Question
The change in free energy of a certain process is equal to 4.000 kJ at 400.K...
The change in free energy of a certain process is equal to 4.000 kJ at 400.K and is equal to 2.000 kJ at 350.K. Assuming both the enthalpy and the entropy changes for the process are independent of temperature between 200.K and 400.K, determine the temperature in kelvin where this process occurs reversibly.
Answers
Substract equation (2) from equation (1)
....(3)
Substitute equation (3) in equation (1)
Hence, the temperature at which process occurs reversibly is 300 K.
Delta G = Delta H - T Delta S Delta G = 4000 J T = 400. K Delta G = Delta H - T Delta S 4000 J = Delta H - (400. K) times Delta S ... (1) Delta G = 2000 J T = 350. K Delta G = Delta H - T Delta S 2000 J = Delta H - (350.K) times Delta S ....(2) Subtract equation (2) from equation (1) 4000 J - 2000 J = Delta H - (400. K) times Delta S - [Delta H - (350.K) times Delta S] 2000 J = -(400. K) times Delta S + [(350.K) times Delta S] 2000 J = -(50.K) times Delta S Delta S = -40 J/K .... (3) Substitute equation (3) in equation (1) 4000 J = Delta H - (400.K) times Delta S 4000 J = Delta H - (400.K) times (- 40 J/K) 4000 J = Delta H + 16000. J Delta H = -12000.J
![\\ \: \\ \: 4000 \: J-2000 \: J = \Delta H - (400. K) \times \Delta S - \left [ \Delta H - (350. K) \times \Delta S \right ]](http://img.homeworklib.com/questions/597ee430-8504-11eb-969b-7d4cebf7301b.png?x-oss-process=image/resize,w_560)
![\\ \: \\ \: 2000 \: J = - (400. K) \times \Delta S+ \left [ (350. K) \times \Delta S \right ]](http://img.homeworklib.com/questions/5c7b6760-8504-11eb-b1e2-45539a42bb72.png?x-oss-process=image/resize,w_560)
....(3)
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