6. For the conversion of reactant A to product B, the change in enthalpy is 80 i

Question

6. For the conversion of reactant A to product B, the change in enthalpy is 80 id/moll and the change in entropy is 40 JK mol. Below what temperature (in Celsius) does the reaction become non-spontaneous? (5 pts) 7. The enzyme mannose-6-p isomerase catalyzes the reversible reacton: Mannose6P (MGP)Fructose 6-P (FGP). The fructose-6-P (F6P) formed can then enter glycolysis. The K -3.1 A) Calculate &G;, for the reaction above at 2S (Use 2 significant figures. What does your answer tell you regarding spontane ty? (3pts) B) Assume you incubate 30 ml of a solution containing 0 075 M M6P overnight at 25 "C with an excess of the enzyme mannose 6-FP somerase. 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 n the cel are IMGPI . 3 M and ISP-141. Calculate AG at 25 C. (Use 2 significant figures) In which direction will the reaction proceed to reach equilibrum? Explain. (S pts)

Explanation / Answer

1. given Change in enthalpy, deltaH= 8 Kj/mole, 1 Kj= 1000J, 8Kj/mole= 8000 J/mole

entropy change, deltaS= 40 J/kmole,

A reaction whether is spontaneous or not depends on Gibbs free energy change, deltaG= deltaH-T*deltaS

if Gibbs free energy is -ve, the reaction is non spontaneous. if deltaG is +ve, the reaction is non-spontaneous.

deltaG=0 means the reaction is at equilibrium.

let us determine the temperature at which detlaG=0, 8000-40T=0, T= 8000/40= 200K. Any temperature above makes deltaG -ve. So the reaction is non spontanous at any temperature below 200K.

2. The reaction is M6P<----> F6P, Keq= 3.1

Gibbs free energy change = deltaG= -RT lnKeq, R= gas constant= 8.314 J/mole.K and T is temperature in K= 25+273= 298K

deltaG=-8.314*298*ln(3.1)= -2803 joules/mole

for the given reaction, M6P<--->F6P, Keq= [F6P]/[M6P]

let x= drop in concentration of M6P to reach equilibrium from initial concentration of 0.075M

at equilibrium, [M6P]= 0.075-x, and [F6P]=x

Keq= x/0.075-x)= 3.1

x= 3.1*0.075-3.1x

4.1x=3.1*0.075, x= 0.057M

this is present in 30ml of solution. hence moles of F6P remaining =Molarity* Volume in L= 0.057*30/1000

millimole of M6P remaining = 0.057*30*1000/1000= 1.71

4. With the given values of F6P= 14uM =14*10-6M and M6P= 3*10-6M

Reaction coefficient= [F6P]/[M6P]= 14/3=4.7 this value is <3.1 So more of M6P has to form to reach equilibrium. Hence the reaction proceeds backwards.

deltaG=-8.314* 298*ln(4.7)= -3834 J/mole

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