6. For the conversion of reactant A to product B, the change in enthalpy is 8.0

Question

6. For the conversion of reactant A to product B, the change in enthalpy is 8.0 k/mol and the change in entropy is 40 VK mol. Below what temperature (in Celsius) does the reaction become non-spontaneous? (5 pts) 7. The enzyme mannose-6 P isomerase catalyzes the reversible reaction: Mannose-6-P (MGP) Fructose 6.P (6P) The fructose 6P (F6P) formed can then enter gycolysis. The K 3.1 A) Calculate a for the reaction above at 25C (Use 2 sisnificant iqures). What does your answer tell you regarding spontaneity? (3pts) Itt will pvoceed forward with a na hre dSaurin, Hue nibus. B) Assume you incubate 30 ml of a solution containing 0,075 M MGP overnight at 25"C with an excess of the enzyme mannose-6P isomerase. How many millimoles of F6P will be in the 30 ml solution the next morning? (5 pts) C) Experimental values for the actual steady-state concentrations of these compounds in the cell are [M6Pl® 3 M and 1F6P-14w. Calculate &G;, at 2S (Use 2 significant feres). In which drection wil the reaction proceed to reach equilbrium? Explain. (5 pts)

Explanation / Answer

6) In order to reaction become non-spontaneous (endothermic)

change in Gibbs energy should be greater than positive. i.e. deltaG>0

Now we know the relation between Gibbs free energy G, enthalpy H, entropy S, and Temperature T. That is

deltaG =delta H - T delta S

for a non-spontaneous reaction '

deltaG > 0    or

delta H - T delta S>0 or

delta H > T delta S

therefore T< (delta H) / (deltaS)

given data are delta H = 8KJ/mol = 8000J/mol and delta S = 40J/K.mol

therefore T< (8000/40) K

T< 200K or T< -73.15OC

That means below -73.15OC temperature, the reaction will become non-spontaneous.

7) in this problem they have given Keq = 3.1, reaction temperature = 25OC and they are looking for the change in standard Gibbs free energy which can be represented as

deltaG0 = -RTln Keq , where

R represents gas constant = 8.314 J K-1mol-1, T represents temperature in Kelvin and Keq represents equilibrium constant

at 250C ( 298.15K) deltaG0 for the enzymatic reaction will be

deltaG0 = -8.314 x 298.15 x ln(3.1)

deltaG0 = -2804.54Jmol-1 = -2.81 KJmol-1

negative deltaG0 indicate that the reaction will be an exothermic (Spontanous).

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