- #1
tbn032
- 34
- 7
- Homework Statement
- for the reaction
SO2(g)+(1/2)O2(g)⇌SO3(g);ΔH=-98.32KJ/mole,ΔS=-95J/(mole-K).
find Kp at 298 Kelvin?
- Relevant Equations
- SO2(g)+(1/2)O2(g)⇌SO3(g);ΔH=-98.32KJ/mole,ΔS=-95J/(mole-K).
SO2(g)+1/2O2(g)⇌SO3(g);ΔHo=-98.32KJ/mole,ΔSo=-95J/(mole-K).
find Kp at 298 Kelvin?
In given question at first Δ G will be calculated using formula ΔG = Δ H – T x ΔS, by putting the given values in formula we get ΔG = -70.01 kJ/mol.
Then Keq will be calculated using equation = Δ G = -RT ln Keq ,ln Keq = -70010J/-8.314 x 298 = 28.25 .or Keq = e28.25 = 1.86x1012.
Now my confusion is Keq=Kpor Keq=Kc(depending on if we measure the equilibrium constant in terms of pressure or molarity, gases can be measured in both).and
Kp = Kc(RT)Δn and Δn=-1/2(for this reaction).
No information about the pressure or concentration is given so how do I conclude that Keq=Kp or Keq=Kc and thus calculate the correct value of Kp
find Kp at 298 Kelvin?
In given question at first Δ G will be calculated using formula ΔG = Δ H – T x ΔS, by putting the given values in formula we get ΔG = -70.01 kJ/mol.
Then Keq will be calculated using equation = Δ G = -RT ln Keq ,ln Keq = -70010J/-8.314 x 298 = 28.25 .or Keq = e28.25 = 1.86x1012.
Now my confusion is Keq=Kpor Keq=Kc(depending on if we measure the equilibrium constant in terms of pressure or molarity, gases can be measured in both).and
Kp = Kc(RT)Δn and Δn=-1/2(for this reaction).
No information about the pressure or concentration is given so how do I conclude that Keq=Kp or Keq=Kc and thus calculate the correct value of Kp
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