Need help with a topic discussed in class

  • Thread starter Thread starter albaboi123
  • Start date Start date
  • Tags Tags
    Class Topic
AI Thread Summary
Soft drinks maintain their carbonation best when bottled under high pressure and stored in a refrigerator because these conditions enhance the solubility of carbon dioxide in the liquid. The equilibrium equation for this process is CO2 (g) + H2O (l) <--> H2CO3 (aq). At higher pressures and lower temperatures, more CO2 dissolves, forming carbonic acid, which is less likely to escape as gas. Conversely, higher temperatures and lower pressures lead to the decomposition of carbonic acid back into CO2 gas, resulting in a flat beverage. Maintaining optimal pressure and temperature is crucial for preserving the beverage's fizziness.
albaboi123
Messages
5
Reaction score
0
we were given this topic about soft drinks. The question is:

Ideally, soft drinks are bottled under high pressures and are stored in a refrigerator in a closed container. Explain why these conditions ensure the best (i.e. least flat) beverage. Write an equilbrium equation in your answer.

Any help would be appreciated.
 
Physics news on Phys.org
The carbonation in soft drinks is due to Carbon Dioxide gas dissolved in the liquid, as the CO2 escapes from solution, the soda goes flat.

The solubility of Carbon Dioxide in water (or any gas in a liquid for that matter) is influenced by the temperature and pressure. Gasses are more soluble in liquids at lower temperatures and at higher pressures.

CO2 (g) + H2O (l) <---> H2CO3 (aq)

When carbon Dioxide dissolves in water it forms Carbonic acid. Carbonic acid is not very stable and will decompose into Carbon Dioxide gas and water if given the chance.
An equilibrium is established between the Carbon Dioxide above the liquid (out of solution) and the Carbon Dioxide which is dissolved in the liquid. Under higher pressures and/or lower temperatures, the equlibrium is shifted to the products side of the reaction (formation of H2CO3), and under higher temperatures and/or lower pressures, the equilibrium moves more toward to the reactant side (formation of CO2 gas).
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top