Consider the reaction.
PbCO3(s)<--> PbO(s) + CO2(g).
calculate the equilibrium pressure of CO2 in the system at the following temperatures.(IN atm)
(a) 270°C
(b) 480°C
..... delta Hf(KJ/mol).. d.Gf(KJ/mol).. S(J/mol-K)
PbCO3(s)..-699.1 .. -625.5 .. 131.0
PbO(s) .. -217.3 .. -187.9 .. 68.70
CO2(g) ... -393.5 .. -394.4 .. 213.6
Please help me!!! U could help me on the first one then i'll do the second one myself
I think you can calculate dGf for the reaction (n*dGf products)-n*dGo reactants) and from there dGo = -RTlnK. Solve for ln K at each T.
Then lnK = pCO2
So do i plug in 270C in kelvin as my temp in the first formula???
yes
i still don't get the right answer
i have like 1.54*10-14
DURRRRR
To calculate the equilibrium pressure of CO2, we need to use the equilibrium constant expression, Kp, which relates the partial pressures of the reactants and products in the reaction. The equilibrium constant expression for the given reaction is:
Kp = (P_CO2)eq / (P_PbO)eq
where (P_CO2)eq and (P_PbO)eq are the equilibrium partial pressures of CO2 and PbO, respectively.
To solve this problem, we will use the following steps:
1. Convert the given temperatures in Celsius to Kelvin.
(a) 270°C = 270 + 273 = 543 K
(b) 480°C = 480 + 273 = 753 K
2. Calculate the change in Gibbs free energy (ΔG) for the reaction at each temperature using the equation:
ΔG = ΔGf_products - ΔGf_reactants
where ΔGf_products and ΔGf_reactants are the standard Gibbs free energies of formation for the products and reactants, respectively.
(a) ΔG = (-394.4) - (-625.5 + (-217.3)) = -394.4 + 842.8 = 448.4 kJ/mol
(b) ΔG = (-394.4) - (-625.5 + (-217.3)) = -394.4 + 842.8 = 448.4 kJ/mol
3. Calculate the equilibrium constant (Kp) using the equation:
Kp = exp(-ΔG/RT)
where R is the ideal gas constant (0.0821 atm⋅L/mol⋅K) and T is the temperature in Kelvin.
(a) Kp = exp(-448.4 / (0.0821 * 543)) = exp(-0.01385) ≈ 0.986
(b) Kp = exp(-448.4 / (0.0821 * 753)) = exp(-0.00722) ≈ 0.993
4. Use the equilibrium constant (Kp) to calculate the equilibrium pressure of CO2 (P_CO2)eq, assuming the pressure of PbO (P_PbO)eq is 1 atm.
(a) (P_CO2)eq = Kp * (P_PbO)eq = 0.986 * 1 atm ≈ 0.986 atm
(b) (P_CO2)eq = Kp * (P_PbO)eq = 0.993 * 1 atm ≈ 0.993 atm
Therefore, the equilibrium pressure of CO2 at 270°C is approximately 0.986 atm, and at 480°C is approximately 0.993 atm.