What is the value of the equilbrium constant (Keq) for the reaction: A <--> B of

Question

What is the value of the equilbrium constant (Keq) for the reaction: A <--> B of the following reaction provided below?

Please provide a correct, detailed, and legible answer to obtain positive feedback. Thank you!

Question 7 1 pts If one starts with 10 mM of metabolite Aand mixes it with the enzyme that converts metabolite A to metabolite B, the equilibrium concentration of metabolite A at 37 OC is 8.3 mM. Given this information, what is the value of the equilbrium constant (Keq) for the reaction: A B? Report your answer to the nearest hundredths.

Explanation / Answer

K = [B]/[A]

T = 37°C = 310 K

[A] = 10 mM = 10*10^-3 M

[B] = 0

after reaction

[A] = 10*10^-3 - x = 8.3*10^-3

x = 1.7*10^-3

[B] = x = 1.7*10^-3

solve for K

K = [B]/[A]

K = (1.7*10^-3)/(8.3*10^-3) = 0.2048

K = 0.2048

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