Asked by Ezra
Acetylene, C2H2, can be converted to ethane, C2H6, by a process known as hydrogenation. The reaction is:
C2H2(g) + 2H2(g) === C2H6(g)
Given the following, what is the Kp for the reaction?
C2H2(g): 209.2 ΔG˚f (kJ/mol)
H2(g): 0 ΔG˚f (kJ/mol)
C2H6(g): -32.89ΔG˚f (kJ/mol)
ΔG˚ = -RTlnK
ΔG = ΔG˚RTlnQ
I don't what the answer is.
ΔG˚ for reaction = - RT lnK
242,090 J/mol = - (8.314 J/mol-K)(298.15 K) lnK
K = Kp = 3.85E-43
That's what I got but it wouldn't go through. Although that seems to be an extremely small number.
Answers
Answered by
Oliver
the delta G of the whole reaction you need to use is found by subtracting the delta G for your reactants from the delta G of your product
-32.89 (c2h6) - 209.2 (c2h2) + 0 (h2) = -242.09
then you have to plug it into the equation deltaG = -RTlnK, making sure to convert R to kJ by dividing it by 1000
8.314/1000 = 0.008314
and use the standard temp, 298.15K for T
-242.09 = -0.008314*298.15*lnK
this can be rewritten as
K=e^(G/-RT)
so
K=e^(-242.09/-0.008314*298.15)
K=e^97.713
the answer is 2.73E42
looks like your only problem was keeping your deltaG positive
-32.89 (c2h6) - 209.2 (c2h2) + 0 (h2) = -242.09
then you have to plug it into the equation deltaG = -RTlnK, making sure to convert R to kJ by dividing it by 1000
8.314/1000 = 0.008314
and use the standard temp, 298.15K for T
-242.09 = -0.008314*298.15*lnK
this can be rewritten as
K=e^(G/-RT)
so
K=e^(-242.09/-0.008314*298.15)
K=e^97.713
the answer is 2.73E42
looks like your only problem was keeping your deltaG positive
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.