A 2.4156 g sample ofPCl5 was placed into a 2.000 l flask and allowed to decompos

Question

A 2.4156 g sample ofPCl5 was placed into a 2.000 l flask and allowed to decompose at250.0 C. PCl5 (g) PCl3 (g) +Cl2(g) It is found that theequilibrium pressure in the flask was observed to be 358.7torr.
a)What is the value for Kp?
b)What is the partial pressure of each gas at equilibrium? c)If 100.0 torr of Cl2 is added to this equilibrium mixture,calculate the concentrations o all gases once equilibrium isreestablished. A 2.4156 g sample ofPCl5 was placed into a 2.000 l flask and allowed to decompose at250.0 C. PCl5 (g) PCl3 (g) +Cl2(g) It is found that theequilibrium pressure in the flask was observed to be 358.7torr.
a)What is the value for Kp?
b)What is the partial pressure of each gas at equilibrium? c)If 100.0 torr of Cl2 is added to this equilibrium mixture,calculate the concentrations o all gases once equilibrium isreestablished. c)If 100.0 torr of Cl2 is added to this equilibrium mixture,calculate the concentrations o all gases once equilibrium isreestablished.

Explanation / Answer

Initial partial pressure of PCl5 = 2.4156 g * 0.0821 L. atm / mol .K * 523 K / 2.0 L * 208.239 g/ mol                                               =0.249 atm At equilibrium pressure of the flask = 358.7torr                                                        = 0.471 atm                                             PCl5(g)        PCl3(g)    +    Cl2(g)                            I(atm)     0.249                    0                  0                           C(atm)         -x                    +x                +x                            E(atm)    0.249 -x              +x                +x                                                           Kp   = (x)(x) / ( 0.249 - x)                                        At equilibrium 0.249 - x +x +x = 0.471                                                                         0.249 +x = 0.471                                                                                      x =0.471 - 0.249                                                                                        = 0.222                                                                              Kp     = ( 0.222 ) 2 / (0.027)                                                                                         =1.83 b) Partial pressure of Cl2(g) =0.222 atm      Partial pressure of PCl3(g)  = 0.222 atm      Partial pressure of PCl5(g)  = 0.027 atm c) According to the Le Chatelier's principle whenCl2 is added , equilibrium shifts to the left. Nowby using the Kp for the revese reaction partial pressures andfinally concentrations can be found. b) Partial pressure of Cl2(g) =0.222 atm      Partial pressure of PCl3(g)  = 0.222 atm      Partial pressure of PCl5(g)  = 0.027 atm c) According to the Le Chatelier's principle whenCl2 is added , equilibrium shifts to the left. Nowby using the Kp for the revese reaction partial pressures andfinally concentrations can be found.                                                                              

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