- #1

tbn032

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- 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).

SO

find K

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 K

Now my confusion is K

K

No information about the pressure or concentration is given so how do I conclude that K

_{2}(g)+1/2O_{2}(g)**⇌**SO_{3}(g);ΔH^{o}=-98.32KJ/mole,ΔS^{o}=-95J/(mole-K).find K

_{p}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 K

_{eq}will be calculated using equation = Δ G = -RT ln K_{eq},ln K_{eq}= -70010J/-8.314 x 298 = 28.25 .or K_{eq}= e^{28.25}= 1.86x10^{12}.Now my confusion is K

_{eq}=K_{p}or K_{eq}=K_{c}(depending on if we measure the equilibrium constant in terms of pressure or molarity, gases can be measured in both).andK

_{p}= K_{c}(RT)^{Δn}and Δn=-1/2(for this reaction).No information about the pressure or concentration is given so how do I conclude that K

_{eq}=K_{p}or K_{eq}=K_{c}and thus calculate the correct value of K_{p}
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