1s22s22p 3s23p64823d3
1 answer:
You might be interested in
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:
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
Explanation:
Reaction equation is as follows.
Here, 1 mole of produces 2 moles of cations.
= 1.16 M
= 0.58 M
The sulphite anion will act as a base and react with to form
and
.
As,
=
=
According to the ICE table for the given reaction,
Initial: 0.58 0 0
Change: -x +x +x
Equilibrium: 0.58 - x x x
So,
x = 0.0003 M
So, x = = 0.0003 M
= 0.58 - 0.0003
= 0.579 M
Now, we will use = 0.0003 M
The reaction will be as follows.
Initial: 0.0003
Equilibrium: 0.0003 - x x x
=
=
Therefore,
As, x <<<< 0.0003. So, we can neglect x.
Therefore,
=
x =
x = =
=
= M
Thus, we can conclude that the concentration of spectator ion is M.