Calcium Oxide is used to remove pollutant SO2 from smokestack gases. The ΔG° of the overall reaction
CaO (s) + SO2 (g) + ½ O2 (g) ⇔ CaSO4 (s)
is -418.6 kJ. What is PSO2 in equilibrium with air (PO2 = 0.21 atm) and solid CaO?
dGo = -RTln*K
Solve for Keq.
Then set up Keq expression (remember solids don't enter the expression). You have Keq from above and pO2 is given in the problem, solve for pSO2. Post your work if you get stuck.
I'm still confused what numbers go where?
dGo is -418600
R is 8.314
T is not given but since they say dGo, I assume it is 298 K.
To find the equilibrium partial pressure of SO2 (PSO2), you can use the equation for Gibbs free energy (ΔG°) and the equation relating ΔG° to the equilibrium constant (Kc).
The equation for Gibbs free energy is:
ΔG° = -RT ln Kc
Where:
ΔG° is the standard Gibbs free energy change
R is the gas constant (8.314 J/(K·mol))
T is the temperature in Kelvin
ln is the natural logarithm
Kc is the equilibrium constant
First, convert the given ΔG° value from kJ to J:
-418.6 kJ = -418,600 J
Next, we need to determine the equilibrium constant (Kc) using the ΔG° value. Rearranging the equation, we have:
Kc = e^(-ΔG°/RT)
Substituting the values, we get:
Kc = e^(-(-418,600 J) / (8.314 J/(K·mol) * T))
Assuming a temperature of 298 K (room temperature), the equation becomes:
Kc = e^(-(-418,600 J) / (8.314 J/(K·mol) * 298 K))
Simplifying further, we have:
Kc = e^(159.355 / T)
Now, we need to calculate the equilibrium partial pressure of SO2 (PSO2). The equation for the reaction is:
CaO (s) + SO2 (g) + ½ O2 (g) ⇔ CaSO4 (s)
Since the reaction involves stoichiometric coefficients of 1 for both CaO and SO2, the partial pressure of SO2 (PSO2) in equilibrium is equal to the equilibrium constant (Kc). Therefore:
PSO2 = Kc
Finally, substitute the value of Kc into the equation:
PSO2 = e^(159.355 / T)
By plugging in the temperature (T), you can calculate the equilibrium partial pressure of SO2.