6. General Chemistry 2 Flashcards

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1
Q

convection

A

movement of heat from hotter to colder portions

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2
Q

radiation

A

electromagnetic waves emitted from a hot body into the surrounding environment,

light colors radiate and absorb less, dark colors radiate and absorb more

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3
Q

black body radiator

A

theoretical perfect body that absorbs all energy incident upon it, and then emits 100% of this energy as electromagnetic radiation

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4
Q

conduction

A

molecular collisions carry heat along a conduit

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5
Q

heat capacity

A

about of energy (in joules or Calories) a system must absorb to give a unit change in temperature (J/K or cal/C)

C=q/delta T

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6
Q

constant volume

A

systems confined into rigid walls

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7
Q

constant pressure

A

open to atmosphere

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8
Q

What happens when heat enters a system? Does the temperature always increase? Is any increase in temperature always exactly proportional to the heat absorbed by that system? (Hint: Think of adding energy to a sealed steel container vs. adding energy to a balloon; remember that temperature is the average kinetic energy of the molecules, but an increase in temperature is NOT the only “place” where added energy can go.)

A

This is our first foray into pv work. PV work is the work necessary to produce an increase in volume. For example, when a sealed balloon is heated, the gases inside the balloon will expand and must do work on the rubber walls of the balloon and the air around it to accomplish this expansion. Because some of the heat energy added to the balloon was used for pv work, only the remaining portion of the heat will go toward increasing the average kinetic energy of the molecules (i.e., temperature). So, when heat enters a system, if the system is capable of volume change, heat can go to either pv work, increased temperature, or both. For this reason, the addition of a certain amount of heat will NOT necessarily be exactly proportional to the resultant increase in temperature. If the system is not capable of changing volume then no pv work can be done, so all of the added heat will go toward an increase in temperature.

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9
Q

For the same system, which heat capacity will be greater, the constant volume heat capacity or the constant pressure heat capacity?

A

If the volume is held constant, then 100% of the energy added will go toward an increase in temperature. If the pressure is held constant the volume can still change and therefore some of the added heat will go toward pv work. If we think of heat capacity as “the amount of energy we can add before the system increases by one temperature unit,” it is fairly easy to see that the system capable of pv work will be able to absorb more heat before increasing by one degree Celsius or Kelvin. It is much like asking how many gallons of water can be added to Tank A vs. Tank B? Tank A and Tank B are both 5-gallon tanks, but Tank B is connected via a hose to a reserve tank that holds 2 gallons. So, you can add 5 gallons to Tank A before it is “full” (analogous to a one unit increase in temperature). However, you can add 7 gallons to Tank B before it is full (the reserve tank being analogous to pv work). We would therefore say that Tank B has the higher “water capacity” in terms of our analogy. This indicates that the constant pressure heat capacity (allows for pv work; i.e., includes the 2-gallon reserve tank) will always be more than the constant volume heat capacity (does not allow for pv work; i.e., no reserve tank) for the same system.

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10
Q

specific heat capacity

A

Above we defined Heat Capacity as the energy absorbed/unit change in temperature for a system. This “system” could include a solution, the container holding the solution, and even a thermometer or stirring rod. Specific Heat Capacity, however, describes energy absorption for one individual substance only and is defined per unit mass. Little “c” is used instead of big “C”.

c=q/mdeltaT ==> q=mcdeltaT

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11
Q

specific heat of water

A

1.0 cal/gCelsius or 4.18 J/gCelsius

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12
Q

Beakers 1 and 2 contain 0.25 L of water and 0.5 L of water, respectively. How does the heat capacity of the water in Beaker 1 compare to that of the water in Beaker 2? How do the specific heat capacities compare?

A

The heat capacity of Beaker 2 will be greater than that of Beaker 1 because there is more water available to absorb heat in Beaker 2. However, the specific heat capacity of water in both beakers will be identical – specific heat capacity is an intensive property.

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13
Q

calorimeters

A

device used to calculate enthalpy change (deltaH), we assume q is equal to delta H

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14
Q

coffee cup calorimeter

A

q=mcdeltaT

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15
Q

bomb calorimeter

A

solve using q=CdeltaT, does NOT give enthalpy, but change in internal energy, deltaU or deltaE, ising heat capacity (C) instead of specific heat capacity (c)

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16
Q

Which type of calorimeter provides constant pressure and which type provides constant volume?

A

The bomb calorimeter is a sealed steel container, meaning volume cannot change. However, because volume is constant any gases produced or consumed during the reaction will change the pressure. The coffee cup calorimeter allows for an increase in volume of the solution inside the coffee cup, but remains at atmospheric pressure throughout. The bomb is constant volume, the coffee cup constant pressure.

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17
Q

Pressure-Volume (PV) work

A

work is energy transfer via force (physics) or via change in volume at constant pressure (chemistry)

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18
Q

PV Work = P delta V

A

requires constant pressure, any change in volume tells you there is pv work

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19
Q

pressure vs. volume graph

A

exponential graph, with area under the curve being work

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20
Q

Which of the calorimeters described above allows for pv work?

A

The coffee cup allows for pv work because it allows for a change in volume.

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21
Q

the First Law of Thermodynamics

A

energy cannot be created or destroyed (work done ON a system is POSITIVE, work done BY a system is NEGATIVE) (note:in physics we note that force and displacement is in the same direction then work is positive, if they are opposing than it is negative)

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22
Q

deltaE = q +w

A

energy change in a closed system is equal to the heat absorbed by that system plus any work done on that system by its surroundings

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23
Q

the Second Law of Thermodynamics

A

heat cannot be changed completely into work in a cylindrical process, entropy in an isolated system can never decrease.

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24
Q

the Third Law of Thermodynamics

A

Pure crystalline substances at absolute zero have an entropy of zero

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25
Q

the Zeroth Law of Thermodynamics

A

If object A is in thermal equilibrium with object B, and object C is also in thermal equilibrium with
object B, then objects A and C must be in thermal equilibrium with each other. (A = B and B = C, therefore A = C) (note: Everything tends to move toward thermal equilibrium with everything else. Objects with higher temperatures will always equilibrate over time with their surroundings, including other objects with which they are in contact. Finally, if two objects are in thermal equilibrium, by definition they have the same temperature.)

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26
Q

temperature

A

average kinetic energy of the molecules

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27
Q

kinetic energy

A

KE = 3/2(kb) T, kb= boltzmann’s constant, shows direct relationship between temperature and KE)

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28
Q

temperature conversions

A

F -> C = (5/9)(F-32)
C -> F = (C+(9/5))+32
C- > K = C+273.15

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29
Q

Enthalpy (∆H)

A

the energy contained within chemical bonds (Joules)

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30
Q

standard state

A

A set of specific conditions chosen as the reference point for measuring and reporting enthalpy, entropy, and Gibbs free energy. You may remember seeing these conditions specified at the top of a table of ∆H values. When values are given for standard state conditions a superscript, called “naught,” is added that resembles a degree symbol (∆H°). standard temperature and pressure, 25C or 298K

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31
Q

STP

A

standard temperature and pressure, 0C or 273K

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32
Q

Elements in their standard state have

A

Hformation° = zero. i.e., Gformation° = zero for elements, but the standard state entropy of elements is not zero.

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33
Q

ΔHrxn

A

the enthalpy change for a reaction

34
Q

ΔHformation

A

The enthalpy value for the formation of a compound from its elements in their standard states. If the number is negative, formation is an exothermic process, if it is positive, the process is endothermic.

35
Q

ΔHcombustion

A

The enthalpy value for the combustion of a compound with O2 to form CO2 and water. A high heat of combustion is associated with an unstable molecule and a low heat of combustion with a stable molecule.

36
Q

ΔHsolution

A

The enthalpy value associated with the dissolution of a species into solution. We’ll discuss this in more detail when we cover solution chemistry.

37
Q

ΔHvaporization

A

The enthalpy value associated with the phase change from liquid to gas. The reverse process (condensation) simply interchanges products and reactants and thus the sign is just changed.

38
Q

ΔHfusion

A

The enthalpy value associated with the phase change from liquid to solid. The sign changes for the reverse process (melting).

39
Q

Entropy (∆S)

A

Entropy = a measure of the randomness or disorder in a system. Reactions and other processes will be more likely to be spontaneous if they increase entropy. Joules/K.  As a reaction proceeds, if randomness increases, energy will be released and thus be available to do work. If randomness decreases, energy is required to “create” this increased order and that amount of energy will thus be unavailable to do work.
 positive ∆S = increased randomness, and thus more energy available to do work.
 negative ∆S = decreased randomness, and thus less energy available to do work.
 Reactions at equilibrium are at maximum entropy.

40
Q

increase entropy

A

1) increased number of items/particles/etc. (Caveat: The number of moles of gas trumps the
number of moles of species in any other phase. Thus, even if a reaction turns 2 moles of reactants into 1 mole of product, if that one mole of product is a gas and the reactants are not, entropy has increased and S will therefore be positive.)
2) increased volume
3) increased temperature
4) increased disorder (a crystal is changed to an amorphous material) or complexity (S is greater
for C2H6 than for CH4)

41
Q

Gibbs Free Energy (∆G)

A

∆G = the amount of “free” or “useful” energy available to do work (excluding pv work; as a result of running an isothermal, isobaric reaction).o If energy is available and the system can do work, Gibbs Free Energy is negative. If energy must be added to the reaction (e.g., heat must be added to the system) to make it proceed, Gibbs Free Energy is positive.
o Units: Joules
 negative ∆G = Spontaneous process; free energy available to do work.
 positive ∆G = Non-spontaneous process; no free energy available; energy is required.

42
Q

isobaric

A

constant pressure

43
Q

isothermal

A

constant temperature

44
Q

thermodynamic relationship

A

∆G = ∆H - T∆S, Recall that enthalpy is basically the change in bond energy from reactants to products. If there were no change in randomness during the reaction, the amount of energy available to do work (ΔG) would be exactly equal to enthalpy (∆H). As described above, if randomness increases (positive ΔS), energy will be released and that energy (in addition to ΔH) will also be available to do work (creating a larger, more negative ΔG). By contrast, if randomness decreases, energy will be “used” to create this order, decreasing the amount of energy available to do work. The “T” term in the equation converts entropy into joules (J/K*K = J). You may recall that energy can also be used to increase temperature or to expand the volume (PV work), but neither occurs here because the system is both isothermal and isobaric.

45
Q

∆G° = - RTlnKeq

A

relates the equilibrium constant to the Gibbs free energy (Note: Remember that the ln of any positive number less than 1 is negative).
note: Keq = e^(-ΔG/RT)

46
Q

The hydration of ammonium nitrate is a highly exothermic dissolution reaction. Which of the following statements is NOT true of this process?

A

The reaction must be spontaneous because it is both exothermic and has a favorable entropy change. Answer A states in words what is demonstrated by the fundamental thermodynamic relationship: ∆G = ∆H - T∆S. Namely, if the enthalpy term is negative and the entropy change is positive (as it must be for a dissolution reaction), and thus the -T∆S term is negative, the reaction must be spontaneous.

47
Q

Arrhenius

A

Acids produce H+ ions in solution; bases

produce OH- ions in solution.

48
Q

Bronsted-Lowry

A

Acids donate protons (H+); bases accept protons (H+).

49
Q

Lewis

A

Acids accept a pair of electrons; bases donate a pair of electrons.

50
Q

Acid/Conjugate Acid and Base/Conjugate Base

A

he “conjugate base” of an acid is the acid minus its
hydrogen (e.g., HCl = acid; Cl- = conjugate base). The “conjugate acid” of a base is the base plus a hydrogen (e.g., NH3 = base; NH4+ = conjugate acid).

51
Q

Amphoteric

A

substances can act as either an acid or a base (e.g., H2O).

52
Q

pH Scale

A

The pH scale is a mathematical ranking system for the acidity or basicity of aqueous solutions. The solution being ranked could be water only, water + acid, water + base, or water + base + acid. Notice, however, that water is always there. In fact, as you’ll see in the next section, the pH scale ranks solutions based not so much on the acids or bases themselves, but on how those acids or bases influence the equilibrium for the ionization of water.

53
Q

pH = -log[H+]

A

pH = -log[H+], pHofpureH2Oat25°C=7.0,andthe[H+]=[OH-];pH>7=basic,andthe[H+] [OH-]; pH = 7 is defined as “neutral”

54
Q

pOH = -log[OH-]

A

pOH = -log[OH-]

55
Q

pH+pOH=14

A

pH+pOH=14

56
Q

Acid-Base Equilibria

A

All equilibrium constants (Keq, Ka, Kb. Kw or Ksp) are written via the law of mass action, with pure liquids (l) and solids (s) omitted.

57
Q

ionization of water

A

H2O + H2O H3O+ + OH- (H3O+ is the same as H+)
Kw = [H3O+][OH-] = 10-14
pKw =pH+pOH=14
pKa+pKb=14

58
Q

acid dissociation

A

HA + H2O H3O+ + A- (H3O+ is the same as H+)
Ka = [H+][A-]/[HA]
KaKb = Kw = 10-14 (at 25°C); because ([H+][A-]/[HA])([OH-][HA]/[A-]) = [H+][OH-] = Kw

59
Q

base dissociation

A

A- + H2O OH- + HA
Kb = [OH-][HA]/[A-]
KaKb = Kw = 10-14 (at 25°C); because ([H+][A-]/[HA])([OH-][HA]/[A-]) = [H+][OH-] = Kw

60
Q

Acid HX dissociates 80% in water. Would you expect its Ka to be greater than, less than, or equal to one?

A

Because the acid almost fully dissociates, we know that the ratio of products over reactants would have to be greater than one. In general terms, an acid with Ka greater than one or a pKa less than zero is considered “strong,” so this acid would clearly qualify as a strong acid.

61
Q

Kw = Ka*Kb

A

Kw = Ka*Kb, Ka and Kb are inversely related

62
Q

equilibrium summary

A

 The equilibrium for the ionization of water is always present in aqueous solutions.
 Adding an acid shifts the reaction to the left, increasing the relative [H+]
 Adding a base shifts the reaction to the left, increasing the relative [OH-]
 Kw, Ka and Kb are used to describe these three equilibriums.
 At 25°C, Kw = Ka*Kb;

63
Q

calculating pH/pOH strong acids/bases

A

for a strong acid (pH= - log [strong acid] ),
for a strong base (pOH= - log [strong base]),
for weak acids.

note: assume 100% dissociation for strong acids and strong bases

64
Q

calculating pH weak acids

A

ICE TABLE

1) Write out the equilibrium equation (HA  H+ + A-)
2) Use x to represent the concentration of each of the two products (or 2x, 3x etc. depending
on the coefficients in the balanced equation).
3) Use “[HA] – x” for the concentration of the original acid.
4) If this results in a quadratic equation, assume that x is much smaller than [HA] (in step #3
above) and omit it.
5) Solve for x from Ka = (x)(x)/[HA - x].
6) Use –log[H+] (i.e., -log[x] to find the pH).

65
Q

strong acids

A

HI, HBr, HCl, HNO3, HClO4, HClO3, H2SO4, H3O+ (HF is NOT a strong acid)

66
Q

strong bases

A

Group IA hydroxides (NaOH, KOH, etc.), NH2-, H-, Ca(OH)2, Sr(OH)2, Ba(OH)2, Na2O, CaO.

67
Q

weak acids

A

Weak acids and bases do not dissociate readily in solution. As a general rule, an acid with a pKa greater than zero, or a Ka less than one, can be considered a weak acid. H2O, H2S, NH4+, HF, HCN, H2CO3, H3PO4, acetic acid, benzoic acid, etc.

68
Q

weak bases

A

Weak acids and bases do not dissociate readily in solution. Similarly, a base with a pKb greater than zero, or a Kb less than one, can be considered a weak base. H2O, NH3, R3N, pyridine, Mg(OH)2, etc.

69
Q

Impact of Salts on the Dissolution of Weak Acids and Weak Bases

A

 The percent dissociation of benzoic acid (weak acid) decreases in a sodium benzoate
solution.
 The percent dissociation of ammonium hydroxide (weak base) decreases in an ammonium chloride solution.
23. Both observations are easily explained by Le Chatelier’s principle. In case one, sodium benzoate dissociates to release benzoate ions, which shift the acid dissociation equilibrium for benzoic acid to the left. Similarly, ammonium chloride dissociates to release ammonium ions, which shift the base dissociation equilibrium for ammonium hydroxide to the left.

70
Q

hydrolysis of salts

A

This terminology can be a bit tricky: The “salt of a weak acid” refers to the conjugate base
of that weak acid combined with a cation to form a salt. HCO3- = “weak acid”; CO32- = “conjugate base”; Na2CO3 = “salt of a weak acid”

71
Q

Hydrolysis of which of the following salts in solution will increase the pH of the solution? a) NaNO2, b) NH4Cl, c) NaF, d) NaClO2, e) CH3COONa, f) NaCl.

A

a) Hydrolysis of NaNO2 will result in the reaction of a nitrite ion with water to form HNO2 and hydroxide ion, increasing pH; b) hydrolysis of NH4Cl will result in reaction of NH4+ with water to form NH3 and H3O+, DECREASING pH; c) hydrolysis of NaF will result in fluoride ion reacting with water to form HF and OH-, increasing pH; d) hydrolysis of NaSO2 will result in reaction of ClO2- with water to form HClO2 and OH-, increasing pH; e) hydrolysis of CH3COONa will result in
reaction of acetate with water to form CH3COOH and OH-, increasing pH. In summary, all of the
options will increase pH except for option b).

72
Q

The conjugate base of a strong acid is usually a weak base. Therefore, it can also be said that the Kb of the conjugate base of a strong acid compares in what way to the Kb of a strong base?

A

The Kb of the conjugate will be less than the Kb of the strong base. Because Kb times Ka is equal to a constant, if one is high, the other must be low, and vice versa. A strong base must have a high Kb and thus has a low Ka. The weaker base would have a lower Kb, also allowing it to have a relatively higher Ka. Answer C is thus correct.

73
Q

Titration

A

Drop by drop mixing of an acid and a base with an indicator.

74
Q

If a question states: “A strong base is titrated with a strong acid,” which one is being added drop-wise and which one is in the beaker? Which solution is referred to as the titrant? Which solution is referred to as the anylate?

A

The terminology used in the question infers that the strong base is in the beaker, which makes it
the analyte. The base is “titrated with” the strong acid, meaning the acid is being added
dropwise and is therefore the “titrant.”

75
Q

titration curves

A

strong acid and strong base = steep titration curve (looks linear with a cliff), strong acid/base with weak acid/base = partly linear and partly sloped, weak acid and weak base = almost fully sloped

76
Q

Equivalence Point/Stoichiometric Point

A

midpoint of the nearly vertical section of the graph. Again, at this point [titrant] = [analyte].

77
Q

Half-Equivalence Point

A

midpoint of the nearly horizontal section of the graph, pH = pKa. [HA] = [A-] at the half-equivalence point. note: SA/SB titrations do NOT have half-equivalence points.

78
Q

For which of the following titrations will the [OH-] = [H+] at the equivalence point? For which titrations will [titrant] = [analyte] at the equivalence point? a) SA with SB, b) WB with SA, c) WA with SB, d) WA with WB.

A

a) At the equivalence point of the titration of a strong acid with a strong base the [OH-] will equal the [H+] AND the [analyte] will equal the [titrant] in the flask; b) At the equivalence point of the titration of a strong acid with a weak base the [OH-] does NOT equal the [H+], but the [analyte] will equal the [titrant]; c) At the equivalence point of a titration between a weak acid and a strong base the [OH-] will NOT equal [H+], but the [analyte] will equal the [titrant]; d) At the equivalence point of a titration between a weak acid and a weak base the [OH-] will NOT equal [H+], but the [analyte] will equal the [titrant] (remember WA/WB titrations are rarely attempted or useful). The pattern is that the hydroxide and hydrogen ions will be equal at the equivalence point for any “strong/strong” titration, but NOT for any other titrations. The concentration of the analyte will equal the concentration of the titrant at the equivalence point for all titrations.

79
Q

Indicators

A

Indicators are weak acids that change color as they dissociate from HA into H+ and A-. To set up a titration, you must know beforehand the approximate pH of your equivalence point; you then select an indicator that will change color at that approximate pH.

80
Q

End Point

A

The end point is simply the point when the indicator causes the color change

81
Q

Buffers

A

a buffer solution contains a weak acid and a weak base, often the conjugates of each other, in a buffer there is an equilibrium between a weak acid and its conjugate base, or between a weak base and its conjugate acid

82
Q

henderson-hasselbach equation

A
pH = pKa + log[A-]/[HA],
pH = pKa – log[HA]/[A-]