The standard molar Gibbs free energy of formation of NO2 (g) at 298 K is 51.30 kJ · mol
−1
and that of N2O4 (g) is 97.82 kJ · mol
−1
.
What is the equilibrium constant at 25
◦
C for
the reaction
2 NO2(g) <=> N2O4(g) ?
dGrxn = (n*dGproducts) - (n*dGreactants)
Substitute and solve for dGrxn, then
dG = -RT*lnK
Substitute and solve for K
Ah, the equilibrium constant, the mathematician's favorite constant! Now, to calculate the equilibrium constant at 25°C for this reaction, we can use the formula K = e^(-ΔG°/RT), where K is the equilibrium constant, ΔG° is the standard Gibbs free energy of the reaction, R is the ideal gas constant, and T is the temperature in Kelvin.
Plugging in the values, we have:
ΔG° = (2 * 51.30) - 97.82 = 4.78 kJ · mol⁻¹ (because 2NO₂(g) - N₂O₄(g) is the reaction equation)
R = 8.314 J · mol⁻¹ · K⁻¹ (the ideal gas constant)
T = 25°C = 298 K
Now let's plug these values into the formula:
K = e^(-ΔG°/RT)
K = e^(-(4.78 * 10³)/(8.314 * 298))
And voila! You now have the equilibrium constant. I hope it brings you more joy than having to calculate it did!
To solve this problem, we need to use the equation relating the equilibrium constant (K) to the standard Gibbs free energy of reaction (∆G°):
∆G° = -RT ln(K)
Where:
∆G° = Standard Gibbs free energy of reaction
R = Gas constant (8.314 J/(mol·K))
T = Temperature (in Kelvin)
K = Equilibrium constant
To convert the temperatures from 298 K to 25°C, we use the following relationship:
T(°C) = T(K) - 273.15
Let's substitute the given values into the equation:
∆G° = -8.314 J/(mol·K) × (298 K) × ln(K)
Simplifying the equation further:
51.30 kJ/mol = -8.314 J/(mol·K) × (298 K) × ln(K)
Converting kJ to J:
51.30 kJ/mol = -8.314 × 10^3 J/(mol·K) × (298 K) × ln(K)
Simplifying:
51.30 = -2.476572 × 10^6 ln(K)
Dividing by -2.476572 × 10^6:
ln(K) = 51.30 / -2.476572 × 10^6
Using the natural logarithm:
K = e^(51.30 / -2.476572 × 10^6)
Now, let's calculate the value of K:
To find the equilibrium constant at 25°C for the reaction, we can use the equation:
ΔG° = -RT ln(K)
Where:
ΔG° is the standard Gibbs free energy change of the reaction,
R is the ideal gas constant (8.314 J K-1 mol-1),
T is the temperature in Kelvin,
ln is the natural logarithm function,
K is the equilibrium constant.
First, let's convert the given temperature from degrees Celsius to Kelvin. The conversion formula is:
T(K) = T(°C) + 273.15
So, 25°C + 273.15 = 298.15 K
Next, we need to calculate the standard Gibbs free energy change of the reaction (ΔG°) using the standard molar Gibbs free energy of formation values for NO2(g) and N2O4(g). The equation for ΔG° is:
ΔG° = Σ(nΔG°f(products)) - Σ(nΔG°f(reactants))
Where:
n is the stoichiometric coefficient of each species in the reaction,
ΔG°f is the standard molar Gibbs free energy of formation.
The balanced equation for the reaction is:
2NO2(g) ↔ N2O4(g)
Applying the equation, we have:
ΔG° = (2 * ΔG°f(N2O4(g))) - (2 * ΔG°f(NO2(g)))
ΔG° = (2 * 97.82 kJ/mol) - (2 * 51.30 kJ/mol)
ΔG° = 195.64 kJ/mol - 102.6 kJ/mol
ΔG° = 93.04 kJ/mol
Now, we can substitute the values into the equilibrium constant equation:
ΔG° = -RT ln(K)
93.04 kJ/mol = -(8.314 J K-1 mol-1) * (298.15 K) * ln(K)
Now, let's solve for K:
ln(K) = (93.04 kJ/mol) / (-(8.314 J K-1 mol-1) * (298.15 K))
ln(K) = -0.1208
Taking the exponential of both sides:
K = e^(-0.1208)
Using a scientific calculator or any calculator with an exponential function (e^x), we find:
K ≈ 0.886
Therefore, the equilibrium constant at 25°C for the reaction 2NO2(g) ↔ N2O4(g) is approximately 0.886.