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Sladkaya [172]
2 years ago
15

Consider the following isomerization reactions of some simple sugars and values for their standard Gibbs free energy ΔGo:reactio

n A: glucose-1-phosphate rightarrow glucose-6-phosphate, ΔGo = -7.28 kJ/mol reaction B: fructose-6-phosphate rightarrow glucose-6-phosphate, ΔGo = -1.67 kJ/molCalculate for the isomerization of glucose-1-phosphate to fructose-6-phosphate.Calculate the equilibrium constant K for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 K.
Chemistry
1 answer:
koban [17]2 years ago
3 0

Answer:

Explanation:

glucose-1-phosphate  →  glucose-6-phosphate,  ΔGo = -7.28 kJ/mol

fructose-6-phosphate → glucose-6-phosphate,    ΔGo = -1.67 kJ/mol

subtracting the equation

glucose-1-phosphate - fructose-6-phosphate  = 0 , ΔGo =  -7.28  - ( -1.67 ) kJ / mol

glucose-1-phosphate  = fructose-6-phosphate      ΔGo  = - 5.61 kJ / mol

- ΔGo = RT lnK

5.61 x 10³ = 8.31 x 298 x lnK

lnK = 2.265

K = 9.63 .

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3 years ago
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<u>Answer:</u>

"Boyle's Law" is based on the graph that is shown below.

<u>Explanation:</u>

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