help please What is delta G degree at 1000 degree C for the following reaction?
ID: 919836 • Letter: H
Question
help please What is delta G degree at 1000 degree C for the following reaction? (Use data from Appendix C)CaCO3(s) CaO (s) + CO2(g) Is the reaction spontaneous at 1000 degree and 1 atm? What is the value of k at 1000 degree C for this reaction? What is the partial pressure of CO ? A solution contains 0.10 M each of Ag (aq) and 0.1 M pb (aq). Solid Nacl is added, i.e.[CI ] increase, slowly without changing the total volume. Given k (AgCl) = 1.6 times 10-10 and k (PbCl2) = 2.4 times 10^-4, when PbCl2 starts to precipitate: What is the [Cl ] that remains in the solution? Based on your answer to A, What concentration of Ag+ will remain in solution?Explanation / Answer
Answer – 5) a) We are given reaction
CaCO3(s) <--->CaO(s) + CO2(g)
T = 1000 + 273 = 1273 K
We need to calculate the Ho and So for this reaction
We know,
Ho = sum of the Hof of product – sum of the Hof of reactant
= [Hof CaO(s) + Hof CO2(g)] – [HofCaCO3(s)]
= (-635.13 kJ + (-393.51)) – ( -1207.13)
= 178.5 kJ
So = sum of the So of product – sum of the So of reactant
= [So CaO(s) + So CO2(g)] – [SoCaCO3(s)]
= ( 38.20 + 213.68) – ( 88.70)
= 163.18 J
We know,
Go = Ho - T So
= 178.5 kJ – 1273 K *0.16318 kJ
= -29.22 kJ/mol
We know when Go is negative then process is spontaneous at forward reaction. So this reaction is spontaneous at 1000 oC and at 1.00 atm
b) We know formula
Go = -RTlnKp
So, -29220 J/mol = -8.314 J/mol.K*1273 K * ln K
ln K = -29220 J/mol / -8.314 J/mol.K*1273 K
= 2.76
So taking antiln form both side
So, Kp = 15.82
We know, Kp for this reaction is
Kp = P(CO2)
So partial pressure of CO2 is 15.82 atm
6) a) Given, [Ag+] = 0.10 M , [Pb2+] = 0.10 M
Ksp for AgCl = 1.6*10-10 , Ksp for PbCl2 = 2.4*10-4
The initial concentration of Cl- is not give
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