the standard free energies of formation of SO2(g) and SO3(g) are -300.4 and -370.4kJ mol-1 respectively. Calculate the value of the equilibrium constant at 25'C for the reaction 2SO2(g)+O2(g)==>2SO3(g)
dGrxn = (n*dGf products) - (n*dGf reactants), then
dGrxn = -RT*lnK
Solve for K.
Well, let's calculate it using a really sophisticated and advanced method called "math"! Don't worry, it's not as scary as it sounds.
To find the equilibrium constant (K) at 25°C, we'll use the formula:
ΔG° = -RT ln(K)
Where ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J/(mol⋅K)), T is the temperature in kelvin (298 K for 25°C), and ln represents the natural logarithm.
First, let's convert the standard free energies of formation from kJ to J:
ΔG°(SO2) = -300.4 kJ/mol = -300,400 J/mol
ΔG°(SO3) = -370.4 kJ/mol = -370,400 J/mol
Now, let's plug the values into the formula:
ΔG° = -RT ln(K)
-300,400 J/mol - 298 K = 8.314 J/(mol⋅K) ln(K)
Simplifying:
ln(K) = -(300,400 J/mol - 298 K) / (8.314 J/(mol⋅K))
ln(K) = -360,698.56 / (8.314)
ln(K) ≈ -43420.383
Now, let's solve for K:
K = e^(ln(K))
K ≈ e^(-43420.383)
K ≈ 7.55 x 10^(-18924)
But hold on, that's a ridiculously small number! Let's face it, that reaction won't be happening much. It's more like finding a unicorn riding a rainbow while juggling flaming marshmallows. So, good luck finding any significant amounts of SO3!
To calculate the equilibrium constant (K) at 25°C for the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g), we need to use the standard free energies of formation (∆G°f) for each compound.
The equation for calculating K using standard free energies of formation is given as:
∆G° = ∑∆G°f(products) - ∑∆G°f(reactants)
First, let's calculate the ∆G° for the reactants and products.
∆G° (SO2) = -300.4 kJ/mol (Given)
∆G° (SO3) = -370.4 kJ/mol (Given)
Since the reaction involves 2 moles of SO2 and 1 mole of O2 forming 2 moles of SO3, we will multiply the respective ∆G° values by their stoichiometric coefficients.
∆G° (reactants) = 2*(∆G° (SO2)) + ∆G° (O2)
= 2*(-300.4 kJ/mol) + 0 kJ/mol (O2 is an element and its ∆G°f is zero)
= -600.8 kJ/mol
∆G° (products) = 2*(∆G° (SO3))
= 2*(-370.4 kJ/mol)
= -740.8 kJ/mol
Now, substitute the values into the equation:
∆G° = ∑∆G°f(products) - ∑∆G°f(reactants)
∆G° = ∆G° (products) - ∆G° (reactants)
∆G° = -740.8 kJ/mol - (-600.8 kJ/mol)
∆G° = -740.8 kJ/mol + 600.8 kJ/mol
∆G° = -140.0 kJ/mol
Since we're calculating the equilibrium constant at 25°C, we also need to convert the temperature from Celsius to Kelvin: T = 25°C + 273.15 = 298.15 K.
The equation relating K and ∆G° at a given temperature is:
∆G° = -RT ln(K)
Where:
R = Ideal Gas Constant = 8.314 J/(mol·K)
T = Temperature in Kelvin
Substituting the values:
-140.0 kJ/mol = -(8.314 J/(mol·K))(298.15 K) ln(K)
To convert kJ to J, multiply by 1000:
-140,000 J/mol = -(8.314 J/(mol·K))(298.15 K) ln(K)
Simplifying:
140,000 J/mol = (8.314 J/(mol·K))(298.15 K) ln(K)
Divide both sides by (8.314 J/(mol·K))(298.15 K):
140,000 J/mol / (8.314 J/(mol·K))(298.15 K) = ln(K)
Calculating the left side of the equation:
140,000 J/mol / (8.314 J/(mol·K))(298.15 K) ≈ 57.1
Now, taking the natural logarithm (ln) of both sides:
ln(K) ≈ 57.1
Finally, solving for K:
K ≈ e^(57.1)
K ≈ 5.8 x 10^24
Therefore, the value of the equilibrium constant (K) at 25°C for the reaction 2SO2(g) + O2(g) ⇌ 2SO3(g) is approximately 5.8 x 10^24.
To calculate the value of the equilibrium constant (K) at 25°C for the given reaction, we need to use the standard free energies of formation.
First, let's write the balanced equation for the reaction:
2SO2(g) + O2(g) → 2SO3(g)
According to the definition of standard free energy change (ΔG°), it is related to the equilibrium constant (K) as follows:
ΔG° = -RT ln(K)
Where:
ΔG°: Standard free energy change
R: Gas constant (8.314 J/mol·K)
T: Temperature in Kelvin (25°C = 298.15 K)
K: Equilibrium constant
Now, we can calculate the value of the equilibrium constant using the standard free energies of formation:
ΔG° = ∑nΔG°f(products) - ∑nΔG°f(reactants)
Given:
ΔG°f(SO2(g)) = -300.4 kJ/mol
ΔG°f(SO3(g)) = -370.4 kJ/mol
Using the equation, we can calculate ΔG°:
ΔG° = (2 × ΔG°f(SO3(g))) - (2 × ΔG°f(SO2(g))) - ΔG°f(O2(g))
ΔG° = (2 × -370.4 kJ/mol) - (2 × -300.4 kJ/mol) - 0 kJ/mol
ΔG° = -740.8 kJ/mol + 600.8 kJ/mol
ΔG° = -140 kJ/mol
Next, we need to convert kJ/mol to J/mol:
ΔG° = -140 kJ/mol × 1000 J/1 kJ
ΔG° = -140,000 J/mol
Now, we can substitute this value into the equation to calculate the equilibrium constant (K):
-RT ln(K) = -140,000 J/mol
R = 8.314 J/mol·K
T = 298.15 K
-8.314 J/mol·K × 298.15 K ln(K) = -140,000 J/mol
ln(K) = -140,000 J/mol / (-8.314 J/mol·K × 298.15 K)
ln(K) ≈ 20.62
Finally, we can solve for K using the natural logarithm function:
K ≈ e^(ln(K))
K ≈ e^(20.62)
K ≈ 2.43 x 10^8 (approximately)
Therefore, the value of the equilibrium constant at 25°C for the reaction 2SO2(g) + O2(g) → 2SO3(g) is approximately 2.43 x 10^8.