When the equilibrium constant K at 25oC for
2 O3(g) = 3 O2(g)
is calculated, using thermodynamic data(use delta G values), the value obtained is
a. ) 1.5 x 1057
b. ) 4.1 x 1028
c. ) 2.3 x 10-27
d. ) 130
e. ) 5.6 x 10-58
delta Go = -RTlnK
Look up values of delta Go for O3 and O2,
delta Go rxn = delta Goproducts - delta Go reactants.
To calculate the equilibrium constant (K) using thermodynamic data, you will need to use the equation:
ΔG = -RTlnK
Where:
ΔG is the change in Gibbs free energy
R is the gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
ln is the natural logarithm
First, let's look for the ΔG values for the reaction 2 O3(g) = 3 O2(g) at 25°C. Based on the given options, we can assume that the ΔG values for the reaction are provided in the standard Gibbs free energy of formation (ΔGf°) table.
To calculate the ΔG for the reaction, we need to use the formula:
ΔG = ΣnΔGf°(products) - ΣnΔGf°(reactants)
Where:
n is the stoichiometric coefficient of each species
ΔGf° is the standard Gibbs free energy of formation for each species
Let's calculate the ΔG for the given reaction.
For reactant O3:
nO3 = 2
ΔGf°(O3) = 0 (assumed since O3 is in its standard state)
For product O2:
nO2 = 3
ΔGf°(O2) = -298.15 kJ/mol (typical value from thermodynamic tables)
ΔG = (3 * -298.15 kJ/mol) - (2 * 0 kJ/mol)
ΔG = -894.45 kJ/mol
Now, let's calculate K using the ΔG value:
ΔG = -RTlnK
Since the temperature is 25°C, we need to convert it to Kelvin:
T = 25°C + 273.15 = 298.15 K
Plugging in the values:
-894.45 kJ/mol = -(8.314 J/(mol·K)) * 298.15 K * lnK
Solving for lnK:
lnK = -894.45 kJ/mol / (-(8.314 J/(mol·K)) * 298.15 K)
lnK = 38.70
To find K, we need to exponentiate the lnK value:
K = e^(lnK)
K = e^(38.70)
Using a calculator, the calculated K value is approximately 1.5 x 10^16. So, the correct answer would be:
a. ) 1.5 x 10^16