   Chapter 12, Problem 12.87QP General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343

Solutions

Chapter
Section General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343
Textbook Problem

A gaseous mixture consists of 87.0 mole percent N2 and 13.0 mole percent O2 (the approximate composition of air). Suppose water is saturated with the gas mixture at 25°C and 1.00 atm total pressure, and then the gas is expelled from the water by heating. What is the composition in mole fractions of the gas mixture that is expelled? The solubilities of N2 and O2 at 25°C and 1.00 atm are 0.0175 g/L H2O and 0.0393 g/L H2O, respectively.

Interpretation Introduction

Interpretation:

Water is saturated with mixture of Nitrogen and Oxygen gases.  The mole fraction of the gases expelled when the solution is heated has to be determined.

Concept Introduction:

Solubility of a gas in liquid is explained by Henry’s law which states –

“At a constant temperature, the amount of a gas dissolved in a given volume of liquid is directly proportional to the partial pressure of the gas that is in equilibrium with the liquid”.

It is expressed as,

S α P

Where,

S = SolubilityP = Partial pressure

Introducing proportionality constant,

S = kHP

Where,

kH is Henry’s constant.

A solution is at least made up of two components.  Mole fraction of a component in the solution correlates to the ratio of number of moles of that component to the total number of moles.  It is expressed as,

Mole fraction = number of moles of a componenttotal number of moles in the solution

Explanation

Determine the solubility of each gas at its partial pressure using Henry’s law.

Given that mole fraction of N2 and O2 are 87% and 13% respectively.  The partial pressure of each gas (P2) is calculated as follows –

partial pressure of N2 = 1.00 atm×0.870 = 0.870 atmpartial pressure of O2 = 1.00 atm×0.130 = 0.130 atm

The solubility of N2 and O2 at 1 atm is 0.0175 g/L and 0.0393 g/L respectively.

Solubility of a same gas at two different pressures can be calculated as,

S2S1 = kHP2kHP1 = P2P1                                            ......(1)

Let S2 be the solubility of Nitrogen and Oxygen in water, S1 at its partial pressure P2. Calculate S2 by rearranging equation (1).

Then, solubility of N2 at its partial pressure P2,

S2 = S1×P2P1 = (0

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