2 times the value calculated in Part (a) and the percentage of the entering propane absorbed is the same (98.5\%). Calculate the mole fraction of propane in the exiting liquid.(c) What are the costs and benefits associated with increasing \(\left(\dot{n}_{L_{1}} / \dot{n}_{G_{2}}\right)\) from its minimum value [the value calculated in Part (a)]? What would you have to know to determine the most cost-effective value of this ratio?
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