A carbonated beverage is made by saturating water with carbondioxide at 0 degree

Question

A carbonated beverage is made by saturating water with carbondioxide at 0 degrees Celsius and a pressure of 3.0atm. The bottleis then opened at room temperature (25 degrees C), and comes toequilibrium with air in the room containing CO2(PCO2 = 3.4x10-4 atm). The Henry'slaw constant for solubility of CO2 in water is 0.0769M/atm at 0 degrees Celsius and 0.0313 M/atm at 25degrees Celsius. (a) What is the concentration of carbon dioxide in the bottlebefore it is opened? (b) What is the concentration of carbon dioxide in the bottleafter it has been opened and come to equilibrium with air?
(a) What is the concentration of carbon dioxide in the bottlebefore it is opened? (b) What is the concentration of carbon dioxide in the bottleafter it has been opened and come to equilibrium with air?

Explanation / Answer

a) Concentration of carbon dioxide in the bottlebefore it is opened :    At 0 0C         The pressure of carbon dioxide   (CO2) P CO2 = 3.0atm       The Henry's law constant forCO2 in water kH = 0.0769M/atm we know the relation for gas    Sgas = kH * Pgas                               Thereforeconcentration of CO2 at 0 0C i,ebefore opened the bottle                                                               SCO2 = (0.0769M/atm)* (3.0atm)                                                               = 0.2307M              The concentration of  CO2   inthe bottle before it is opened   = 0.2307M b) Concentration of carbon dioxide in the bottle afterit has been opened and come to equilibrium with air :          At 250C (room temperature )   CO2 gas is equilibrium with air after opened thebottle                         The pressure of carbon dioxide  (CO2) P CO2 =3.4*10-4atm                           TheHenry's law constant for CO2 in water kH = 0.0313 M/atm                                                  Fromthe eqution         Sgas = kH * Pgas                                                                                       = (0.0313 M/atm ) * (3.4*10-4atm )                                                                          = 1.06*10-5 M               Theconcentration of carbon dioxide equilibrium with air =1.06*10-5 M        So Now we calculate theconcentration of    CO2 gas in thebottle ater opened                                                                        = Conc.of CO2 gas before - Conc.of CO2 gas equilibrium with air                                                   = 0.2307M - 1.06*10-5 M                                                  = 0.2306 M              The concentration of    CO2gas in the bottle ater opened = 0.2306 M                                                                                      

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---