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Maslowich
3 years ago
10

In the enzyme reaction above, catechol is the substrate, oxygen is a reactant, catechol oxidase is the enzyme, benzoquinone is t

he product, and water (H2O) is a by-product. Which of the following will decrease the rate of this reaction? a. Reducing the amount of oxygenb. Reducing the amount of catechol c. Increasing the amount of catechol oxidase d. Increasing the amount of benzoquinone
Chemistry
1 answer:
hammer [34]3 years ago
3 0

Answer:  d. Increasing the amount of benzoquinone

Explanation:  The given situation can be written in terms of the equation as follows-

Catechol + Oxygen ⇆ Benzoquinone + water (catechol oxidase is the catalyst)

Le chatleir Principle for the system of equilibrium explains that whenever a reaction is being applied to the stress , the reaction will move in that direction in which the effect of stress is being lessened.

For example if there is a decrease in the concentration of the reactants, the equilibrium will shift to that direction in which the reactants concentration is being increased in order to maintain the equilibrium.

Thus the correct option is (D) as increasing the concentration of the benzoquinone, the equilibrium will shift in the direction in which its concentration is getting decreased, that is in backward reaction, thus decreasing the rate of reaction.

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The reaction between zinc (Zn) and hydrogen chloride (HCl) produces zinc chloride (ZnCl), hydrogen (H2) gas, and heat. If more h
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3 years ago
The equilibrium constant for the gas-phase isomerization of borneol (c10h17oh) to isoborneol at 503 k is 0.106. a mixture consis
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The molar mass of isoborneol/borneol is 154.25 g/mol

Mol isoborneol = 15 g/154.25 = 0.0972 mol
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Use the ICE approach

        borneol  →  isoborneol
I         0.0972           0.0486
C         -x                     +x
E     0.0972 - x        0.0486 + x

Total moles = 0.1458

Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P

0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832 

Thus,
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<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
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3 years ago
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