Chemical Mechanisms of the Fizziness of Soda in Foods

Learning Objectives

This Exemplar will teach the following concept(s) from the ACS Examinations Institute General Chemistry ACCM:

• IV. A. 1. d. Gases in mixtures generally act as independent species that can be quantitatively described using partial pressure. 

• IV. D. 2. d. Gases are capable of dissolving in liquids and the extent of solubility is governed by the nature of the molecules involved and the pressure.  

• VIII. C. 1. b. The equilibrium constant, K, is a function of temperature.

Fizzy beverages have existed in nature far before the invention of soda. For example, the mineral water in natural springs was found to contain carbon dioxide after it dissolved limestone. However, it was not until the late 18th century that a man-made apparatus was created to infuse drinks with carbon dioxide gas. Joseph Priestley (who also discovered oxygen) generated carbonated water using chalk and sulfuric acid.⁶ 

Figure 1. The first carbonation apparatus, as designed by Joseph Priestly in 1772. 

 

Carbonation Reactions

When carbon dioxide exists in water, it becomes part of an equilibrium reaction. As shown below, the carbon dioxide reacts with water to form carbonic acid (\(H_2 CO_3 \)). 

\[\ce{CO2(g) + H2O(l) <=> H2CO3(aq)}∖n\nonumber \]

When carbon dioxide is in its gaseous form, it is visible as the bubbles one can see in fizzy soda. On the other hand, the carbonic acid becomes part of the solution and provides soda with a slightly sour burning sensation (which is what makes fizzy drinks so appealing). The main factor that determines how fizzy a soda may be is the state of equilibrium this reaction is in. At higher pressures, the reaction shifts to the right, trapping the carbon dioxide in aqueous (\(H_2 CO_3 \)), but at lower pressures, the carbon dioxide is free to exist as bubbles or escape the solution entirely.⁷

Solubility of CO₂

The solubility of carbon dioxide in a soda can depend on multiple factors. However, the major factors that determine how fizzy your soda is, and how fast it will go flat are the pressure of the surroundings and the thermodynamic favorability (which is dependent on temperature).⁷

Partial Pressure and Henry's Law

The solubility of a gas is defined as the concentration of that gas that will react with the solvent to form a solution, in this case carbonic acid (as shown above). One of the factors that plays a role in determining the solubility of a gas is the partial pressure of the gas in the surrounding environment. As a rule, the higher the partial pressure of a gas in the surrounding atmosphere, the more soluble it is in solution. This relationship is described in Henry’s Law³:

\[ C_g = k \cdot P_g \]

In this equation the solubility of the gas, C_g, is directly related to the partial pressure of the gas, P_g, in the immediate environment. k is a constant that is determined by the identities of the solute and solvent, as well as the temperature of the solution, and is commonly expressed in mol/L.³ Using this equation therefore, we can clearly see that when the solution is within an unopened can (which is sealed at high pressure in a factory), the partial pressure is high, causing the solubility of the gas to be very high. Before the can is opened, most of the carbon dioxide exists as carbonic acid in solution. 

Once the can is opened, however, the atmosphere surrounding the solution drastically changes. The partial pressure of carbon dioxide in the Earth’s atmosphere is much lower than that of the highly pressurized gas in the can. For this reason, the solubility of the gas rapidly becomes far lower, and the aqueous carbonic acid once again becomes gaseous carbon dioxide and liquid water. Because the carbon dioxide is less dense than the water, it will naturally rise and diffuse out of the solution. Here's an example problem:

Example 1: Henry's Law in a Soda Can

Problem

At 298 K, a unopened can of LaCroix (a popular sparkling water brand) contains CO₂ with a concentration (solubility) in solution of \( 1.8\cdot 10^-3 mol L^-1\). The partial pressure of the CO₂ above the liquid inside the can is 1.18 atm. After the can is open, and allowed to reach equilibrium with the surrounding air, what will be the new concentration of CO₂ in solution? (The partial pressure of CO₂ in the atmosphere is 0.0004 atm)

Solution

First, by starting with Henry's Law, we can plug in the known constants (equation 2): the concentration and partial pressure of the CO₂. Then, as shown in equation 3, we can solve for the value of k. 

\[ C_{g} \, = \, kP_{g} \]

\[ 1.8 \cdot 10^{-3}mol \cdot L^{-1} = k \cdot (1.18\,atm) \]

\[k = \frac{1.8 \cdot 10^{-3}mol \cdot L^{-1}}{1.18\,atm} = 0.001525\,mol\cdot L^{-1}\cdot atm^{-1}\]

Next, using the new partial pressure of CO₂ and our known value of k, we can now reuse Henry's Equation, this time to solve for the new concentration. It becomes clear that the concentration/fizziness of the soda has significantly decreased after being exposed to the air. 

\[  C_{new} = 0.001525 \cdot P_{new} = 0.001525 \cdot 0.0004 = 6.1 \cdot 10^{-7} mol\cdot L^{-1} \]

Thermodynamic Favorability and Le Chatelier's Principle

An additional factor that affects the solubility of a gas in a solution is temperature. Temperature is an important factor in determining the change in Gibbs Free Energy (ΔG₀) for a reaction. The smaller ΔG₀, the more favorable the reaction. The full equation for Gibbs Free Energy is as follows⁷:

\[   \Delta G_{0}=\Delta H_{0}-\Delta S_{0}   \]

In this equation, ΔH₀ is the change in enthalpy, T is the temperature in Kelvin, and ΔS₀ is the change in entropy. If a reaction is not thermodynamically favorable, such as in cases where ΔG₀ is positive, the reaction will not proceed. If it is a reversible reaction, then according to Le Chatelier’s Principle, the equilibrium will shift to the left⁷. 

In the case of the dissolution of gasses, solubility will almost always decrease as temperature increases. Gasses tend to have a much greater entropy than aqueous solutions, therefore ΔS₀ for a dissolution reaction will be negative. A high temperature will therefore cause ΔG₀ to be higher, which decreases the favorability of the reaction⁷. 

Example 2: Thermodynamic Favorability for "Sulfurous Soda"

Problem

Lord Voldemort pours a glass of Diet Coke from the fridge (4°C). He takes it outside where the temperature is 25°C, and Voldemort’s Diet Coke soon reaches thermal equilibrium with the atmosphere. When Voldemort returns to his drink, he is shocked and appalled to find that his Diet Coke has gone flat! He is convinced that Harry Potter has cast a nasty spell to make it so, but Snape informs him that it is a result of a change in thermodynamic equilibrium. While Snape is very good at potions, his chemistry isn’t quite up to scratch, and he decides to create a new kind of fizzy drink using sulfur dioxide rather than carbon dioxide in the hopes that it won’t go flat as quickly. The equation for the reaction of sulfur dioxide to form sulfurous acid is

\[  \ce{SO2(g) + H2O(l) -> H2SO3(aq)}  \]

Given the following data, determine which reaction is less thermodynamically favorable at 4°C. What about 25°C?

\(\ce{CO2(g)}\) Enthalpy of Formation^2	-393.5 kJ/mol
\( \ce{CO2(g)}\) Standard Entropy	213.8 J/mol K
\(\ce{H2CO3(aq)}\) Enthalpy of Formation^2	-677.1 kJ/mol
\(\ce{H2CO3(aq)}\) Standard Entropy	-56.9 J/mol K
\(\ce{SO2(g)}\) Enthalpy of Formation^5	-296.8 kJ/mol
\(\ce{SO2(g)}\) Standard Entropy	248.2 J/mol K
\(\ce{H2SO3(aq)}\) Enthalpy of Formation^5	-635.6 kJ/mol
\(\ce{H2SO3(aq)}\) Standard Entropy	-29.3 J/mol K
\(\ce{H2O(l)}\) Enthalpy of Formation^2	-285.8 kJ/mol
\(\ce{H2O(l)}\) Standard Entropy	70.0 J/mol K

Solution

To figure out which reaction is less thermodynamically favorable, we must calculate ΔG₀ for each reaction. Recall that the equation for ΔG₀ is ΔH₀ - TΔS₀. Therefore, to calculate ΔG₀ we must first calculate ΔH₀ and ΔS₀. We can do this by subtracting Enthalpy and Entropy values of the reactants from the products. These calculations are shown below.

\[\Delta H_0 (CO_2) = (-677.1 - (-393.5) - (-285.8)) kJ/mol = 2.2 kJ/mol\]

\[\Delta S_0 (CO_2) = (-56.9 - 213.8 - 70.0) J/mol K * (1 kJ/ 1000 J) = -0.3407 kJ/mol K\]

\[\Delta G_0 (CO_2) \mathrm{\:at\: 4\: Degrees\: C}= 2.2 kJ/mol - (277 K * -0.3407 kJ/mol K) = \textbf{96.57 kJ/mol}\]

\[\Delta G_0 (CO_2) \mathrm{\:at\: 25\: Degrees\: C}= 2.2 kJ/mol - (298 K * -0.3407 kJ/mol K) = \textbf{103.73 kJ/mol}\]

\[\Delta H_0 (SO_2) = (-635.6 - (-296.8) - (285.8)) kJ/mol = -53.0 kJ/mol\]

\[\Delta S_0 (SO_2) = (-29.3 - 248.2 - 70.0) J/mol K * (1 kJ/ 1000 J) = -0.3475 kJ/mol K\]

\[\Delta G_0 (SO_2) \mathrm{\:at\: 4\: Degrees\: C}= -53.0 kJ/mol - (277 K * -0.3475 kJ/mol K) = \textbf{43.26 kJ/mol}\]

\[\Delta G_0 (SO_2) \mathrm{\:at\: 25\: Degrees\: C}= -53.0 kJ/mol - (298 K * -0.3475 kJ/mol K) = \textbf{50.56 kJ/mol}\]

Evidently, Snape has succeeded; the change in Gibbs Free Energy for the SO₂ reaction is lower than that of the CO₂ reaction at both temperatures, so the dissolution of SO₂ is more favorable. It ought to be noted that as temperature increases, so too does the Gibbs Free Energy, which aligns with our previous understanding of the solubility of gasses and Henry’s Law.

In conclusion, the solubility of a gas in a liquid is a very complicated subject that requires the application of many different areas of chemistry to fully understand. External partial pressure and temperature are just two of the many factors that produce this phenomenon of 'fizziness', with these principles being applied in many different branches of science. Just to be clear though, you should NOT try the sulfuric acid soda!

References

1. Carbon dioxide - thermophysical properties https://www.engineeringtoolbox.com/C...es-d_2017.html (accessed Nov 10, 2022).

2. CO2 (G) + H2O (l) → H2CO3(AQ). http://chemistry-reference.com/react...asp?rxnnum=237 (accessed Nov 10, 2022).

3. Libretexts. 13.4: Solutions of gases in water- how soda pop gets its fizz https://chem.libretexts.org/Courses/..._Gets_Its_Fizz (accessed Nov 10, 2022).

4. Lv, J.; Ren, K.; Chen, Y. CO2 Diffusion in Various Carbonated Beverages: A Molecular Dynamics Study https://pubs.acs.org/doi/10.1021/acs.jpcb.7b10469 (accessed Nov 10, 2022).

5. SO2 (G) + H2O (l) → H2SO3(AQ). http://chemistry-reference.com/react...asp?rxnnum=589 (accessed Nov 10, 2022).

6. Wolinski, Cat. “The 18th Century Chemist Who Discovered Oxygen and Changed Champagne and Beer Forever.” VinePair, VinePair, 4 Aug. 2022, https://vinepair.com/articles/champa...20soda%20water.

7. Zumdahl, S. S.; DeCoste, D. J. In Chemical Principles; Cengage Learning: Boston, MA, 2017.