For the reaction A rightarrow B, delta G^10 = -60 kJ/mol. The reaction Is starte

ID: 956965 • Letter: F

Question

For the reaction A rightarrow B, delta G^10 = -60 kJ/mol. The reaction Is started with 10 mmol of A; no B is Initially present. After 24 hours, analysis reveals the presence of 2 mmol of B, 8 mmol of A. Which is the most likely explanation? A and B have readied equilibrium concentrations. An enzyme has shifted the equilibrium toward A. B formation is kinetically slow; equilibrium has not been readied by 24 hours. Formation of B is thermoddynamically unfavorable. The result described is impossible, given the fact that deltaGdegree is -60 kJ/mol.

Explanation / Answer

First, identify K

K = exp(G/(-RT)) = exp(60000/(8.31*298)) = 3.33*10^11

meaning it favours extremly products

then

this is clearly not the 2:8 ratio expected

the rate msut be very slow, we need more time to make all a react to B

choose optino C

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For the reaction A rightarrow B, delta G^10 = -60 kJ/mol. The reaction Is starte

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