By what factor will a reaction at 25 degree C be accelerated if a catalyst reduc
ID: 1065407 • Letter: B
Question
By what factor will a reaction at 25 degree C be accelerated if a catalyst reduces the free energy of its activated complex by 1 kJ * mol^-1; by 10kJ * mol^-1? In the following lactonization reaction, the relative reaction rate when R = CH_3 is 3.4 times 10^11 times that when R = H. Explain. Explain, in thermodynamics terms, why an "enzyme" that stabilizes its Michaelis complex (i.e.. E middot S) as much as its transition state (i.e.. E middot S) does not catalyze a reaction. An energy diagram should be included in your answer.Explanation / Answer
#3
(i) When the free energy of activation is reduced by a factor of 1 kJ/mol, the factor by which reaction is accelerated is 3 times
when the free energy of activation is reduced by 10 kJ/mol, the factor by which reaction is accelerated is 10 times
(ii) In the given reaction, when R = CH3, the carboxylate anion would be present in minimal amount as the Me groups are electron donating and would make the substrate least acidic in nature than when R = H. This thus makes the carbonyl carbon of acid more reactive and reaction proceeds faster.
(iii) An enzyme which stabilizes Michealis constant and transition state does not break into free enzyme and catalyst easily and thus low conversion of ES complex to product is observed. In the reaction the rate of enzyme dissociation reduces and this gives no catalytic reaction by the said enzyme.
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