Gf of H2CO3 = -616.1 KJ/mol
Gf of H20 = -237.1 KJ/mol
Gf of CO2 = - 394.4 KJ/mol
3.
A. Carbon dioxide dissolves in water to form carbonic acid. Estimate the thermodynamic equilibrium constant for this reaction using the ΔGf° values shown here.
B. Carbonic acid then ionizes in water (Ka1 = 4.5× 10-7). Ignoring Ka2, estimate K for the overall process by which CO2 and H2O form H and HCO3-.
C. What is the pressure of CO2 in equilibrium with carbonated water at 25 °C and pH = 4.57?
To solve these problems, we need to understand the concept of thermodynamic equilibrium constant and how it relates to standard Gibbs free energy. Let's break down each part of the question and explain how to find the answers.
A. The thermodynamic equilibrium constant (K) can be calculated using the ΔGf° values of the reactants and products. The equation for ΔG° is:
ΔG° = ΣnΔGf°(products) - ΣmΔGf°(reactants)
In this case, the reaction is CO2 (g) + H2O (l) -> H2CO3 (aq). The ΔGf° values for H2CO3, H2O, and CO2 are given:
ΔGf°(H2CO3) = -616.1 KJ/mol
ΔGf°(H2O) = -237.1 KJ/mol
ΔGf°(CO2) = -394.4 KJ/mol
To find ΔG° for this reaction, we substitute the values into the equation:
ΔG° = ΔGf°(H2CO3) - (ΔGf°(H2O) + ΔGf°(CO2))
ΔG° = -616.1 - (-237.1 - 394.4) = -616.1 + 237.1 + 394.4 = 15.6 KJ/mol
The equilibrium constant (K) can be calculated using the equation:
ΔG° = -RT ln(K)
Where R is the gas constant (8.314 J/(K·mol)) and T is the temperature in Kelvin. We can convert the temperature from Celsius to Kelvin by adding 273:
T = 25 + 273 = 298 K
Now we can solve for K:
15.6 = -8.314 * 298 * ln(K)
ln(K) = 15.6 / (-8.314 * 298) = -0.00757
K = e^(-0.00757) ≈ 0.992
Therefore, the estimated thermodynamic equilibrium constant for the reaction CO2 (g) + H2O (l) -> H2CO3 (aq) is approximately 0.992.
B. The overall process by which CO2 and H2O form H and HCO3- involves the ionization of carbonic acid in water. The equation is:
H2CO3 (aq) + H2O (l) -> H3O+ (aq) + HCO3- (aq)
The equilibrium constant (K) for this process is related to the acid dissociation constant (Ka1) of carbonic acid. The equation for K is:
K = [H3O+][HCO3-] / [H2CO3][H2O]
Since we are ignoring Ka2, we assume that the concentration of HCO3- is equal to the concentration of H2CO3. Therefore, we can simplify the equation:
K = [H3O+][HCO3-] / [H2CO3][H2O]
= [H3O+][H2CO3] / [H2CO3][H2O]
= [H3O+]/[H2O]
Since [H2O] is constant, we can rewrite the equation as:
K = [H3O+]
Now we know that K is equal to the concentration of H3O+. At pH = 4.57, we can calculate the concentration of H3O+:
[H3O+] = 10^(-pH) = 10^(-4.57) ≈ 2.99 × 10^(-5)
Therefore, the estimated equilibrium constant (K) for the overall process is approximately 2.99 × 10^(-5).
C. The pressure of CO2 in equilibrium with carbonated water can be calculated using Henry's law, which states that the concentration of a gas dissolved in a liquid is directly proportional to its partial pressure. The equation is:
C = k * P
Where C is the concentration of the dissolved gas, P is the partial pressure of the gas, and k is the Henry's law constant.
Since we know the concentration of CO2 (H2CO3) is equal to [H2CO3], which is the same as [HCO3-], we can rewrite the equation as:
[HCO3-] = k * P
From part B, we know that K is equal to [HCO3-]. Therefore, we can substitute:
K = k * P
Solving for P:
P = K / k
Now we need to determine the Henry's law constant (k) for CO2 in water. The value of k depends on temperature and varies with different gases. For CO2 in water at 25°C, the Henry's law constant is approximately 3.3 × 10^(-2) M/atm.
Substituting the values into the equation:
P = (2.99 × 10^(-5)) / (3.3 × 10^(-2))
≈ 9.06 × 10^(-4) atm
Therefore, the estimated pressure of CO2 in equilibrium with carbonated water at 25°C and pH = 4.57 is approximately 9.06 × 10^(-4) atm.