Ch 1 - Problems

Problems

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1. Identify the circled c ircled functional groups and linkages linkages in the compound c ompound below.

Answer: A. A Thiol (sulfhydryl) group B. Carbonyl group C. Amide linkage D. Phosphoanhydride (pyrophosphoryl) (pyrophosphoryl) linkage linkage E. Phosphoryl group (P i) F. Hydroxyl group 2. Why is the cell membrane not an absolute barrier between the cytoplasm and the external environment? Answer:

The cell membrane must be semipermeable so that the cell c ell can retain reta in essential compounds while while allowing allowing nutrients to enter and wastes to exit. 3. A spheroidal spheroidal bacterium bacte rium with with a diameter of 1 μm contains two molecules molecules of a pa rticular protein. What is

the molar concentration of the protein? Answer:

Concentration Concentrat ion = (number of moles)/(volume) moles)/(volume)

Problems


4. How many glucose glucose molecules does the cell in Problem 3 contain when its internal glucose glucose concentrat co ncentration ion is

1.0 mM? Answer:

5. (a) Which has greater entropy, liquid water at 0°C or ice at 0°C? (b) How does the entropy of ice at –5°C

differ, if at all, from its entropy at –50°C? Answer: (a) Liquid water; (b) ice has less entropy at the lower temperature. 6. Does entropy increase or de crease in the following following processes? (a) (b)

(c)

(d)

Answer: (a) Decreases; (b) increases; (c) increases; (d) no change.

Problems


–1 7. Consider a reaction with Δ H = 15 kJ and ΔS = 50 J · K . Is the reaction spontaneous (a) at 10°C, (b) at

80°C? Answer: (a)

ΔG is greater than zero, so the reaction is not spontaneous. (b)

ΔG is less less than zero, so the reaction is spontaneous. 8. Calculate the equilibrium constant for the reaction –1

at pH 7.0 and 25°C (ΔG°′ = –20.9 kJ · mol ). Answer:

9. Calculate ΔG°′ for the reaction A + B C + D at 25°C when the equilibrium equilibrium concentrations concent rations are [A] = 10 μM, [B] = 15 μM, [C] = 3 μM, and [D] = 5 μM. Is the reaction exergonic or endergonic under standard

conditions? Answer:

Since ΔG°′ is positive, positive, the reaction is endergonic under standard conditions c onditions.. 10. ΔG°′ for the isomerization isomerization reaction –1

is –7.1 kJ · mol . Calculate the equilibrium equilibrium ratio of [G1P] to [G6P] at 25°C. Answer:

From Eq. 1-17,

Problems


11. For the reaction A → B at 298 K, the change cha nge in enthalpy is –7 kJ · mol –1 and the change in entropy is –1

–1

–25 J · K · mol . Is the reaction spontaneous? If not, should the temperature be increased or decreased to make the reaction spontaneous? Answer:

The reaction is not spontaneous because ΔG > 0. The temperature must be decreased in order to decrease the value of the T ΔS term. 12. For the conversion of reactant A to product B, the change in enthalpy is 7 kJ · mol –1 and the change in –1

–1

entropy is 20 J · K · mol . Above what temperature does the reaction become spontaneous? Answer:

In order for ΔG to have a negative value (a spontaneous reaction), T ΔS must be greater than Δ H .

following statements statements true t rue or false: fa lse: 13. Label the following (a) A reaction is said to be spontaneous when it can proceed in either the forward or reverse

direction. (b) spontaneous process always happens very quickly. (c) A nonspontaneous nonspontaneous reaction re action will proceed spontaneously in the reverse direction. (d) A spontaneous process can occur with a large decrease in entropy. Answer: (a) False. False. A spontaneous reaction re action only occurs occu rs in one direction. (b) False. False. Thermodynamics does not specify the rate of a reaction. rea ction. (c) True. (d) True. A reaction rea ction is spontaneous so long as ΔS > Δ H /T . 8 14. Two biochemical reactions have ha ve the t he same K eq temperature T 1 = 298 K. However, Reaction eq = 5 × 10 at temperature –1 –1 1 has Δ H ° = –28 kJ · mol and Reaction 2 has Δ H ° = +28 kJ · mol . The two reactions utilize utilize the same reactants. Your lab partner has proposed that you can get more of the reactants to proceed via Reaction 2 rather than t han Reaction Rea ction 1 by lowering the temperature of the t he reaction. rea ction. Will Will this this strategy work? Why or why not? How much would the temperature have to be raised or lowered in order to change the value of

Problems


K 2/K 1

from 1 to 10?

Answer:

This strategy strategy will NOT NOT work because Reaction 1 has a negative enthalpy entha lpy change, releasing heat, and a nd will therefore become more favorable with decreasing temperature, whereas Reaction 2, which has a positive enthalpy change, will become less favorable. Thus decreasing the temperature te mperature will favor favor Reaction Rea ction 1, not Reaction 2. In order to make Reaction 2 more favorable, the temperature must be raised. To calculate the amount that th at the temperature must be raised, Equation 1-18 may be used as follows: follows:

On subtraction of the previous two equations, and taking into account that

We would like

, we get

. Substituting in all values value s and solving for T 2 we get

Solving for T 2 we get

Hence to increase K 2/K 1 from 1 to 10, the temperature must be raised from 298 K to 332 K.

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Ch 1 - Problems

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