We have that the value of Ki for the <em>inhibited </em>reaction
- The place the enzyme inhibitor binds solely to the complicated fashioned between the enzyme and the substrate.
- KI = 3.551mM
<h3>Chemical Reaction</h3>
a)
Generally, if the fee of Vmax binds, the place the enzyme inhibitor binds solely to the complicated fashioned <em>between </em>the enzyme and the substrate.
b) V diminished through 70%
Thereofore
Alternate = 1 - 0.7 = 0.3V
Hence
0.3 =1/ ( 1 + [I]/KI)
1= 0.3(1 + 0.657/KI )
0.3 + 0.1971KI = 1
KI = 3.551mM
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CO₂ ₍g₎ + 2H₂ ₍g₎ ---------> CH₃OH ₍g₎
So in order to answer this question you have to have a good understanding of moles in an equation. We know that mole often exist in ratios that is denoted by the value used to balance the equation ('mole ratio'). Thus from the equation we can tell that H₂ has twice the amount of moles CO₂ or CH₃OH does. As such:
- reactant with most moles is H₂
- raectant with least moles is CO₂
- the limiting reagent is H₂
