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
Answer:
Living organisms need water to survive. Many scientists even believe that if any extra-terrestrial exists, water must be present in their environments. All oxygen-dependent organisms need water to aid in the respiration process. Some organisms, such as fish, can only breathe in water. Other organisms require water to break down food molecules or generate energy during the respiration process. Water also helps many organisms regulate metabolism and dissolves compounds going into or out of the body.
Explanation:
3! You have to ensure balance of all the different elements.
Answer: 9.3 x 10^ 18 g CO
Explanation:
Start by knowing that carbon monoxide is the compound CO. To convert molecules to grams, you first need to convert molecules to moles. This can be done using the conversion factor for Avogadro's Number:
(2.0 x 10^5 molecules CO) x 1 mol CO / 6.02 x 10^23 molecules CO
This cancels molecules CO.
Then, you can convert moles to grams, which is your desired quantity. You can find the number of grams for CO by looking at the periodic table and adding together their masses. C = 12 g and O = 16 g. Total of 28 g CO:
(1 mol CO) x 28 g CO / 1 mol CO
This cancels mol CO, which leaves grams CO.
Answer:
Explanation:
The expression for the work done is:
Where,
W is the amount of work done by the gas
R is Gas constant having value = 8.314 J / K mol
T is the temperature
P₁ is the initial pressure
P₂ is the final pressure
Given that:
T = 300 K
P₁ = 10 bar
P₂ = 1 bar
Applying in the equation as:
