I\'m having difficulty trying to figure out the delta G in the lab for KHT. from

Question

I'm having difficulty trying to figure out the delta G in the lab for KHT.
from making a scattar plot in excell, i obtained [y = -7595.3x + 20.09]
my data looks like this:
1/T (1/K) lnKsp
0.003019 -2.887
0.003076 -3.25
0.003125 -3.56
0.003147 -3.83
0.003180 -4.06
0.003201 -4.27

from what they told me in lab, lnKsp = (-deltaH/RT) + deltaS/R, where -deltaH/R is M, 1/T is x, and delta S/R is b.
i got that deltaH = 76036.14 J and delta S = 167.02826
i'm not sure if delta G is spontaneous or not nonspontaneous. from what others in lab got, it seems that they're getting that KHT is spontaneous, but i seem to be getting that delta G is nonspontaneous. please help.

Explanation / Answer

I plotted the regression in Excel. Here's my bestfit equation from my Excel:

y = -7595.3x + 20.09

My S value agrees with you, but my H would be equal to -7595.3 * -8.314 = 63,147 Joules.

S is 167.02826J/K

G = H - TS = 63,147 Joules - T*167.02826J/K

This reaction's spontaneity is TEMPERATURE DEPENDENT. It is spontaneous at HIGH temperaures. Reactions are spontaneous when G<0

If we set G =0, we can solve for T, the minimum temperature one must reach before this reaction becomes spontaneous.

0 = 63,147 Joules - T*167.02826J/K

T = 63,147J/167.02826J/K = 354.7 K

Spontaneous around 82 Celsius or 355 K

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