Chemistry-
General
Easy

Question

Following is the graph between left parenthesis a minus x right parenthesis to the power of negative 1 end exponent and time t for second order reaction. theta equals t a n to the power of negative 1 end exponent invisible function application left parenthesis 1 divided by 2 right parenthesis semicolon O A equals 2 L m o l to the power of negative 1 end exponent hence rate at the start of reaction will be:

  1. 1.25 L m o l to the power of negative 1 end exponent m i n to the power of negative 1 end exponent    
  2. 0.125 mol L to the power of negative 1 end exponent m i n to the power of negative 1 end exponent    
  3. 0.5 mol L to the power of negative 1 end exponent m i n to the power of negative 1 end exponent      
  4. 1.25 mol L to the power of negative 1 end exponent m i n to the power of negative 1 end exponent    

The correct answer is: 0.5 mol L to the power of negative 1 end exponent m i n to the power of negative 1 end exponent  

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