Answer:
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Explanation:
From the given information:
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.
In this same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;
Rate factor in the absence of catalyst:
Rate factor in the presence of catalyst:
Assuming the catalyzed reaction and the uncatalyzed reaction are taking place at the same temperature :
Then;
the ratio of the rate factors can be expressed as:
![\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}](https://tex.z-dn.net/?f=%5Cdfrac%7Bk_2%7D%7Bk_1%7D%3D%7B%20%20%5Cdfrac%20%7Be%5E%7B%5B%20%20Ea_1%20-%20Ea_2%20%5D%20%7D%7D%7BRT%7D%20%7D%7D)
Thus;
Let say the assumed temperature = 25° C
= (25+ 273)K
= 298 K
Then ;
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
The part of an atom that is actively exchanged or shared in a chemical bond is ELECTRON.
An atom is made up of three sub particles, which are electron, proton and neutron. The proton and the neutron are located in the nucleus of the atom and they make up the major mass of the atom. The electron is located outside of the nucleus and it orbit around the nucleus; it has negligible mass. The electron is negatively charged and because it is located outside of the nucleus, it is the one that is always involved in chemical reactions. There are different types of chemical bonds in chemical compounds and it is electrons that are normally used to form these bonds. During bond formation, electrons can either be donated or shared.
Answer:
c2h4 (etthane........
bexcuse it has 6 bond of hydrogen
Answer: D
Explanation:
the atom of the metal loses one electron which becomes delocalised and is attragted by the positive nucleus leading to formation of metallic bond.
