Practice Exams

Question 1

Which has the highest entropy (per mole)?

a. H2O (s)
b. H2O (g)
c. H2O (l)
d. All the same

Answer 1

b. H2O (g)

Question 2

When liquid water freezes below 0 °C…

a. the entropy of the system increases and the entropy of the surroundings increases.
b. the entropy of the system increases and the entropy of the surroundings decreases.
c. the entropy of the system decreases and the entropy of the surroundings decreases.
d. the entropy of the system decreases and the entropy of the surroundings increases.

Answer 2

d. the entropy of the system decreases and the entropy of the surroundings increases.

Question 3

You (a human being) are what kind of system?

a. Open
b. Isolated
c. Closed
d. It depends on whether you are awake or asleep.

Answer 3

a. Open

Question 4

What is the boiling point of methanol, if ΔH° for vaporization is 35.3 kJ/mol, and ΔS° for vaporization is 105 J/mol K?

a. 336 K
b. 2.97 K
c. 373 K
d. 0.33 K

Answer 4

a. 336 K

Question 5

Which of the following would you predict to have the smallest ΔHvap? (The heat required to change one mole of substance from a liquid phase to a gas phase [vaporize]?)

a. CH3OH
b. CH4
c. OF2
d. NaF

Answer 5

b. CH4

Question 6

If you wanted to prepare 500 mL of a 0.20M solution of dextrose (C6H12O6) in water, how much dextrose would you need?

a. 180 g
b. 90 g
c. 36 g
d. 18 g

Answer 6

d. 18 g

Question 7

How would you prepare 500 mL of a 0.20M solution of dextrose (C6H12O6) in water?
a .Put the dextrose in a beaker and add 500 mL of water.
b. Put the dextrose into 0.5 L volumetric flask and add 500 mL of water.
c. Put the dextrose into 0.5 L volumetric flask and add water until the volume of the solution is 500 mL.
d. Fill a 0.5L volumetric flask with water and then add in the dextrose.

Answer 7

c. Put the dextrose into 0.5 L volumetric flask and add water until the volume of the solution is 500 mL.

Question 8

Is Ne or Xe more soluble in liquid water and which of the following best explains why it is the more soluble?

a. Ne, because it is smaller than Xe so it can fit between the water molecules easier which results in stronger interactions.
b. Ne, because it is similar in size to water so that it can replace water molecules easily due to its polarizablity.
c. Xe, because it is larger than Ne and the larger the atom the more soluble it is in water.
d. Xe, because it is more polarizable than Ne resulting in stronger interactions with the water molecules.

Answer 8

d. Xe, because it is more polarizable than Ne resulting in stronger interactions with the water molecules.

Question 9

Which compound do you predict would be most soluble in hexane C6H14?

a. H2O
b. CH3OH
c. CH3CH2OH
d. CH3CH2CH2OH
e. It is not possible to predict solubility from the information given.

Answer 9

d. CH3CH2CH2OH

Question 10

Which of the following must be true about a process where the sign of ΔG is negative?

a. It must be exothermic
b. The entropy of the system must increase
c. The entropy of the universe must increase
d. All are true

Answer 10

c. The entropy of the universe must increase

Question 11

When calcium nitrate is placed in water, the solution becomes clear and the temperature increases. What are the values of ΔH, ΔS and ΔG for this process?

  ΔH    ΔS 	 ΔG

a. + + +
b. – – +
c. – + –
d. + – –

Answer 11

c. – + –

Question 12

Which would be the best solvent for grease (a mixture of hydrocarbons)?

a. toluene (C7H8)
b. ammonia (NH3)
c. molten sodium chloride
d. apple juice

Answer 12

a. toluene (C7H8)

Question 13

Which compounds can form hydrogen bonds with water but not with itself? (Hint: draw out the Lewis structures)

a. CH3OCH3
b. CH3CH2OH
c. CH3CO2H
d. none of them form hydrogen bonds

Answer 13

a. CH3OCH3

Question 14

A high blood alcohol concentration is considered to be 0.25 g/dL. What is the molarity of this blood alcohol (CH3CH2OH) level?

a. 5.4 x 10^–2 M
b. 5.4 x 10^–4 M
c. 115 M
d. 11.5 M

Answer 14

a. 5.4 x 10^–2 M

Question 15

Rank the following compounds in increasing order of solubility when placed in ammonia (NH3):

a. H2O, CH2Cl2, CH4
b. H2O, CH4, CH2Cl2
c. CH4, CH2Cl2, H2O
d. CH4, H2O, CH2Cl2

Answer 15

c. CH4, CH2Cl2, H2O

Question 16

When a solution of CaCl2 is dissolved in water the temperature rises. Which of these statements can definitely be inferred from this observation?

I. The entropy change for this process is negative.
II. The entropy change for this process is positive.
III. The interactions between the species in solution are stronger compared to those in the separate solute and solvent.
IV. The interactions in the separate solute and solvent are stronger compared to those between the species in solution.

a. I and III
b. I and IV
c. II and III
d. II and IV
e. III only

Answer 16

e. III only

Question 17

The entropy change when carbon dioxide (CO2(g)) dissolves in water is:

a. Positive, because mixing things up always increases the entropy since there are more available positions.
b. Negative, because the number of positions available to the carbon dioxide molecules decreases.
c. Zero, because carbon dioxide does not dissolve in water.
d. Either positive or negative depending on the temperature.

Answer 17

b. Negative, because the number of positions available to the carbon dioxide molecules decreases.

Question 18

Ethanol has a boiling point of 78°C and a melting point of -114°C. What happens when ethanol goes from -84°C to 90°C?

a. the entropy of the system decreases and the entropy of the surroundings decreases.
b. the entropy of the system decreases and the entropy of the surroundings increases.
c. the entropy of the system increases and the entropy of the surroundings increases.
d. the entropy of the system increases and the entropy of the surroundings decreases.

Answer 18

d. the entropy of the system increases and the entropy of the surroundings decreases.

Question 19

Consider the following reaction for the synthesis of ammonia.

N2+ 3 H2→ 2 NH3	           
∆H = -92.2 kJ/mol 
∆S = -72.6 J/K

At 25 °C, would this reaction be thermodynamically spontaneous?

a. No because ∆G is positive
b. No because ∆G is negative
c. Yes because ∆G is positive
d. Yes because ∆G is negative

Answer 19

d. Yes because ∆G is negative

Question 20

Why don’t oil and water mix?

a. Because oil is nonpolar and water is polar.
b. There are no attractive forces between oil molecules and water molecules; therefore, the hydrogen bonds between water molecules would require too much energy to break.
c. The entropy of the system is higher in the unmixed state because non-polar molecules cause water molecules to cluster around them.
d. Oil and water repel each other because they both contain negative polar regions.

Answer 20

c. The entropy of the system is higher in the unmixed state because non-polar molecules cause water molecules to cluster around them.

Question 21

26.	Calcium phosphate (Ca3(PO4)2) is insoluble in water.  The ΔH for solution is about zero.  Predict the signs of ΔS and ΔG.
		ΔS 	 ΔG
a.		+	–
b.		–	–
c.		+	+
d.		–	+

Answer 21

d. – +

Question 22

If you place a container of a mixture of liquid water and water vapor (the system) on a hotplate:
When the hotplate temperature is above 100 °C what change will occur in the system?

Answer 22

The liquid water will vaporize since the temperature is above the boiling point of water.

Question 23

What is the sign of these thermodynamic functions for this change:

a. ΔH system:
b. ΔH surroundings:
c. ΔS system:
d. ΔS surroundings:
e. ΔG:

Answer 23

a. ΔH system: (+)
b. ΔH surroundings: (-)
c. ΔS system: (+)
d. ΔS surrounding: (-)
e. ΔG: (-)

Question 24

If you place a container of a mixture of liquid water and water vapor (the system) on a hotplate:
At T= 100 °C (373K) what is ΔG?

Answer 24

ΔG = 0 (because the system will be at equilibrium)

Question 25

When a hot block is placed next to a cold block the two blocks eventually end up at the same temperature. Explain both how this happens (what processes allow this to occur) and why this happens (and why it will not reverse).

Answer 25

Energy is transferred from the hot block to the cold block until the two have the same number of energy quanta. HOW: Energy is transferred by collisions of metal atoms. When the blocks touch energy can be transferred from hot to cold, then collisions within the block allow the energy quanta to disperse. WHY: There are more energy states available (arrangements of energy quanta) when the two blocks are at the same temperature (have equal #s of energy quanta) than if one block is hot and one is cold. That is the entropy is higher (more possibilities for energy states) when the temperatures are the same.

Question 26

Describe how you would make 250 mL of a 1.25 M solution of sodium chloride, given a 250 mL volumetric flask, water, sodium chloride and a balance.

Answer 26

Put 14.6 g NaCl into a 250 mL volumetric flask and fill to the measuring line with water until the bottom of the meniscus reaches the line.

Question 27

Calculate the amount of sodium chloride needed to make 250 mL of a 1.25 M solution of sodium chloride.

Answer 27

250 mL x (1.25 mol NaCl/1000mL) x (58.5 g NaCl/1mol NaCl) = 14.6 g NaCl in 250 mL of water.

Question 28

When magnesium chloride is added to water, the white solid seems to disappear and the temperature increases. What information can you deduce from these observations? For example about the thermodynamic functions ΔH, ΔS and ΔG, and the relative strengths of interactions

Answer 28

MgCl2 + H2) -> MgCl2(aq) Magnesium chloride dissolves. Before solution there are ionic interactions in MgCl2 and H-bonding, dipole-dipole, and LDFs in the water. After the solution forms, there are ion-dipole interactions. Sine the temperature goes up, energy is released to the surroundings. Therefore the interactions after the solution is formed bust be stronger than the interactions before. ΔH: (-) ΔG: (-) since MgCl2 dissolved ΔS: don't know, probably (+)

Question 29

Imagine a reaction in which compound A (which is a blue compound) reacts with compound B (which is a yellow compound) to give AB. A2 + B2 -> 2AB What can you say about the color of compound AB? A. It is blue B. It is yellow C. It is green D. It is impossible to predict what the color of AB is from the information given.

Answer 29

D. It is impossible to predict what the color of AB is from the information given.

Question 30

Which acid is the strongest? a. CH4 b. OH2 c. NH3 d. HF

Answer 30

d. HF

Question 31

When a bond forms between two atoms A. energy is released to the surroundings B. energy is absorbed from the surroundings C. energy is neither created nor destroyed D. the kinetic energy of the system falls

Answer 31

A. energy is released to the surroundings

Question 32

Which is the strongest base? a. -CH3 b. -OH c. -NH2 d. F-

Answer 32

a. -CH3

Question 33

Explain why -CH3 is the strongest base out of | -CH3, -OH, -NH2, and F-

Answer 33

-CH3 is the strongest base b/c its the least stable and therefore the most reactive. The negative charge on the carbon is not very electronegative. Carbon has the lowest effective nuclear charge and therefore the least able to stabilize the extra negative charge by interactions with the positive charge by interactions with the positive charge on the nucleus. CH4 (conjugate acid) is the weakest.

Question 34

Which is a stronger acid, NH3 or H2O? A. NH3, because N is more electronegative than O, and therefore is better able to stabilize the charge on the conjugate base B. NH3, because N is less electronegative than O, and therefore is better able to stabilize the charge on the conjugate base C. H2O, because O is more electronegative than N, and therefore is better able to stabilize the charge on the conjugate base D. H2O, because O is less electronegative than N, and therefore is better able to stabilize the charge on the conjugate base

Answer 34

C. H2O, because O is more electronegative than N, and therefore is better able to stabilize the charge on the conjugate base

Question 35

When this reaction occurs : H2O + NH3 ->  –OH + NH4+ Do you think the final reaction mixture contains: A. More reactants than products because H2O + NH3 are weaker acids and bases than –OH + NH4+ B. More reactants than products because H2O + NH3 are stronger acids and bases than –OH + NH4+ C. More products than reactants because H2O + NH3 are weaker acids and bases than –OH + NH4+ D. More products than reactants because H2O + NH3 are stronger acids and bases than –OH + NH4+

Answer 35

A. More reactants than products because H2O + NH3 are weaker acids and bases than –OH + NH4+

Question 36

``` What is the pH of a 0.10 M solution of HNO3? A. 0.10 B. 13.0 C. 2.0 D. 1.0 ```

Answer 36

D. 1.0

Question 37

``` What is the pH of a 0.10 M solution of KOH at 25°C? A. 0.10 B. 13.0 C. 2.0 D. 10 ```

Answer 37

B. 13.0

Question 38

``` If the pH of a 0.020 M acid is 1.70 what does that tell you about the acid? I. The acid is completely ionized II. The acid is strong III. The acid is not completely ionized IV. The acid is weak ``` A. I B. III C. I, II D. III, IV

Answer 38

C. I, II

Question 39

``` What is the percent ionization in a 0.10 M solution of an acid that has a pH of 3.75? A. 1.78 x 10^–4 % B. 17.8 % C. 0.178 % D. 1.78 x 10^–3 % E. 1.78 % ```

Answer 39

C. 0.178 %

Question 40

``` What is the oxidation number of C in H2CO3? A. +4 B. +2 C. -4 D. -3 ```

Answer 40

A. +4

Question 41

``` In the reaction Zn + Cu2+ -> Cu + Zn2+ The zinc a. is oxidized in the reaction b. is reduced in the reaction c. is metabolized d. is a catalyst ```

Answer 41

a. is oxidized in the reaction

Question 42

Which reaction is NOT a redox reaction: a. 2Na + 2H2O -> 2NaOH + H2 b. H2 + O2 ->  H2O c. H+ + H2O -> H3O+ d. 2Fe + 3S -> Fe2S3

Answer 42

c. H+ + H2O -> H3O+

Question 43

Explain for each reaction how you know it is, or is not, a redox reaction. a. 2Na + 2H2O -> 2NaOH + H2 b. H2 + O2 ->  H2O c. H+ + H2O -> H3O+ d. 2Fe + 3S -> Fe2S3

Answer 43

a. Is a redox reaction because oxidation number of sodium changes from 0 (Na) to +1 in NaOH and H changes from +1 in H2O to 0 in H2 b. Oxidation numbers change: H from 0 to +1 and O from 0 to -2 c. No oxidation changes. H stays +1 and O stays -2 d. Fe changes from 0 to +3 and S from 0 to -2

Question 44

``` In nitric acid, HNO3, the formal charge and oxidation number on the N are: A. +1, +5 B. +1, –3 C. +1, +3 D. –1, +5 ```

Answer 44

B. +1, –3

Question 45

If the bond dissociation energies of the products are less than the bond energies of the reactants what does this mean in terms of the energy change in the reaction. A. The reaction is exothermic B. The reaction is endothermic C. There is no overall energy change because energy can neither be created nor destroyed D. It depends on the temperature

Answer 45

B. The reaction is endothermic

Question 46

What is the enthalpy change for the reaction: CH4 + 2Br2 -> CH2Br2 + 2 HBr A. – 184 kJ B. + 184 kJ C. –76 kJ D. + 76 kJ

Answer 46

C. –76 kJ

Question 47

What mass of sodium hydroxide would be required to react with 50 mL of 0.15 M HCl? NaOH(s) + HCl(aq)  -> NaCl(aq) + H2O(l) A. 7.5 x 10–3 g B. 7.5 g C. 3.0 g D. 0.30 g

Answer 47

D. 0.30 g

Question 48

Is Ne or Xe more soluble in liquid water and which of the following best explains why it is the more soluble? A. Ne, because it is smaller than Xe so it can fit between the water molecules easier which results in stronger interactions. B. Ne, because it is similar in size to water so that it can replace water molecules easily due to its polarizablity. C. Xe, because it is larger than Ne and the larger the atom the more soluble it is in water. D. Xe, because it is more polarizable than Ne resulting in stronger interactions with the water molecules.

Answer 48

D. Xe, because it is more polarizable than Ne resulting in stronger interactions with the water molecules.

Question 49

The entropy change when carbon dioxide (CO2(g)) dissolves in water is: A. Positive, because mixing things up always increases the entropy since there are more available positions. B. Negative, because the number of positions available to the carbon dioxide molecules decreases. C. Zero, because carbon dioxide does not dissolve in water. D. Either positive or negative depending on the temperature.

Answer 49

B. Negative, because the number of positions available to the carbon dioxide molecules decreases.

Question 50

For the reaction of acetic acid (CH3COOH) and water. What products do you expect?

Answer 50

CH3COOH + H2O -> CH3COO- + H3O+

Question 51

If the pH of a 0.10 M solution of acetic acid is about 2.9, what does that tell you about the extent of the reaction between acetic acid and water?

Answer 51

If acetic acid were a strong acid (fully ionized in water), the pH would be 1 for a 0.1 M solution. Since the pH is 2.9, this means not all of the acetic acid is ionized. [H30+] = 10^-pH [H3O+] = 10^-2.9 = 0.00125 M = 1.3 x 10^-3M

Question 52

If the pH of a 0.10 M solution of acetic acid is about 2.9, what is the % ionization of the acetic acid in water?

Answer 52

% ionization = (amount ionized / total concentration) x 100 Amount ionized = [H3O+] (10^-2.9 / 0.1) x 100 = 1.258%

Question 53

Why do we use multiple models of acid-base reactions? Why don’t we just pick one and stick with it?

Answer 53

We use models that can help us predict and explain outcomes. Generally we use the simplest model that will work in a given situation. So while all acid base reactions can be described using the Lewis model (a base is an electron pair donor and an acid is an electron pair acceptor), sometimes it is simpler and easier to use the Bronsted-Lowry acid base model (where an acid is a proton donor and a base is a proton acceptor). This is particularly true for reactions in aqueous solution where proton transfer from acid to base is the easiest way to think about the reaction. (Arrhenius acid base models are pretty useless because they are not widely applicable.)

Question 54

What is the difference between an acid-base reaction and a redox reaction? Use examples of each to illustrate your answer.

Answer 54

Acid base reaction: Bonds are formed and broken and a proton is transferred, but NO electrons are transferred between atoms. The oxidation number of each atom remains the same. In contrast in 2Na(s) + H2O  2Na+(aq) + -OH (aq) + H2(g) the Na starts with an electron and a zero oxidation number, but ends up as Na+ (+1 oxidation number). It has lost an electron and been oxidized. The hydrogen has been reduced (from +1 in H2O to zero in H2).

Question 55

Calculate the enthalpy change for this reaction C2H5OH + 3 O2 -> 2 CO2 + 3 H2O.

Answer 55

Bonds broken = 1 C-C, 5 C–H, 1 C–O, 1 O–H, 3 O=O Bonds formed = 4 C=O (in CO2), 6 O-H Bonds broken = (347 + 5(414) + (351) + (464) + 3(498)) = 4726 kJ Bonds formed = (4(803) + 6(464) = 5996 kJ Enthalpy change = Σ (bonds broken) – Σ (bonds formed) = – 1270 kJ

Question 56

Explain the energy change that occurs in an exothermic reaction in part a by discussing the strength of forces and interaction present in the reactants versus products.

Answer 56

The reaction is exothermic – which means that energy is released to the environment. The bonds in the products (CO2 and H2O) are stronger than those in the reactants. That is, less energy is required to break the bonds in the reactants than is released in the products. The bonds are formed by electrical attractions of electrons of one atom to the nucleus of the other – this means that the forces of attraction between the atoms in the products are higher than the forces of attraction between the reactants (meaning that the bonds in the products stronger)

Question 57

The actual enthalpy change is somewhat different to the one that could be calculated – why do you think that might be?

Answer 57

Bond energies are calculated or measured in the gas phase. Most reactions do not take place in the gas phase – usually they are in a solvent – and interactions with the solvent can make bond breaking easier. Solvents may also stabilize either the product or the reactants by interactions (solvation)

Question 58

Which reaction do you think has a larger rate constant (faster rate)? I. HCl + H2O -> H3O+ + –Cl II. CH3Br + –OH  -> CH3OH + Br– A. I, because acid base reactions don’t have a high activation energy, since the H being transferred is partially ionized. B. I, because HCl is a gas, and therefore can react very quickly C. II, because there is a C-Br bond breaking in the rate determining step D. II, because hydroxide is a strong base

Answer 58

A. I, because acid base reactions don’t have a high activation energy, since the H being transferred is partially ionized.

Question 59

What could affect both the rate of a reaction and the equilibrium constant? I. change in temperature II. change in concentrations of reactants III. presence of a catalyst a. I and II b. I c. I and II d. All of them

Answer 59

b. I

Question 60

Explain what effect each answer I. change in temperature II. change in concentrations of reactants III. presence of a catalyst has on the rate and equilibrium constant.

Answer 60

I. increasing T will increase the rate (more energetic collisions) and the value of K. II. will only increase the rate (it will change the position of equilibrium – but NOT the K) III. increases rate – but not K. A catalyst provides an alternate path.

Question 61

``` For a reaction A + 2B -> C the rate of reaction Δ[A]/Δ[t] is related to the rate in terms of B by A. Δ[B]/Δ[t] B. 2 Δ[B]/Δ[t] C. ½ Δ[B]/Δ[t] D. it is not possible to determine ```

Answer 61

C. ½ Δ[B]/Δ[t]

Question 62

For the reaction A + B -> C the rate equation is: rate = k[A]2[B] if the concentration of A is doubled and the concentration of B is doubled, by what factor will the rate increase? A. 4 B. 2 C. 8 D. 16

Answer 62

C. 8

Question 63

``` A radioactive tracer Tc has rate constant for decay of 1.05 x 10–2 hr–1. What is the half-life for the decay? A. 66 hours B. 66 days C. 0.693 hrs D. 0.480 hrs ```

Answer 63

A. 66 hours

Question 64

What is the equilibrium constant expression for the following equilibrium?       3Fe(s)  +  4H2O(g)  ->   Fe3O4(s)  +  4H2(g) A. K  =  [H2]4 B. K  =   [H2] / [H2O] C. K  =  [H2]^4 / [H2O]^4 D. K  =  [Fe2O3][H2]^4 / [Fe]^3[H2O]^4

Answer 64

C. K  =  [H2]^4 / [H2O]^4

Question 65

If the pH of a 0.15 M acid is 3.6, what is the Ka? a. 1.2 x 10^-3 b. 2.5 x 10^-4 c. 4.2 x 10^-7 d. 1.0 x 10^-10

Answer 65

c. 4.2 x 10^-7

Question 66

``` If the pH of a 0.15 M acid is 3.6, what is the % ionization of this acid? A. 0.80% B. 0.17% C. 99.9% D. 1.07% ```

Answer 66

B. 0.17%

Question 67

``` For a system at equilibrium A. ΔG = 0 B. ΔG° = 0 C. Both ΔG and ΔG° = 0 D. None are true ```

Answer 67

A. ΔG = 0

Question 68

The equilibrium constant for a reaction A + B -> 2C is 1.0. If a 1 L solution containing 10 mol A, 2 mol B and 10 mol C is allowed to react, will the concentration of A when equilibrium is reached be: A. > 10 M B.

Answer 68

A. > 10 M

Question 69

Calculate the pH of a solution of 0.10 M hydrogen cyanide HCN? (Ka = 4.9 x 10-10) a. 9.31 b. 5.15 c. 4.16 d. 1.00

Answer 69

b. 5.15

Question 70

Consult the table of pKas to help you determine whether this reaction would have more products or reactants at equilibrium NH4+ Cl– + H2O -> H3O+ Cl– + NH3 A. Reactants B. Products C. Equal amounts D. Can’t tell from this information

Answer 70

A. Reactants

Question 71

Which statement is true for the following reaction at equilibrium? 3H2(g) + N2(g) ⇌ 2NH3(g) A. All the reactants have been converted to products. B. The forward reaction has stopped so no more ammonia is formed. C. The rate of the forward reaction is equal to the rate of the back or reverse reaction. D. There is no more energy for molecules to get over the activation energy barrier.

Answer 71

C. The rate of the forward reaction is equal to the rate of the back or reverse reaction.

Question 72

What is the age of bottled wine that has a tritium (H-3) content that is 60% of the tritium in freshly bottled wine. Tritium decays by beta decay and has a half-life of 12.3 yr. 31H → 32He + 0-1β ``` A. 0.029 yr B. 7.4 yr C. 9.1 yr D. 16 yr E. 35 yr ```

Answer 72

C. 9.1 yr

Question 73

The reaction of CH3Br with -OH has a rate equation: rate = k[CH3Br][-OH]. What does this tell you about the mechanism of the reaction? I.e. the sequence of events at the molecular level that leads from reactants to products.

Answer 73

Both CH3Br and -OH are in the rate determining step of the reaction. Therefore, it is likely that this reaction has only one step that involves both reactants.

Question 74

Explain why reaction rates depend on concentration of reactants.

Answer 74

For a reaction to occur, the reactant molecules must collide The more molecules there are per unit volume, the more likely the are to collide

Question 75

Explain why reaction rates depend on the temperature.

Answer 75

The reactant molecules must have enough energy to get over the activation energy (Ea) barrier. This is where bond breaking happens, which requires energy). The high the temperature, the more kinetic energy the molecules have and therefore the more energy of collision.

Question 76

Explain why reaction rates depend on the type of reaction

Answer 76

The rate depends on the activation energy which depends on bonds broken and formed. For example, acid-base reactions are fast because the bond to the acidic H is already partially broken. But, to break a C-C bond, a lot of energy is needed, so the reaction is slow.

Question 77

Explain why reaction rates depend on the presence of a catalyst

Answer 77

A catalyst provides an alternate pathway with (usually) lower activation energy. For example, in enzyme-substrate complexes the bonds are "loosened" so they take less energy to break

Question 78

For the reaction: N2(g) + 3H2(g) -> 2NH3(g) ΔH = +65 kJ What effect would adding ammonia have on the position of the equilibrium.

Answer 78

Would shift the equilibrium to the left. | Would increase the rate of the reverse reaction so more reactants would form.

Question 79

For the reaction: N2(g) + 3H2(g) -> 2NH3(g) ΔH = +65 kJ What effect would increasing the pressure have on the position of the equilibrium.

Answer 79

Would shift the equilibrium to the right. Same effect as increasing the concentration of reactants. So there are more collisions and the rate of the forward reaction is increased.

Question 80

For the reaction: N2(g) + 3H2(g) -> 2NH3(g) ΔH = +65 kJ What effect would increasing the temperature have on the position of the equilibrium.

Answer 80

Would shift the reaction to the right. This reaction is endothermic. Adding energy will increase the probability of the reactant molecules having enough energy to get over the activation energy barrier.

Question 81

For the reaction: N2(g) + 3H2(g) -> 2NH3(g) ΔH = +65 kJ What effect would adding a catalyst have on the position of the equilibrium.

Answer 81

No change. | A catalyst provides an alternate pathway but the energies of the reactants and products stays the same.

Question 82

For the reaction: N2(g) + 3H2(g) -> 2NH3(g) ΔH = +65 kJ What effect would adding helium have on the position of the equilibrium.

Answer 82

No change. | Helium is not a component of the reaction.

Question 83

When acetic acid (CH3COOH) reacts with water what is happening, at the molecular level.

Answer 83

The OH bond in the acetic acid is breaking. The H from the OH bond in the acetic acid, goes to the water molecule and accepts the electrons making water have a positive charge. The electrons from the OH bond in acetic acid remains on the O creating a negative charge on the oxygen.

Question 84

Why does acetic acid (CH3COOH) reacts with water.

Answer 84

The acetic acid acts as an acid and donates a proton (accepts electrons). The water acts as a base and accepts a proton (donates electrons). Collisions have to occur in order to the reaction to happen.

Question 85

Bromoacetic acid (BrH2CCOOH) ionizes in water. What is the percent ionization in a 0.25 M solution of bromoacetic acid that has a pH of 1.6?

Answer 85

% ionized = (amount ionized / concentration) x 100 [H3O+] = 10^-1.6 = 0.02511 100 x (0.02511/0.25) = 10.0475 10.05%

Question 86

Bromoacetic acid (BrH2CCOOH) ionizes in water. The percent ionization is 10.05%. It is a 0.25 M solution of bromoacetic acid that has a pH of 1.6. Is this a weak or strong acid? Why?

Answer 86

Weak acid. | A strong acid would have a 100% ionization.

Question 87

Consider the reaction A -> B with an initial concentration of [A] = 6.0 M and an equilibrium constant K = 0.5. What are the concentrations of A and B at equilibrium?

Answer 87

``` K = [B] / [A] K = [2] / [4] = 0.5 ``` ``` [A] = 4 [B] = 2 ```

Question 88

The Ka for formic acid is 1.7 x 10^-4. What does the Ka tell you about the relative amounts of formic acid and formate ion (the conjugate base) at equilibrium?

Answer 88

K

Question 89

What should happen to the pH of a 1.0 M formic acid solution if you added some sodium folate (the conjugate base of formic acid) to the solution? Why?

Answer 89

Increase. The reaction would go to the left because Le chatelier's principle. Less hydronium ions would make the solution less acid, the pH would go up. pH = -log[H3O+] decrease of [H3O+] ions