Question
Calculate the temp at which the reaction (2POCL3(g) -> 2PCl3(g)+O2(g))would become spontaneous. Given: DH= +572kJ and DS= 179 J/K.
Would the answer be 298 K? I found DG to be 518.658 kJ, and then plugged in all the values to DG= DH-TDS.
Or, should I ignore the DG, set DH-TDS to = 0, and find the answer to be about 3195.53 K?
Which is the proper way to go about answering this question?
Would the answer be 298 K? I found DG to be 518.658 kJ, and then plugged in all the values to DG= DH-TDS.
Or, should I ignore the DG, set DH-TDS to = 0, and find the answer to be about 3195.53 K?
Which is the proper way to go about answering this question?
Answers
I think you do the latter. Set DG = 0 (or -1 since, technically, DG = 0 at equilibrium). I think these are handled by setting DG = 0, then stick an appendix to the answer that says "anything above this T."
Related Questions
Calculate the delta H and delta S for the reaction. From there, calculate the delta G at 25 degrees...
At 298K,
2PCl3(g)+O2(g)---> 2POCl3(l)
Delta H= -620.2kJ
for the reaction and the standard mo...
If delta G (Gibbs) is positive and you are trying to find at what temperature the reaction is sponta...
calculate heat transferred when 2.5g pocl3 is formed at constant pressure for the reaction 2pocl3-2...