jnney
lemon
- Joined
- Mar 3, 2011
- Messages
- 1,437
- Gender
- Female
- HSC
- 2012
Using LCP, account for the sharp acidic taste of soft drinks.
Soft drinks contain carbon dioxide dissolved in water.
To begin, CO2(g) is converted into CO2(aq) under pressure [as it is in a closed system]. CO2(g) <---> CO2(aq). By LCP, the equilibrium will favour the forwards reaction and shift right to the side with the fewer moles as to counteract the increased pressure. Hence, [CO2(aq)] increases.
In turn, the [CO2(aq)] also increases in the reaction between water. CO2(aq) + H2O(l) <---> H2CO3-(aq). By LCP, the equilibrium favours the forwards reaction and shifts to the right to decrease the increased [CO2(aq)]. This causes CO2(aq) to dissolve more readily, increasing the [H2CO2(aq)], producing the sharp acidic taste.
Is this correct? I'm kind of confused.
Soft drinks contain carbon dioxide dissolved in water.
To begin, CO2(g) is converted into CO2(aq) under pressure [as it is in a closed system]. CO2(g) <---> CO2(aq). By LCP, the equilibrium will favour the forwards reaction and shift right to the side with the fewer moles as to counteract the increased pressure. Hence, [CO2(aq)] increases.
In turn, the [CO2(aq)] also increases in the reaction between water. CO2(aq) + H2O(l) <---> H2CO3-(aq). By LCP, the equilibrium favours the forwards reaction and shifts to the right to decrease the increased [CO2(aq)]. This causes CO2(aq) to dissolve more readily, increasing the [H2CO2(aq)], producing the sharp acidic taste.
Is this correct? I'm kind of confused.
Last edited: