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51. At elevated temperatures, SbCl5 gas decomposes to SbCl3 gas and Cl2 gas, as shown by the following equation.
SbCl5(g)    SbCl3(g) + Cl2(g)
    
a. An 89.7 g sample of SbCl5 ( MM 299.0) is placed in an evacuated 15.0 L container at 182oC.
    
1. What is the concentration in moles per liter of SbCl5 in the container before any decomposition occurs?   Periodic Chart
    
1 mol = 300. g                                    Molar Mass Equality
    

                1 mol
89.7 g x ----------- = 0.299 mol
               300. g

   

                     n       0.299 mol
Molarity = ----- = -------------- = 0.0199 Molar
                   
V        15.0 L 

     
    
2. What is the pressure in atm of SbCl5 before any decomposition occurs?
     
PV = nRT                              Ideal Gas Law
    

        nRT      0.299 mol x 0.0821 L atm mol -1 K -1 x 455 K
P = -------- = ------------------------------------------------------------- =  0.745 atm
         V                               15.0 L

    
    
b. If the SbCl5 is 29.2 percent decomposed when equilibrium is established at 182oC, calculate the value for either equilibrium constant, Kp or Kc, for this decomposition reaction. Indicate whether you are calculating Kp or Kc.   Equilibrium Constant    Kp and Kc
      
100 M SbCl5 initial = 29.2 M SbCl5 /\                                  Percent Equality
    

                                             29.2 M SbCl5 /\
0.0199 M SbCl5 initial x ---------------------------- = 0.00581 M SbCl5 /\
                                          100 M SbCl5 initial 
     

           SbCl5(g)    SbCl3(g)    +     Cl2(g)
i          0.0199 M               0                     0                    
All have the same delta because they have the
/\         0.00581 M       0.00581 M        0.00581 M       
same coefficients.  Because the products start at
e        0.0141 M          0.00581 M        0.00581 M       
0, reactant decreases and products increase.
    

          [ SbCl3][Cl2]        (0.00581 M) 2 
Kc = ------------------- = ---------------------- =  0.00239 M
               [SbCl5]               0.0141 M

     
     
c. In order to produce some SbCl5, a 1.00 mol sample of SbCl3 is first placed in an empty 2.00 L container maintained at a temperature different from 182oC. At this temperature, Kc equals 0.117. How many moles of Cl2 must be added to reduce the number of moles of SbCl3 to 0.700 mole at equilibrium?   Equilibrium Constant
    
                     n                     1.00 mol
Molarity = ---------------- = ------------- = 0.500 M SbCl3  initial
                  
Vsolution(L)        2.00 L 
  

                     n                     0.700 mol
Molarity = ---------------- = --------------- = 0.500 M SbCl3  equilibrium
                
   Vsolution(L)        2.00 L 
   

           SbCl5(g)    SbCl3(g)    +     Cl2(g)
i              0                   0.500 M           X + 0.150 M       
All have same /\ because of same coefficients
/\         0.150 M           0.150 M           0.150 M              
SbCl5 must increase from 0.
e         0.150 M           0.350 M                X                     
Cl2 ends at X and must start 0.150 M larger
    

          [ SbCl3][ Cl2]        0.350 X
Kc = ------------------- = --------------- =  2.33 X           Remember to only use equilibrium [ ]'s
           
   [ SbCl5]             0.150

   

        Kc             0.00239 M                                        Kc from previous step. 
X = ---------- = ------------------ = 0.00103 M
        2.33                2.33
     

[Cl2]i = 0.00103 M  + 0.150 M = 0.151 M