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
4-ethyl-3-methyl 1 hexane
Answer:
See explanation below
Explanation:
In order to calculate this, we need to use the following expression to get the concentration of the base:
MaVa = MbVb (1)
We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:
Atomic weights of the elements to be used:
K = 39.0983 g/mol; H = 1.0078 g/mol; C = 12.0107 g/mol; O = 15.999 g/mol
MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol
Now, let's calculate the mole of KHP:
moles = 0.5053 / 204.2189 = 0.00247 moles
With the moles, we also know that:
n = M*V (2)
Replacing in (1):
n = MbVb
Now, solving for Mb:
Mb = n/Vb (3)
Finally, replacing the data:
Mb = 0.00247 / (13.4473/1000)
Mb = 0.184 M
This would be the concentration of NaOH
pH is the measure of the hydrogen ion concentration while pOH is of hydroxide ion concentration in the solution. The pH is 0.939 and pOH is 13.061 pOH.
pH is the concentration of the hydrogen ion released or gained by the species in the solution that depicts the acidity and basicity of the solution.
pOH is the concentration of the hydroxide ion in the solution and is dependent on the pH as an increase in pH decreases the pOH and vice versa.
Both HCl and HBr are strong acids and gets ionized 100 % in the solution. If we let 1 L of solution for the acids then the concentration of the hydrogen ion will be 0.100 M.
Since both completely dissociate we would just add the molarities of each of the H+ ions together and then calculate the PH and POH from that :
HCL(0.040M)----> H+(0.040M) +CL-(0.040M)
HBr(0.075M)----> H+(0.075M) +Br-(0.075M)
so 0.040M (H+ from HCL) + 0.075M (H+ from HBr) = 0.115M H+ in total.
pH is calculated as:
pH = -log[H+]
Substituting values in the equation:
log(0.115M)= 0.939 pH
pOH is calculated as:
14 - pH = pOH
Substituting values in the equation above:
14 - 0.939= 13.061 pOH
Therefore, pH is 0.939 and pOH is 13.061.
Learn more about pH and pOH here:
brainly.com/question/2947041
#SPJ4
