   Chapter 12, Problem 12.88QP 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 natural gas mixture consists of 88.0 mole percent CH4 (methane) and 12.0 mole percent C2 H6 (ethane). Suppose water is saturated with the gas mixture at 20°C and 1.00 atm total pressure, and the gas is then expelled from the water by heating. What is the composition in mole fractions of the gas mixture that is expelled? The solubilities of CH4 and C2 H6 at 20°C and 1.00 atm are 0.023 g/L H2O and 0.059 g/L H2O, respectively.

Interpretation Introduction

Interpretation:

Water is saturated with mixture of Methane and Ethane 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 Methane and ethane are 88% and 12% respectively.  The partial pressure of each gas (P2) is calculated as follows –

partial pressure of methane = 1.00 atm×0.880 =  0.880 atmpartial pressure of ethane = 1.00 atm×0.120 =  0.120 atm

The solubility of methane and ethane at 1 atm is 0.023 g/L and 0.059 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 Methane and Ethane in water, S1 at its partial pressure P2. Calculate S2 by rearranging equation (1).

Then, solubility of methane at its partial pressure P2,

S2 = S1×P2P1 = (0.023 g/L H2O)×0.880 atm1.00 atm = 0.0202 g/L H2O

Solubility of ethane at its partial pressure P2,

S2 = S1×P2P1 = (0.059 g/L H2O)×0

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts 