<em>An example of a</em><em>n</em><em> </em><em><u>exothermic</u></em><em> </em><em>reaction is when metals react with oxygen to form metal</em><em> </em><em><u>Oxides</u></em>
<em><u>Hope</u></em><em><u> </u></em><em><u>this</u></em><em><u> </u></em>helped you- have a good day bro cya)
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
X= 101.325 kPa
The absolute pressure of the air in the balloon is 101.325 kilopascals.
I believe the correct answer from the choices listed above is option B. A chemical formula written above or below the yield sign indicates <span>that the substance is used as a catalyst. I am certain with this answer. Hope this helps. Have a nice day.</span>