18. The following equation represents the chemical equilibrium in an unopened pop can: CO2(g) + H₂O → H₂CO3(aq) a. Explain what is happening in terms of equilibrium when you open a pop can, and why it produces effervescence (fizz). b. Explain in terms of equilibrium why pop can stay carbonated (fizzy) longer in a re-closed bottle than in an open pop can. C. Bonus: In Gr. 11 chemistry, you learned about how temperature affects the solubility of gases. Which direction of this equilibrium would you expect to be the exothermic direction? Why?

Introductory Chemistry: A Foundation
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Chapter17: Equilibrium
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18. The following equation represents the chemical equilibrium in an unopened pop can:
CO2(g) + H₂O(1) ↔ H₂CO3(aq)
a. Explain what is happening in terms of equilibrium when you open a pop can, and why it produces
effervescence (fizz).
b. Explain in terms of equilibrium why pop can stay carbonated (fizzy) longer in a re-closed bottle than in
an open pop can.
C.
Bonus: In Gr. 11 chemistry, you learned about how temperature affects the solubility of gases. Which
direction of this equilibrium would you expect to be the exothermic direction? Why?
Transcribed Image Text:18. The following equation represents the chemical equilibrium in an unopened pop can: CO2(g) + H₂O(1) ↔ H₂CO3(aq) a. Explain what is happening in terms of equilibrium when you open a pop can, and why it produces effervescence (fizz). b. Explain in terms of equilibrium why pop can stay carbonated (fizzy) longer in a re-closed bottle than in an open pop can. C. Bonus: In Gr. 11 chemistry, you learned about how temperature affects the solubility of gases. Which direction of this equilibrium would you expect to be the exothermic direction? Why?
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