{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

SI session Sept. 18 Ch. 5 and 6 answers

# SI session Sept. 18 Ch. 5 and 6 answers - Kaiser Imam BIOL...

This preview shows pages 1–4. Sign up to view the full content.

Kaiser Imam September 18, 2007 BIOL 215 SI Session Chapters 5 and 6 Chapter 5: Bioenergetics 1. Write the (P/R) ratio for the following process: 2A + 3B 4C + 5D The (P/R) ratio would be [C] 4 [D] 5 / [A] 2 [B] 3 2. Write an equation to find Δ G’ . Δ G’ = -RT ln(K eq ) + RT ln(([C] c cell )([D] d cell )/([A] a cell )([B] b cell )) Δ G’ = -RT ln(K eq ) + RT ln([P]/[R]) 3. Write an equation to find Δ G°’ . Δ G°’ = -RT ln(K eq ) + RT ln(1) Δ G°’ = -RT ln(K eq ) 4. Write an equation to relate Δ G’ and Δ G°’ . Δ G’ = Δ G°’ + RT ln([P]/[R]) 5. Fill in the following table: Sign of Δ G°’ Spontaneous or Non-spontaneous (P/R) cell ? K eq Direction of Reaction Can work be done? < 0 Spontaneous < Reactants to products Yes; free energy released > 0 Non-spontaneous > Products to reactants Yes if energy is applied = 0 Equilibrium = Equilibrium No; reaction at equilibrium 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6. Using information from the table below determine if each of the following reactions is thermodynamically favorable. Reaction Δ H -TΔ S C 6 H 12 O 6 → 2C 2 H 5 OH + 2CO 2 -82 kJ/mol -136 kJ/mol 2CO 2 + 3H 2 O → C 2 H 5 OH + 3O 2 +1367 kJ/mol -41 kJ/mol N 2 O 5 → 2NO 2 + 1/2O 2 +110 kJ/mol -140 kJ/mol Use the equation Δ G = Δ H – TΔ S when finding the free energy changes for each of these reactions. The first reaction has a Δ G of -218 kJ/mol and is spontaneous as written. The second reaction has a Δ G of +1326 kJ/mol and is nonspontaneous. Finally, the third reaction has a Δ G of -30 kJ/mol and is spontaneous. 7. The interconversion of dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3- phosphate (G3P) is part of the glycolytic pathway for producing energy for the cell. DHAP ↔ G3P The value for Δ G°’ for this reaction is +1.8 kcal/mol at 25 °C. In the glycolytic pathway, this reaction goes to the right, converting DHAP to G3P. a) In which direction does equilibrium lie? What is the equilibrium constant at 25 °C? The Δ G°’ for this reaction is positive so the K eq value is less than one. Δ G°’ = -(1.987 cal/mol-K)(298 K) ln K eq = +1800 cal/mol. 1800/-592 = -3.04, so K eq = 0.048. Therefore, this equilibrium lies far to the left. b) In which direction does this reaction proceed under standard conditions? What is the Δ G’ for the reaction in that direction? Because the Δ G°’ value is the Δ G’ value under standard conditions, the reaction will tend to proceed to the left. The Δ G’ value in that condition is +1.8 kcal/mol. c) In the glycolytic pathway, this reaction is driven to the right because G3P is consumed by the next reaction in the sequence, thereby maintaining a low G3P concentration. What will Δ G’ be (at 25 °C) if the concentration of G3P is maintained at 1% of the DHAP concentration (i.e., if [G3P]/[DHAP] = 0.01)? Δ G’ = Δ G°’ + RT ln ([G3P]/[DHAP]) = +1800 + (1.987 cal/mol-K)(298) ln(0.01) = +1800 + 592(-4.605) = 1800 – 2726 = -926 cal/mol = -0.93 kcal/mol 2
Chapter 6: Enzymes 8. Suppose that the standard free energy change for an uncatalyzed reaction (S ↔ P) is +60 kJ/mol, and a specific enzyme enhances the rate of the reaction by 10 7 -fold. Under standard conditions in the presence of the enzyme, in which direction will the reaction go?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern