For a reaction, 2 S O 2 ( g ) + O 2 ( g ) 2 S O 3 ( g )  , 1.5 moles of S O 2 and 1 mole of O 2 are taken in a 2 L vessel. At equilibrium the concentration of S O 3 was found to be 0.35 mol L - 1 . The for the reaction would be 
  1. 5.1 L  m o l - 1
  2.  0.6 L  m o l - 1
  3.  2.95 L  m o l - 1
  4. 1.4 L  m o l - 1

The given equilibrium is                            2SO2 +O2  2 SO3Initial moles :    1.5           1            -At eqm         :    1.5-2x    1-x    2xno of moles                                                                Given that at equilibrium 2x = 0.35 mol/Lx= 0.175                        1.5-2(0.175)    1-0.175     2×0.175                      =1.15                  0.825             0.35Equilibrium concentration nv                             1.15 2             0.8252          0.352Kc = [SO3]2[SO2]2 [O2]       =  0.3522 1.15 2 20.8252   = 0.2245 L mol-1None of the option matches

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