Chemistry-
General
Easy

Question

Figure shows a graph in l o g subscript 10 end subscript invisible function application K v s fraction numerator 1 over denominator text end text T end fraction where K is rate constant and T is temperature. The straight line BC has slope, T a n invisible function application theta equals negative fraction numerator 1 over denominator 2.303 end fraction and an intercept of 5 on y-axis. Thus E subscript a end subscript, the energy of activation is:

  1. 2.303×2cal    
  2. 2/2.303cal    
  3. 2cal    
  4. None of these    

The correct answer is: 2cal

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