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kolbaska11 [484]
3 years ago
11

During a laboratory experiment you discover that an enzyme-catalyzed reaction has a delta G of -20 kcal/mol. If you double the a

mount of enzyme in the reaction, what will be the delta G for the new reaction
Chemistry
1 answer:
Ilia_Sergeevich [38]3 years ago
3 0

Answer:

-20 kcal/mol

Explanation:

In a reaction, enzymes reduces the energy of activation, but they have no effect on the Gibb's free energy of reaction. The free energy of the product is the same as it would be without the enzyme. Thus, the enzyme does not affect the free energy of the reaction. Mathematically

Gibb free energy = Enthalpy - temperature x change in entropy

Note that from the relation above, enzyme is not a factor nor is activation energy a factor.

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Answer:

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Explanation:

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<em />

That means, the moles of KHP added to the NaOH solution = Moles NaOH at equivalence point. With the moles of NaOH and the volume in liters we can find the molar concentration of NaOH.

<em>Assuming the volume added of KHP was 25.50mL and the solution of NaOH contains 30.0mL (0.0300L), the concentration of the NaOH is:</em>

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<em>Moles KHP = Moles NaOH:</em>

25.50mL = 0.02550L * (0.500mol / L) = 0.01275 moles KHP = Moles NaOH

<em>Molarity NaOH:</em>

0.01275 moles NaOH / 0.0300L =

<h3>0.425M NaOH assuming the volume of KHP was 25.50mL and the volume of the NaOH solution was 30.0mL</h3>
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