Respuesta :
Answer:
The answer to the question is
The enzyme must lower the activation energy by 35624.5 KJ/mol
Explanation:
The relation is
k = [tex]Ae^{\frac{-Ea}{RT} }[/tex]
Where k = rate of chemical reaction
A = Arrhenius constant
Ea = Activation energy
T = Temperature
R = Gas constant
at 37 °C we have T = (37 +273.15) K = 310.15 K
k = [tex]Ae^{\frac{-Ea}{(8.314)(310.15)} }[/tex]
when a catalyst is introduced we have Ea(catalyst)
kc = [tex]Ae^{\frac{-Eacatalyst}{(8.314)(310.15)} }[/tex]
Finding the ratio of the two reaction rate and noting that Kc/K = 1.0 x 10⁶
Thus we have
Kc/k = [tex]\frac{Ae^{\frac{-Eacatalyst}{(8.314)(310.15)} }}{Ae^{\frac{-Ea}{(8.314)(310.15)} }}[/tex] = 1.0 x 10⁶ by taking natural logarithm of both sides, we have
6 × ln (10) = Ln (Kc/k ) = [tex]{\frac{Ea}{(8.314)(310.15)} }- {\frac{Eacatalyst}{(8.314)(310.15)} }[/tex]
so that Ea - Eacatalyst = 8.314 × 310.15 × 13.816 = 35624.5 KJ/mol
Therefore the enzyme must lower the activation energy by 35624.5 KJ/molto achieve a million fold increse in the reaction rate constant
The enzyme must lower the activation energy by 35624.5 KJ/mol
The relation for activation energy and rate constant:
[tex]k = Ae^\frac{-E_a}{RT}[/tex]............(1)
where
k = rate of chemical reaction
A = Arrhenius constant
Ea = Activation energy
T = Temperature
R = Gas constant
Given:
Temperature= 37 °C we have T = (37 +273.15)
In Kelvins, T = 310.15 K
On substituting the values in equation (1)
[tex]k = Ae^\frac{-E_a}{RT}\\\\[/tex]
The ratio's for the two reaction rate will be:
[tex]\frac{k_c}{k}=\frac{Ae\frac{-E_{c}}{RT} } {Ae\frac{-E_a}{RT} }=1.0*10^6[/tex]
Taking log both sides:
[tex]6*ln (10)=ln\frac{k_c}{k} =\frac{E_a}{RT} =\frac{E_c}{RT}[/tex]
On further solving,
[tex]E_a-E_c=8.314*310.15*13.816=35624.5KJ/mol[/tex]
Thus, the enzyme must lower the activation energy by 35624.5 KJ/mol to achieve a million fold increase in the reaction rate constant.
Find more information about "Activation energy" here:
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