2CO (g) + O2 <=> 2CO2(g)
Standard molar energies of formation at 288K is -137 kJ/mol for CO and -450 KJ/mol for CO2, find equilibrium constant.
I know that we’re supposed to multiply n to the Gf of products and take the sum and do the same to the reactants. However O2 is not given.
O2 is zero as are all elements in their standard state.
Okay, so I did it. I got 1174 for the delta G rxn and I used that to the other formula. However the next question is saying I’d the 450 was changed to -390. And I got 0 as K.
Liz, you should proof read your posts. You omitted something in that last sentence. If you clarify that please post your work so I can see what you're doing.
To find the equilibrium constant (K) for the given reaction, you need to use the standard molar energies of formation (ΔGf°) of CO and CO2. The equation you provided represents the equilibrium between carbon monoxide (CO) and carbon dioxide (CO2).
First, let's write the expression for the equilibrium constant (K) using the standard Gibbs free energy change (ΔG°) for the reaction:
ΔG° = ΣnΔGf°(products) - ΣnΔGf°(reactants)
where:
ΔG° = standard Gibbs free energy change
ΔGf° = standard molar energy of formation
n = the coefficient of each species in the balanced equation
In this case, we need to consider the formation of 2 moles of CO2:
ΔG° = (2 × ΔGf°(CO2)) - (2 × ΔGf°(CO))
Given the values you provided:
ΔGf°(CO) = -137 kJ/mol
ΔGf°(CO2) = -450 kJ/mol
Substituting these values into the equation, we get:
ΔG° = (2 × -450 kJ/mol) - (2 × -137 kJ/mol)
Simplifying the equation further:
ΔG° = -900 kJ/mol + 274 kJ/mol
ΔG° = -626 kJ/mol
Now, we can relate the ΔG° to the equilibrium constant (K) by using the following equation:
ΔG° = -RT ln(K)
where:
R = ideal gas constant (8.314 J/(K·mol))
T = temperature in Kelvin (288 K)
ln = natural logarithm
Converting kJ to J and Kelvin (since the ideal gas constant is in J/(K·mol)), we have:
ΔG° = -626,000 J/mol
Next, we solve for K:
-626,000 J/mol = -8.314 J/(K·mol) × 288 K × ln(K)
Simplifying the equation:
ln(K) = (-626,000 J/mol) / (8.314 J/(K·mol) × 288 K)
ln(K) ≈ -88.73
Finally, solve for K by taking the inverse natural logarithm (e^x):
K = e^(-88.73)
Using a scientific calculator or any calculator with exponential function to evaluate e^(-88.73), you'll find:
K ≈ 1.39 × 10^(-39)
Therefore, the equilibrium constant (K) for the given reaction is approximately 1.39 × 10^(-39).