Michaelis constant is known as km which is the substrate concentration that encourages the compound to work at half maximum velocity represented by Vmax/2. Michaelis constant is inversely related to the substrate and the affinity of the enzyme.
Induced fit model: The premise of the purported induced fit hypothesis, which expresses that the attachment or association of a substrate or some other atom to an enzyme causes an adjustment to the enzyme in order to fit or restrain its activity.
In substrate, analog Km or Michaelis constant will be high as the substrate will stay because of analogs inhibit activity.
In the transitional state, analog Km will be in the middle of the substrate and product analogs. Progress state analogs are synthetic mixes with a structure catalyzed reaction that looks like the progressing condition of a substrate atom in a compound enzyme.
1. Balanced equation for the reaction between tin (Sn) metal and aqueous hydrochloric acid (HCl) to produce tin(II) chloride (SnCl₂) and hydrogen gas (H₂).
This is illustrated below:
Sn (s) + HCl (aq) –> SnCl₂ (aq) + H₂ (g)
There are 2 atoms of Cl on the right side and 1 atom on the left side. It can be balance by putting 2 in front of HCl as shown below:
Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)
Now, the equation is balanced
2. Determination of the element that is oxidize and reduced.
This can be obtained as follow:
We shall determine the change in oxidation number of each element.
NOTE:
a. The oxidation number of H is always +1 except in hydrides where it is –1.
b. The oxidation state of Cl is always –1.
Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)
For Tin (Sn):
Sn = 0
SnCl₂ = 0
Sn + 2Cl = 0
Cl = – 1
Sn + 2(–1) = 0
Sn – 2 = 0
Collect like terms
Sn = 0 + 2
Sn = +2
Therefore, the oxidation number of Tin (Sn) changes from 0 to +2
For H:
H = +1
H₂ = 0
The oxidation number of H changes from +1 to 0
For Cl:
Cl is always –1. Therefore no change.
Summary:
Element >>Change in oxidation number
Sn >>>>>>>From 0 to +2
H >>>>>>>>From +1 to 0
Cl >>>>>>>No change
Therefore,
Sn is reduced since its oxidation number increased from 0 to +2.
H is oxidized since it oxidation number reduced from +1 to 0
