Calculating K values using ΔG = ΔG° + RT ln(Q)?
Calculate K at 298 kelvin
2H2S(g) + 3O2(g) = 2H2O(g) + 2SO2(g The answer is K= 6.57x10^173
Somehow I get K = 2.85x10^-5
So at 298 K nonstandard ΔG=0
So 0=ΔG° + RT ln(K)
which means ΔG° = - RT ln(K)
Do I find the standard ΔG° to find ΔG value or the equation for nonstandard ΔG= ΔH -TΔS?
Either way I get K values that are nothing compared to the answers in the book.
Help please, with lots of explaining :)
More Questions From This User See All
Answers & Comments
Verified answer
using the standard value calculate ΔG°
http://en.wikipedia.org/wiki/List_of_standard_Gibb...
ΔG° = [ 2ΔG°f(H2O) +2ΔG°f(SO2) ] - [ 2ΔG°f(H2S) + 3ΔG°f(O2) ]
ΔG° = [ 2 X -228.61 + 2 X -300.13 ] - [ 2 X -33.4 + 3 X 0 ]
similar to ΔH°f ...ΔG°f of elementary substance like O2(g) is taken as zero
ΔG° = [ -457.22 - 600.26 ] + 66.8
ΔG° = -1057.48 + 66.8
ΔG° = -990.68 kj/mole
now ΔG° = -RTln K
-990.68 X 1000 = -8.314 X 298 X ln K
990680 = 2477.572 X ln K
ln K = 990680/2477.572
ln K = 399.859
converting ln to log ...
2.303 X log K = 399.859
log K = 399.859/2.303
log K = 173.626
taking antilog ....
K = 10^173.626 = 4.227 X 10^173
which is almost equal to your answer
there is some difference between your given answer and mine because i dont know what value your book is using for ΔG°f values ....
feel free to ask any question
ln is organic log or log2. in case you do no longer do maths or physics you probable have not been instructed approximately it. that's relatively the different of e or exp. in case you seem at your calculator you are able to locate it. to be honest you do no longer relatively decide to now what it relatively is, merely a thank you to apply it. in case you notice 'ln' in and equation merely press the function on your calculator and you may get a life like answer.