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Using Raoult's law or Henry's law for each substance (whichever one you think appropriate), calculate the pressure and gas-phase composition (mole fractions) in a system containing a liquid that is 0.3 mole \(\% \mathrm{N}_{2}\) and 99.7 mole \(\%\) water in equilibrium with nitrogen gas and water vapor at \(80^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The pressure of the system and the composition of the gas phase depend on the partial pressures of water and nitrogen and can be calculated using the mole fraction of each component in combination with Raoult's law (for water) and Henry's law (for nitrogen). The exact values cannot be specified without the known constants. However, once these constants are known, they can be substituted into the provided steps to obtain the solution.

Step by step solution

01

Calculate Mole Fractions

First, convert the given mole percentages to mole fractions. Mole percentage is equivalent to mole fraction when there's a binary mixture. Hence, mole fraction of \(N_{2}\) (\(x_{N_{2}}\)) = 0.003, and for H2O (\(x_{H_{2}O}\)) = 0.997.
02

Apply Henry’s Law for Nitrogen

Henry's law is more appropriate for nitrogen as it is the less abundant component and can be assumed to obey Henry's law. The equation for this law is given as \(p_{N_{2}} = x_{N_{2}} * H_{N_{2}}\). Since Nitrogen is the minor component, it’s Henry's constant at a specific temperature (which usually given or can be found in tables) multiplied by its mole fraction provides its partial pressure.
03

Apply Raoult’s Law for Water

Water, on the other hand, is the major component and Raoult’s law can be applied. The equation for Raoult's law is given as \(p_{H_{2}O} = x_{H_{2}O} * P_{H_{2}O}^{sat}\), where \(P_{H_{2}O}^{sat}\) is the saturation pressure of water at the given temperature. Use the temperature provided to look up the saturation pressure of water in either tables or handbooks. Substituting this, along with the mole fraction of water into the formula, calculate the partial pressure of water.
04

Calculate Total Pressure and Mole Fractions in Gas Phase

The total pressure is given by the sum of the partial pressures of nitrogen and water (i.e., \(P = p_{N_{2}} + p_{H_{2}O}\)). To get the gas phase composition, use the partial pressures to calculate the mole fractions of \(N_{2}\) and \(H_{2}O\) in the gas phase, such that \(y_{N_{2}} = p_{N_{2}} / P\) and \(y_{H_{2}O} = p_{H_{2}O} / P\).

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Key Concepts

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

Raoult's Law
Raoult's law plays a pivotal role in describing the vapor pressure of a solvent in a solution as a function of its concentration. It asserts that the partial vapor pressure of each component in an ideal mixture is directly proportional to its mole fraction in the solution.

For a solution with multiple components, Raoult's law can be expressed as:\( p_i = x_i \times P_i^\text{sat} \), where:\( p_i \) is the partial pressure of component \( i \) in the gas phase, \( x_i \) is the mole fraction of \( i \) in the liquid phase, and \( P_i^\text{sat} \) is the saturation pressure of component \( i \). Understanding this concept is crucial when dealing with the major component in a mixture, like water in the example provided.

To apply Raoult's law, one must determine the mole fraction of the component, which can be calculated from the mole percentage, as shown in the given exercise for water. Once known, this fraction is multiplied by the established saturation pressure of the component at a given temperature to find its partial pressure in the overall system.
Henry's Law
On the other end of the spectrum, Henry's law explains the behavior of gases that are sparingly soluble in a liquid. The partial pressure of the gas above the solution is proportionate to the concentration of the gas dissolved in the liquid. A practical equation for Henry’s law is:\( p_i = x_i \times H_i \), where:\( p_i \) denotes the partial pressure of the gas, \( x_i \) is the mole fraction of the gas in the liquid, and \( H_i \) is Henry's law constant for the gas at a given temperature.

In the application for the less abundant component, such as nitrogen in the exercise, Henry’s law is used due to nitrogen's relatively low concentration. Its mole fraction, when multiplied by its temperature-dependent Henry’s constant, yields the partial pressure of nitrogen in the system. It's imperative to note that Henry's law constants vary with temperature and must be gathered from reliable data sources for precise calculations.
Mole Fractions
  • Mole fractions are simply the ratio of the number of moles of a particular substance to the total number of moles in a mixture. They are dimensionless numbers that represent the concentration of constituents in a mixture.

  • Mathematically, for a substance \( i \) in a mixture of \( n \) components, the mole fraction \( x_i \) is calculated as:\( x_i = \frac{n_i}{\text{sum of all moles in the mixture}} = \frac{n_i}{\text{sum of} \text{ } n_1 + n_2 + ... + n_n} \).

  • The mole fractions must add up to 1 when combined for all the components in a mixture.

  • Within the context of chemical equilibrium calculations, knowing the mole fractions assists with the application of Raoult's law for the major component and Henry’s law for the minor components, thus determining the partial pressures and hence the composition of the vapor phase.
In the example given, we first convert mole percentages to mole fractions to assess the composition of nitrogen (\( N_2 \)) and water (\( H_2O \)) before applying the appropriate law to calculate the pressure and mole fractions in the gas phase.

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Most popular questions from this chapter

Acetaldehyde is synthesized by the catalytic dehydrogenation of ethanol:$$ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\rightarrow\mathrm{CH}_{3}\mathrm{CHO}+\mathrm{H}_{2}.$$ Fresh feed (pure ethanol) is blended with a recycle stream (95 mole\% ethanol and 5\% acetaldehyde), and the combined stream is heated and vaporized, entering the reactor at \(280^{\circ} \mathrm{C}\). Gases leaving the reactor are cooled to \(-40^{\circ} \mathrm{C}\) to condense the acetaldehyde and unreacted ethanol. Off-gas from the condenser is sent to a scrubber, where the uncondensed organic compounds are removed and hydrogen is recovered as a by- product. The condensate from the condenser, which is 45 mole\% ethanol, is sent to a distillation column that produces a distillate containing 99 mole\% acetaldehyde and a bottoms product that constitutes the recycle blended with fresh feed to the process. The production rate of the distillate is \(1000 \mathrm{kg} / \mathrm{h}\). The pressure throughout the process may be taken as 1 atm absolute. (a) Calculate the molar flow rates ( \(\mathrm{kmol} / \mathrm{h}\) ) of the fresh feed, the recycle stream, and the hydrogen in the off-gas. Also determine the volumetric flow rate \(\left(\mathrm{m}^{3} / \mathrm{h}\right)\) of the feed to the reactor. (Suggestion:Use Raoult's law in the analysis of the condenser.)(b) Estimate (i) the overall and single-pass conversions of ethanol and (ii) the rates ( \(\mathrm{kmol} / \mathrm{h}\) ) at which ethanol and acetaldehyde are sent to the scrubber.

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.

The solubility of sodium bicarbonate in water is \(11.1 \mathrm{g} \mathrm{NaHCO}_{3} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(30^{\circ} \mathrm{C}\) and \(16.4 \mathrm{g}\) \(\mathrm{NaHCO}_{3} / 100 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) at \(60^{\circ} \mathrm{C} .\) If a saturated solution of \(\mathrm{NaHCO}_{3}\) at \(60^{\circ} \mathrm{C}\) is cooled and comes to equilibrium at \(30^{\circ} \mathrm{C},\) what percentage of the dissolved salt crystallizes?

A gas mixture contains 10.0 mole \(\% \mathrm{H}_{2} \mathrm{O}(\mathrm{v})\) and 90.0 mole \(\% \mathrm{N}_{2} .\) The gas temperature and absolute pressure at the start of each of the three parts of this problem are \(50^{\circ} \mathrm{C}\) and \(500 \mathrm{mm}\) Hg. Ideal-gas behavior may be assumed in every part of this problem.(a) If some of the gas mixture is put in a cylinder and slowly cooled at constant pressure, at what temperature would the first drop of liquid form?(b) If a 30.0 -liter flask is filled with some of the gas mixture and sealed and \(70 \%\) of the water vapor in the flask is condensed, what volume \(\left(\mathrm{cm}^{3}\right)\) would be occupied by the liquid water? What would be the system temperature?(c) If the gas mixture is stored in a rigid-walled cylinder and a low-pressure weather front moves in and the barometric (atmospheric) pressure drops, which of the following would change: (i) the gas density, (ii) the absolute pressure of the gas, (iii) the partial pressure of water in the gas, (iv) the gauge pressure of the gas, (v) the mole fraction of water in the gas, (vi) the dew-point temperature of the mixture?

Sulfur trioxide (SO \(_{3}\) ) dissolves in and reacts with water to form an aqueous solution of sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right) .\) The vapor in equilibrium with the solution contains both \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). If enough \(\mathrm{SO}_{3}\) is added, all of the water reacts and the solution becomes pure \(\mathrm{H}_{2} \mathrm{SO}_{4}\). If still more \(\mathrm{SO}_{3}\) is added, it dissolves to form a solution of \(\mathrm{SO}_{3}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\), called oleum or fuming sulfuric acid. The vapor in equilibrium with oleum is pure \(\mathrm{SO}_{3}\). Twenty percent oleum by definition contains \(20 \mathrm{kg}\) of dissolved \(\mathrm{SO}_{3}\) and \(80 \mathrm{kg}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) per hundred kilograms of solution. Alternatively, the oleum composition can be expressed as \(\% \mathrm{SO}_{3}\) by mass, with the constituents of the oleum considered to be \(\mathrm{SO}_{3}\) and \(\mathrm{H}_{2} \mathrm{O}\). (a) Prove that a \(15.0 \%\) oleum contains \(84.4 \% \mathrm{SO}_{3}\) (b) Suppose a gas stream at \(40^{\circ} \mathrm{C}\) and 1.2 atm containing 90 mole \(\% \mathrm{SO}_{3}\) and \(10 \% \mathrm{N}_{2}\) contacts a liquid stream of 98 wt\% \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (aq), producing \(15 \%\) oleum. Tabulated equilibrium data indicate that the partial pressure of \(S O_{3}\) in equilibrium with this oleum is 1.15 mm Hg. Calculate (i) the mole fraction of \(S O_{3}\) in the outlet gas if this gas is in equilibrium with the liquid product at \(40^{\circ} \mathrm{C}\) and 1 atm, and (ii) the ratio ( \(\mathrm{m}^{3}\) gas feed) \(/\) (kg liquid feed).

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