Question

Consider the following reaction

4HCl(g) + O₂(g) 2H₂O(g) + 2Cl₂(g)

	HCl(g)	H2O (g)
Hof (kJ/mol)	- 92.3	- 241.8
Gof (kJ/mol)	- 75.3	- 228.6

a) (12 pts) Predict which direction, forward or reverse, is accompanied by increasing disorder

(Show your calculations).

b) (3 pts) Can this reaction be reversed by changing the temperature? Explain.

c) (3 pts) Calculate K_p for the above reaction at 25^oC?

d) (5 pts) Calculate the free energy, G, for the above reaction at 200C if the initial partial pressures of HCl and O₂ are 0.100 and the initial pressures of Cl₂ and H₂O are 0.150.

Answer 1

뀨 ICI (g) + O2(g) +-21 120(g) + 2Cl2(g) a). AH ( system) = [2x Ho H2O (g) + 2x Ho Cl2 (g)]-[4x Ho HCI (g) + Ho O2 (g)] 12 x (-241.8)+2x01-[4 x (-92.3)+ 0 -1 14.4 kJ/mol AG (system)-[2x Go H2O (g) + 2 × Go Cl2 (g)]-[4x Go HCI (g) + Go O2 (g) 12 x (-228.6+2x0-[4x (75.3)+ 0 =-156 kJ/mol (AG < 0, it is a spontaneous process) it is assumed that the temperature is 298.15 K G (system) AH (system)-TAS (system) AS (system)- 139.3 J/mol/K AS (surroundings) (AH (system AS (universe) AS (system AS (surroundings) 523 J/mol/K, ) /T)-(114.4/298.15)-383.7 J/mol/K Hence AS (universe) > 0, is a spontaneous process as it is increased in disorder and the direction of the reaction will be forward. b) AG (system)AH (system) - TAS (system)- (114.4x10) (Tx139.3) From the above equation it is clear that whatever the value of T may be, AG (system) always be less than zero and forward reaction wi be always spontaneous independent of temperature So, this reaction can't be reversed by changing the temperature c) The standard equation related to AG and Kp is as follows: AG--RTInKp at 250C Or,-RTInKp =-156 kJ/mol Or, Kp 10 at 25°c d) Note: the word "200C" should be replaced by "20"C" 4HCİ (g) + O2(g) ←→ 2H20(g) + 2C12(g) for this reaction Kp-(PH2O3x Pcu22) / ((PHCǐ+, P02) [( 0.1 50fx( 0.150f] / [( 0.100)% ( 0.1 001-50 atm" AG--RTInKp at 20C 8.314 x 293 x In50 9529.7 J/mol 9.5297 kJ/mol