pH dependence of Specific Acid catalyzed reaction
Type the paragraph content hereConsider a specific acid catalyzed reaction
S + H+ →P + H+
Where S is the substrate and P is the Product and the rate constant of this reaction is kH+
The rate of the reaction is given by
-d[S]/dt = kH+ [S] [H+] Since [ H+] is the catalyst used , the concentration of [H+] does not change. So it can be considered as constant and the above equation can be simplified as
-d[S]/dt = kobs [S] kobs is the observed rate constant(pseudo –first order)
kobs = kH+ [H+] Taking logarithm on both sides and simplifying we have log kobs = logkH+ + log [H+]
= logkH+ - pH The variation of kobs with pH and log kobs with pH for a specific acid catalyzed reaction is given below
S + H+ →P + H+
Where S is the substrate and P is the Product and the rate constant of this reaction is kH+
The rate of the reaction is given by
-d[S]/dt = kH+ [S] [H+] Since [ H+] is the catalyst used , the concentration of [H+] does not change. So it can be considered as constant and the above equation can be simplified as
-d[S]/dt = kobs [S] kobs is the observed rate constant(pseudo –first order)
kobs = kH+ [H+] Taking logarithm on both sides and simplifying we have log kobs = logkH+ + log [H+]
= logkH+ - pH The variation of kobs with pH and log kobs with pH for a specific acid catalyzed reaction is given below
pH dependence of Specific Base catalyzed reaction
Consider a specific base catalyzed reaction
S + OH- →P + OH-
Where S is the substrate and P is the Product and the rate constant of this reaction is kOH-
The rate of the reaction is given by
-d[S]/dt = kOH- [S] [OH-] Since [ OH-] is the catalyst used , the concentration of [ OH-] does not change. So it can be considered as constant and the above equation can be simplified as
-d[S]/dt = kobs [S]
kobs is the observed rate constant(pseudo –first order)
kobs = kOH- [OH-]
We know that the ionic product of water is
Kw = [H+] [OH-] = 10-14 (at 25°) [OH-] = Kw/[H+] Substituting the value of [OH-] in the above equation kobs = kOH- (Kw/[H+]) Taking logarithm on both sides logkobs = log kOH- + logKw - log[H+] = log kOH- -14 + pH The variation of kobs with pH and log kobs with pH for a specific base catalyzed reaction is given below
S + OH- →P + OH-
Where S is the substrate and P is the Product and the rate constant of this reaction is kOH-
The rate of the reaction is given by
-d[S]/dt = kOH- [S] [OH-] Since [ OH-] is the catalyst used , the concentration of [ OH-] does not change. So it can be considered as constant and the above equation can be simplified as
-d[S]/dt = kobs [S]
kobs is the observed rate constant(pseudo –first order)
kobs = kOH- [OH-]
We know that the ionic product of water is
Kw = [H+] [OH-] = 10-14 (at 25°) [OH-] = Kw/[H+] Substituting the value of [OH-] in the above equation kobs = kOH- (Kw/[H+]) Taking logarithm on both sides logkobs = log kOH- + logKw - log[H+] = log kOH- -14 + pH The variation of kobs with pH and log kobs with pH for a specific base catalyzed reaction is given below
pH dependence of enzyme catalyzed reaction
Since enzymes are proteins, the pH changes profoundly affect the ionic character of the aminoacid and carboxylic acid groups of the protein, thereby affecting the catlysic site and conformation of the enzyme. pH changes can cause considerable denaturation, leading to the inactivation of the enzyme protein.