chapter one Flashcards

1
Q

True or false The value for equilibrium constants, Keq, depends on the temperature at which the reaction is taking place

A

False

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2
Q

A Keq = 0.1 tell us what about where the equilibrium stands?

A

The reaction equilibrium lies to the left.

A larger Keq means the reaction is more likely to go to completion

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3
Q

Consider the binding of oxygen to haemoglobin: Hb + O2—–> Hb-O2. If we decrease the pressure of the system, what is the effect on this equilibrium reaction?

A

The equilibrium shifts to the left

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4
Q

Exothermic enthalpy changes are shown as

A

negative

exothermic=releasing heat

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5
Q

Change in enthalpy of a system is due to heat supplied at

A

constant pressure

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6
Q

what will happen when the mass-action ratio Q=0.8 and the Keq = 0.8

A

the reaction is at equilibrium

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7
Q

For the reaction in which reactant A is converted to product B, tell whether this process is favorable or unfavorable at 4ºC and at 37ºC ?

H (kJ/mol)	            S (J/mol K) A	                 54	                   22 B	                 60	                   43
A

unfavorable, favorable

A is unfavorable because ΔG= ΔH-TΔ S. ΔG= 6-((277)(0.021))=0.183= positive=unfavorable

B is favorable because ΔG= 6-((310)(0.021))= -0.51

*check units, should be in kJ and Kelvin

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8
Q

In glycolysis, glucose is broken down to yield energy. If the equilibrium constant of the isomerization of glucose-6-phosphate to fructose-6-phosphate is 0.5

would this reactions proceed in the forward or reverse direction when Q= 1.0?

What is the free energy of the reaction in kJ/mol at 37ºC?

In which direction will the reaction proceed if the concentration of glucose-6-phosphate=0.3M and fructose-6-phosphate= 0.15M as an initial state?

A

reaction will move backwards, 1.79, left

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9
Q

volume of a sphere

A

4/3 Pi x r^3

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10
Q

simplified surface to volume ratio formula

A

3/r

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11
Q

what happens to the SA:V ratio as diameter of the cell increase

A

it decreases, smaller SA:V ratio is bad

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12
Q

Calculate the surface-to-volume ratio for a spherical bacterium with a diameter of 0.6 micro m

A

diameter/2 = radius

r= 0.3

3/0.3=10

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13
Q

describe rotation with single bonds as compared to double

A

single bonds allow free rotation unless there are very large or highly charged groups attached. double bonds restrict most rotation.

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14
Q

first law of thermodynamics

A

the energy in the universe is constant but can shift in forms

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15
Q

how do configurational and conformational isomers differ

A

configurational isomers have restricted rotation between covalent bonds. Cis/Trans isomers are configurational and when shifting to another structure must break the double bond.

conformational isomers can quickly convert between forms at room temperature. Exhibit free rotation between carbons at room temperature.

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16
Q

Describe what a stereoisomer is and explain what is the difference between enantiomers and diastereomers

A

Stereoisomers are molecules with the same chemical bonds and formula but different configuration.

enantiomers are stereoisomers that are mirror images of each other. These are non-superimposable. Enantiomers also exhibit optical activity

diastereomers are stereoisomers that are not mirror images

17
Q

Describe optical activity of a racemic mixture

A

A racemic mixture shows no optical rotation. In a racemic mixture, there is equal concentration of both D and L enantiomers which essentially cancel each other’s rotation out

18
Q

configuration versus conformation

A

configuration is the fixed spatial arrangement while conformation is the spatial arrangement of substituent groups that are free to rotate to different positions without breaking bonds

19
Q

what does Keq measure

A

the tendency of a reaction to go to completion

20
Q

What does the value of Keq tell us

A

A larger Keq means the reaction tends to proceed until the reactants are almost completely converted to products

21
Q

stereoisomer formula

A

2^n where n is the number of chiral carbons

22
Q

R and S priority

A

—OCH3 > —OH > —NH2 > —COOH > —CHO > —CH2OH > —CH3 > —H

23
Q

compare an isolated system and a closed system

A

An isolated system does not exchange matter or energy with surroundings. A closed system exchanges energy only.

24
Q

If under a given set of conditions, the reaction A → B occurs with ∆G = –14 kJ/mol and the reaction C → B occurs when ∆G = +16 kJ/mol, then:

A

The conversion of A to C is exergonic

25
mass-action ratio, Q
ratio of product concentrations to reactant concentrations at a given time – can be calculated to determine how far the reaction is from equilibrium
26
compare ΔG and ΔGº
the Gibbs free energy depends on the experimental conditions whereas the standard free energy describes the Gibbs free energy for reactants and products that are in their standard state
27
relationship between ΔGº and Keq
ΔGº= -RT ln Keq
28
catabolism vs anabolism
catabolism = degradative, free-energy-yielding reactions – drives ATP synthesis – produces the reduced electron carriers NAD(P)H anabolism = synthetic pathways that require the input of energy
29
If the equilibrium constant for an equation is 2.22x10^5M, calculate the standard free energy change at 25ºC.
-30.5kJ -(8.315x (25+273)) ln (2.22x10^5)
30
compare transition state and activation energy
transition state = higher free energy than reactant or product activation energy, ∆G‡ = difference in energy between the reactant in its ground state and its transition state
31
what sets limit of upper cell size
rate of transport and need to deliver O2 to all parts of the cell
32
carbon arrangement and rotation
carbon atoms have a tetrahedral arrangement free rotation around each single bond limit rotation around double bond
33
what 2 things make up most of the weight of a cell
water and proteins
34
a molecule can have 2^n stereoisomers where n is the number of what
n is the number of chiral carbons
35
what is the first law of thermodynamics
total amount of energy in the universe is constant, changes forms
36
second law of thermodynamics
entropy of the universe is increasing