Solved papers for JEE Main & Advanced AIEEE Solved Paper-2002

done AIEEE Solved Paper-2002 Total Questions - 5

• question_answer1) For a reaction \[A+2B\xrightarrow{{}}C\], rate is given by \[+\frac{d[C]}{dt}=k[A][B]\], hence the order of the reaction is   AIEEE  Solved  Paper-2002
  A)
  3                                
  done clear
  B)
            2                                
  done clear
  C)
  1                                
  done clear
  D)
            0
  done clear
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• question_answer2) Consider the following two reactions, \[A\xrightarrow{{}}\] Product \[-\frac{d[A]}{dt}={{k}_{1}}{{[A]}^{o}}\] \[B\xrightarrow{{}}\] Product \[-\frac{d[B]}{dt}={{k}_{2}}[B]\] \[{{k}_{1}}\] and \[{{k}_{2}}\] are expressed in terms of molarity(mol \[{{L}^{-1}}\]) and time \[({{s}^{-1}})\] as   AIEEE  Solved  Paper-2002
  A)
  \[{{s}^{-1}},M\,{{s}^{-1}}\,{{L}^{-1}}\]      
  done clear
  B)
            \[M\,{{s}^{-1}},\,\,M\,{{s}^{-1}}\]              
  done clear
  C)
            \[{{s}^{-1}},{{M}^{-1}}\,{{s}^{-1}}\]       
  done clear
  D)
            \[M\,{{s}^{-1}},\,{{s}^{-1}}\]
  done clear
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• question_answer3) For the reaction, \[{{H}_{2}}+{{I}_{2}}\xrightarrow{{}}2Hl\], the differential rate law is   AIEEE  Solved  Paper-2002
  A)
  \[-\frac{d\,[{{H}_{2}}]}{dt}=-\frac{d\,[{{l}_{2}}]}{dt}=2\frac{d\,[Hl]}{dt}\]
  done clear
  B)
  \[-2\frac{d\,[{{H}_{2}}]}{dt}=-2\frac{d\,[{{l}_{2}}]}{dt}=\frac{d\,[Hl]}{dt}\]
  done clear
  C)
  \[-\frac{d\,[{{H}_{2}}]}{dt}=-\frac{d\,[{{l}_{2}}]}{dt}=\frac{d\,[Hl]}{dt}\]
  done clear
  D)
  \[-\frac{d\,[{{H}_{2}}]}{2dt}=-\frac{d\,[{{l}_{2}}]}{2dt}=\frac{d\,[Hl]}{dt}\]
  done clear
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• question_answer4) \[{{H}_{2}}\] gas is absorbed on the metal surface like tungsten. This follows ...... order reaction.     AIEEE  Solved  Paper-2002
  A)
  third          
  done clear
  B)
                            second    
  done clear
  C)
            zero                         
  done clear
  D)
            first
  done clear
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• question_answer5) Rate constant k of the first order reaction when initial concentration \[{{C}_{0}}\] and concentration \[{{C}_{t}}\] at time \[t\] is given by equation \[kt=\log \,{{C}_{0}}-\log {{C}_{t}}\] Graph is a straight line if we plot   AIEEE  Solved  Paper-2002
  A)
  \[t\,\,vs\,\,\log \,\,{{C}_{0}}\]   
  done clear
  B)
            \[t\,\,vs\,\,\log \,{{C}_{t}}\]          
  done clear
  C)
            \[{{t}^{-1}}\,\,vs\,\,\log \,{{C}_{t}}\]       
  done clear
  D)
            \[\log \,\,{{C}_{t}}\,vs\,\,\log \,\,{{C}_{t}}\]
  done clear
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