1. Using the free energy change equation, solve the following questions assuming ∆H = 15,000 J and ∆S =50 J/K [provide solutions, not just answers]
1a. Is this reaction spontaneous at 20oC?
1b. Is the above reaction spontaneous at 40oC?
1c. At what temperature Celsius would the above reaction be at equilibrium?
1d. If the temperature of the reaction was 0oC and the ∆H = 15,000 J, what would the ∆S need to be for the reaction to be at equilibrium?
2. How many reducing ends are in the molecule of glycogen which contains 5,000 residues with branch every 10 residues?
3. Imagine a person that has a genetic defect in the ability of liver to store large amounts of glycogen. In which cases would that person be limited by this physiological defect and can you think of ways to mitigate such limitations.
4. Disaccharides can be linked together in many ways depending on what anomer is used and what carbons are linked together. For the disaccharides trehalose, sucrose and lactose draw (do not copy and paste) the disaccharide and indicate the type of glycosidic bond that is present (i.e., in chitin, two glucose molecules are bonded together via b 1-4 linkages). Furthermore please be sure to label the individual monosaccharides.
5. Explain how chemical coupling of endergonic and exergonic reactions is used to generate ATP during glycolysis.
1) The free energy, enthalpy and entropy of a reaction are related as
ΔG = ΔH – TΔS
where T is the absolute temperature of the reaction.
a) We are given T = 20ºC = (20 + 273) K = 283 K, ΔH = 15000 J and ΔS = 50 J/K.
Plug in values and get
ΔG = (15000 J) – (293 K)*(50 J/K)
= (15000 J) – (14650 J)
= 350 J
Since ΔG > 0, hence, the reaction is not spontaneous at 20ºC (ans).
b) We now have T = 40ºC = (40 + 273) K = 313 K.
Plug in values and get
ΔG = (15000 J) – (313 K)*(50J/K)
= (15000 J) – (15650 J)
= -650 J
Since ΔG < 0, the reaction is spontaneous at 40ºC (ans).
c) At equilibrium ΔG = 0; therefore,
ΔH – TΔS = 0
====> ΔH = TΔS
====> ΔS = ΔH/T
====> 50 J/K = (15000 J)/T
====> T = (15000 J)/(50 J/K) = 300 K
Therefore, the equilibrium temperature is 300 K = (300 – 273)ºC = 27ºC (ans).
d) We now have T = 0ºC = 273 K and ΔH = 15000 J. The reaction is at equilibrium; therefore,
ΔG = 0
and ΔS = ΔH/T
= (15000 J)/(273 K)
= 54.945 J/K
≈ 54.9 J/K (ans).
1. Using the free energy change equation, solve the following questions assuming ∆H = 15,000 J...