1. Calculate delta S at 27*c: 2CH4 (g) --> C2H6 (g)+ H2 (g) 2. Calculate delta S
ID: 982668 • Letter: 1
Question
1. Calculate delta S at 27*c: 2CH4 (g) --> C2H6 (g)+ H2 (g)2. Calculate delta S at 27*c: 2NH3 (g) --> N2H4 (g) + H2 (g)
3. If delta H (+) and delta S (-) is it spontaneous? 1. Calculate delta S at 27*c: 2CH4 (g) --> C2H6 (g)+ H2 (g)
2. Calculate delta S at 27*c: 2NH3 (g) --> N2H4 (g) + H2 (g)
3. If delta H (+) and delta S (-) is it spontaneous? 2CH4 (g) --> C2H6 (g)+ H2 (g)
2. Calculate delta S at 27*c: 2NH3 (g) --> N2H4 (g) + H2 (g)
3. If delta H (+) and delta S (-) is it spontaneous?
Explanation / Answer
1)
delts Sf (CH4) = 182.26 J/Kmol
delts Sf (C2H6) = 229.60 J/Kmol
delts Sf (H2) = 130.68 J/mol K
delta S(rxn) = delta S (H2) + delta S (C2H6) - 2* delta S (CH4)
= 130.68 + 229.60 - 2*182.26
= -4.24 J/K mol
ANSWER: -4.24 J/K mol
2)
delts Sf (NH3) = 192.45 J/K mol
delts Sf (N2H4) = 121.21 J/K mol
delts Sf(H2) = 130.68 J/mol K
delta S(rxn) = delta S (H2) + delta S (N2H4) - 2* delta S (NH3)
= 130.68 + 121.21 - 2*192.45
= -133.01 J/K mol
Answer: -133.01 J/K mol
3)
for sponataneous, delta G should be negative
delta G = delta H - T* delta S
If delta H (+) and delta S (-) then delta G will be come positive
Hence it is not spontaneous
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