a. Suppose you allow the reaction above to procede to equilibrium at 25 C and fi

Question

a. Suppose you allow the reaction above to procede to equilibrium at 25 C and find the concentrations of maltose and glucose to be as shown above. Use these equilibrium conditions above to calculate the Keq of the reaction. b. Use the Keq to calculate the delta G degree for the reaction. c. Given the Keq and delta G degree values calculated above, do you think this reaction would be considered reversible or irreversible? The enzyme that catalyzes this rection is called maltase. Would maltase be used to make maltose or digest maltose or both? Why? d. Write out the reaction above backward, so that when read from left to right it shows the production of maltose from two glucose molecues What is the delta G degree for the reaction as written now? Next couple this reaction to the reaction for the hydrolysis of ATP to create a new reaction that is favorable for the synthesis of maltose from glucose. Write out the coupled reaction and calculate its delta G degree'. If this enzyme really existed, it would be called maltose synthetase (a synthetase enzyme catalyzes the making a of molecule by putting together two other molecules using ATP energy).

Explanation / Answer

1. we know that

Keq = [Glucose]^2/ [Maltose] = 1.2 X 1.2 / 0.0008 =1800

2. we know that

Delta G= -RTlnK =- 8.314 X 297 Xln 1800 = -8.314 X 298X 7.49= - 18.557 KJ

3. The reaction is not reversible as delta G value is negative and value of equilibrium constant is very high.

Maltase = Malta + Ase = lysis of maltose only or digestionof maltoe only

4.a)

2Glucose ---> maltose

b) The Keq =1/18.557= 0.053

so delta G = +18.557 KJ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.