Chem Ii Final

Front 1

11.1 List the different phase transitions that are possible and give examples of each.

Back 1

~Melting: H2O(S)  H2O(L)
~Freezing: H2O (L)  H2O (S)
~Vaporizing: H2O (L)  H2O (G)
~Sublimation: H2O (S)  H2O (G)
~ Condensation: H2O (g)  H2O (l)
~ Deposition: H2O (g)  H2O (s) H2O (S)
~Liquefaction: H2O (g)  H2O (l)

Front 2

Describe vapor pressure in molecular terms. What do we mean by saying it involves a dynamic equilibrium?

Back 2

Vapor Pressure – partial pressure of the vapor over the liquid, measured at equilibrium at a given temp, at the molecular level it refers to the kinetic energies of molecules. Those moving away from the liquid, or toward the vapor phase, has to be greater than the attraction to the molecules in the liquid to vaporize, have to be larger than the liquid potential are, thus giving pressure.

Front 3

Explain why evaporation leads to cooling of the liquid.

Back 3

Evaporation leads to the cooling of a liquid due to loss of latent heat during phase changes, because energy is required to overcome the molecular forces of attraction between the water molecules (surface tension) and because the removal of the high-kinetic-energy, phase-changed molecules leave behind the less energetic molecules leading to cooling.

Front 4

Why does the vapor pressure of a liquid depend on the intermolecular force?

Back 4

Because of the ease of difficulty with which a molecule leaves the liquid. The liquid strength depends on the attraction of it to the molecules.

Front 5

You are presented with three bottles, each containing a different liquid: bottle A, bottle B, and bottle C. Bottle A’s label states that it is an ionic compound with a boiling point of 35*C. Bottle B’s label states that it is a molecular compound with at boiling point of 29.2* C. Bottle C’s label states that it is a molecular compound with a boiling point of 67.1*C.

A. Which of the compounds is most likely to be incorrectly identified?

B. If Bottle A were a molecular compound which of the compounds has the strongest intermolecular compound attractions?

C. If bottle A were a molecular compound, which of the compounds would have the highest vapor pressure?

Back 5

Bottle A

Bottle C; because it has the highest boiling point

Bottle B; because it is the compound with the lowest boiling point

Front 6

As a demonstration, an instructor had a block of ice suspended between two chairs. She then hung a wire over the block of ice, with weights attached to the ends of the wire. Later at the end of the lecture, she pointed out that the wire was lying on the floor apparently having cut its way through the block of ice remained intact. Explain by referring to the phase diagram for water (fig 11.11) what has happened.

Back 6

The pressure of the wire will melt the ice under it. Liquid water will flow over the top of the wire and freeze again, because it is no longer under the pressure of the wire. As a result the wire will cut through the ice and fall to the floor but the block of ice will remain in one piece.

Front 7

Identify the phase transition occurring in each of the following.

Back 7

A. Mothballs slowly become smaller and eventually disappear.
Sublimation
B. Rubbing alcohol spilled on the palm of the hand feels cool as the volume of liquid decreases.
Vaporization
C. A black deposit of tungsten metal collects on the inside of a light bulb whose filament is tungsten metal.
Sublimation and Condensation/Deposition
D. Raindrops hit a cold metal surface, which becomes covered with ice.
Freezing
E. Candle wax turns to liquid under the heat of the candle flame.
Melting

Front 8

A 35.8g sample of cadmium metal was melted by an electric heater providing 4.66 J/s of heat. If it took 6.92 min from the time the metal began to melt until it was completely melted, what is the heat of fusion per mole of cadmium?

Back 8

Power x Time 4.66 x (6.92 x 60) = 1934.8 = 6191.5 J/mol
# of moles of Cd/Atomic weight (35.8/112) .3125
6191.5 J/mol = 6,1915 kj/mol

Front 9

Shown here is the phase diagram for compound X. The triple point of X is -25.1*C at .50 atm and the critical point is 22*C and 21.3 atm.

Back 9

A. What is the state of X at position A?
Gas
B. If we decrease the temperature from the compound at position A to -28.2*C while holding the pressure constant, what is the state of X?
Solid
C. If we take the compound starting under the conditions of part b and increase the temperature to 15.3*C and decrease the pressure to 0.002atm, what is the state of X?
Gas
D. Would it be possible to make the compound starting under the conditions of part c a solid by increasing just the pressure?
No, When the pressure increases, it will move back into the liquid region.

Front 10

11.96 Snow forms in the upper atmosphere in a cold air mass that is supersaturated with water vapor. When the snow later falls through a lower, warm air mass, rain forms. When this rain falls on a sunny spot, the drops evaporate. Describe all of the phase changes that have occurred.

Back 10

Snowing forming in a cold air mass that is supersaturated with water vapor is condensation or deposition
Snow falls and turns to rain: Melting
Rain falls to the sunny spot, the drops evaporate: vaporization.
ANOTHER ANSWER:
Snow forms in the upper atmosphere in a cold air mass that is super saturated with water vapor. Gas (water vapor) solid (snow)
When snow later falls through a lower, warm air mass, rain forms. Solid (snow)  liquid (rain)
When rain falls on a sunny spot, the drops evaporate. Liquid (rain)  gas (water vapor)

Front 11

Give one example of each: a gaseous solution, a liquid solution, a solid solution

Back 11

Gaseous solution: air
Liquid solution: soda water
Solid solution: dental fillings (mercury, silver & small amounts of other metals)

Front 12

Explain in terms of intermolecular attractions why octane s immiscible in water.

Back 12

Octane is immiscible in water because water’s stronger hydrogen bonding (octane does not have hydrogen bonds) keeps the water from mixing with the octane (octane molecules attract one another via weak London forces

Front 13

Explain the process of reverse osmosis to produce drinkable water from ocean water.

Back 13

To produce drinkable H2O from ocean H2O, osmotic pressure must be provided to the higher concentration solution with enough pressure to prevent the H2O from naturally flowing down the concentration gradient. The pressure must indeed push the solvent against the cocentration gradient to a more pure H2O solution.

Front 14

You want to purchase a salt to melt snow and ice on your sidewalk. Which one of the following salts would best accomplish your task using the least amount: KCL, CaCl2, PbS2, MgSO4, AgCl?

Back 14

CaCl2

Front 15

Consider the following dilute NaCl (aq) solutions

Back 15

a. Which one will boil at a higher temperature?
B, It has the highest solute concentration and highest Boiling point elevation
b. Which one will freeze at a lower temperature?
B will freeze at a lower temperature
C. If the solutions were separated b a semi permeable membrane that allowed only water to pass, which solution would you expect to show an increase in the concentration of NaCl?
A would increase in NaCl concentration because the H2O would move from A B leaving the # of NaCl molecules (ions) to the # of H2O greater in A

Front 16

Which of the following ions would b e expected to have the greater energy of hydration, F- or Cl-?

Back 16

F- because it has a smaller ionic radius

Front 17

Caffeine, C8H10N4O2 is a stimulant found in tea and coffee. A sample of the substance was dissolved in 45.0 g of chloroform, CHCl3, to give a .0946 m solution. How many grams of caffeine were in the sample?

Back 17

CHCl3 = 45.0g = 45 x 10-3 kg
/moles of C8H10N4O2 / 45 x 10-3 = 0.0946m
moles of C8H10N4O2 = .0946m x (45 x 10-3)
=0.04257 mol caffeine
C8H10N4O2 = 194g/moL
0.04257moL C8H10N4O2 x 194g/moL(C8H10N4O2) = 8.25858 g C8H10N4O2

Front 18

A sample of aluminum sulfate 18-hydrate, Al2(SO4)3 18H2O, containing 159.3 mg is dissolved in 1.000 L of solution. Calculate the following:

Back 18

Al2(SO4)3 = 342 g/mol
H2O = 18 g/mol x 18 = 324 g/mol
324 + 342 = 666 g/mol (342/666) x 100 = 51%
(SO4)3 = 96 g/mol x 3 = 288 g/mol (288/666) x 100 = 43%
a. Molarity of Al2(SO4)3
(0.1593g Al2(SO4)3 18H2O)/1L x 0.51 = 0.082g/L x 1mol/342g = 2.4 x 10-4 M
b. Molarity of SO42-
(0.1593g x .43 x (1 mol/96g)) = 7.14 x 10-4 M SO42-
c. Molality of Al2(SO4)3 ,assuming that the density of the solution is 1.00g/mL.
(2.4 x 10-4mol Al2(SO4)3)/1000ml = 2.4 x 10-4m Al2(SO4)3

Front 19

Which of the following reactions involve homogeneous equilibria and which involve heterogeneous equilibria? Explain the difference.

Back 19

a. 2NO(g) + O2(g) 2NO(g)
homogeneous: all the same phases in both the reactants and products
b. 2Cu(NO3)2 (s) 2CuO(s) + 4NO2(g) + O2(g)
heterogeneous: the phases change between the reactants and products
c. 2NO(g) 2N(g)+O(g)
homogeneous; all same phase
d. 2NH3(g) + 3CuO(s) 3H2O(g) + N2(g) + 3Cu(s)
heterogeneous; different phases

Front 20

14.9 List the possible ways in which you can alter the equilibrium composition of a reaction mixture.

Back 20

Changing the concentrations by removing products or adding reactants to the reaction vessel
Changing the partial pressure of gaseous reactants and products by changing the volume
Changing the temp.

Front 21

List four ways in which the yield of ammonia in the reaction can be improved for a given amount of H2. Explain the principle behind each way.

Back 21

[NH3]2/[N2][H2]3
The yield of ammonia can be improved by:
a) Remove ammonia. Per Chatelier’s principle, the reaction will resist the lack of product by producing more product.
b) Increase pressure. Chatelier’s principle also indicates that increasing pressure on a reaction that goes from a higher amount of moles on the reactant side and lower on product side will increase the reaction toward the product side.
c) Add more nitrogen to favor reactants switching to products.
d) Decrease temperature. For ∆H˚<0 (exothermic) reactions, the amounts of products are increased at equilibrium by a decrease in temperature.
e) Add a catalyst, if available.

Front 22

When a continuous, stream of hydrogen gas, H2, passes over hot magnetic iron oxide, Fe3O4, metallic iron and water vapor form. When a continuous stream of water vapor passes over hot metallic iron, the oxide Fe 3O4 and H2 form. Explain why the reaction goes in one direction in one case but in the reverse direction in the other.

Back 22

The system must exist as an equilibrium mixture of all four substances.
Fe3O4(s) + 4H2(g)  3Fe(s) + 4H20(g)
If you pass H2(g) over Fe3O4(s) reaction shifts to right, giving 3Fe(s) + 4H20(g)
If you pass H20(g) over Fe(s), Reaction shifts to left and forms Fe3O4(s) and H2(g)

Front 23

For the endothermic reaction AB(g) A(g) + B(g), the following represents a reaction container at two different temperatures. Which one I or II has lower temp

Back 23

I=lower temperature (more reactants)
II=higher temperature (less reactants)
(the amount of products is increased at equilibrium by temperature increase)

Front 24

Write equilibrium-constant expression Kc for each of the following.

Back 24

a. N2(g) +2H2(g) N2H4(g)
[N2H4]
[N2][H2]2
b. 2NOCl(g) 2NO(g) + Cl2(g)
[NO2]2[Cl]
[NOCl]2
c. 2NO(g) + 2H2(g) N2(g) + 2H2O(g)
[N2][H2O]2
[NO]2[H2]2
d. 2HCl (g) + .5O2 (g) Cl2(g) + H2O(g)
[Cl2][H2O]
[HCl]2[O2]1/2

Front 25

A vessel originally contained .2mol iodine monobromide (IBr), .0010mol I2 and .0010mol Br2 The equilibrium constant Kc for the reaction is 1.2 x 102 at 150 degree C. What direction (forward or reverse) needed to attain equilibrium at 150 degree C.
I2(g) +Br2(g) 2IBr(g)

Back 25

Qc = [IBr]2i / [I2]i[Br2]I = (0.2)2/(0.001)(0.001) = 0.04/0.000001 = 40000
Qc = 40000 > Kc = 120 reaction goes to left; therefore it is a reverse reaction

Front 26

Which of the following are strong acids? Which are weak acids? (a) HC 2H 3O2: (b) HClO: (c) HCl: (d) HNO3: (e) HNO2: (f) HCN

Back 26

a. weak
b. weak
c. strong
d. strong
e. weak
f. weak

Front 27

Define an acid and a base according to Bronsted-Lowery concept. Give an acid-base equation and identify each species as an acid or a base.

Back 27

Acid: species donating a proton in a proton-transfer reaction
Base: species accepting the proton in a proton-transfer reaction
NH3 (aq) base + H2O (l) acid NH4+ (aq) acid + OH—(aq) base

Front 28

Compare structures of HNO2 and H2CO3. Which would you expect to be stronger acid? Explain your choice.

Back 28

HNO2 has greater electro-negativity over H2CO3 so it would be a stronger acid

Front 29

A weak acid HA, is dissolved in water. Which one of the following beakers represents the resulting solution?

Back 29

A – Since weak acids partially disassociate, you will see all components of the reaction. H3O, OH-, and acid anions

Front 30

For the following reactions, level each species as an acid or a base. Indicate the species that are conjugates of one another

Back 30

a. HPO42- + HCO3- PO43- + H2CO3
ACID (A) BASE (B) BASE(C) ACID (D)
Congugate: A C , B- D
b. F- + HSO4- HF + SO42-
BASE (A) ACID (B) ACID (C) BASE (D)
Congugate A C, B D

c. HSO4- + H2O SO42- + H3O+
ACID (A) BASE (B) BASE(C) ACID (D)
Congugate: A C , B- D
d. HS- + CN- S2- + HCN
ACID (A) BASE (B) BASE(C) ACID (D)
Congugate: A C , B- D

Front 31

The ball and stick models of the reactions in a Lewis acid base reaction. Write the complete equation for the reaction, including the product. Identify each reactant as a Lewis acid or a Lewis base.

Back 31

AlCl3 (acid) + :NH3 (base) AlCl3:NH3

Front 32

Write a reaction for each of the following in which the species acts as a Bronsted base. The equilibrium should favor the product side

Back 32

a. H2O(l)base + HC2H3O2 (aq) acid C2H3O2-- (aq)base + H3O+ (aq)acid

b. HCO3—(aq)base + HF (aq)acid H2CO3 (aq)acid +F-- (aq)base

c. NH3(aq)base + H2O(l)acid NH4+ (aq)acid + OH--(aq)base

d. H2 PO4-- (aq)base + HNO3 (aq)acid H3PO4 (aq)acid + NO3-- (aq)base

Front 33

Which of the following is the weakest acid: HClO4 : HCN or HC2H3O2

Back 33

HCN then HC2H3O2 then HCLO4

Front 34

Phosphorous acid, H2PHO3, is a diprotic acid. Write equations for the acid ionizations. Write the expressions for Ka1 and Ka2

Back 34

H2PHO3 + H2O  HPHO3- +H3O+
HPHO3- + H2O  PHO32- + H3O+
Ka1= [HPHO3-][H3O+] / [H2PHO3]
Ka2= [PHO32-][H3O+]/ [HPHO3-]

Front 35

Define the term buffer, give an example.

Back 35

A solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. Ex. H2PO4-, HPO42-

Front 36

You have .10 mol samples of three acids identified simply as HX, HY, and HZ. For each acid, you make up .10M solutions by adding sufficient water to each of the acid samples. When you measure the pH of these samples, you find that the pH of HX is greater than the pH of HY, which in turn is greater than the pH of HZ.

Back 36

A – Which is the least ionized? HX represents a high pH which is at best a weak acid, and weak acids only partially ionize

B – Which has the largest Ka? HZ has the lowest pH which means it’s the strongest acid and the strongest concentration of H3O+ ions

Front 37

A friend of yours has performed three titrations, strong acid with a strong base, weak acid with a strong base, and weak base with a strong acid. He hands you the three titration curves, saying he has forgotten which is which. What attributes of the curves would you look at to identify each curve correctly.

Back 37

Weak acid titration starts at a higher pH, and the pH change is shorter than a strong acid. A strong acid starts at the a lower pH and takes longer to change. Generaly the stronger acid will change before the equivalence point while the weak acid changes near the point. A strong base titration will be the same as an acid except that it is flipped. It slowly drops in pH levels, . A weak base drops rapidly in pH levels and is closer to the eq. point.

Another answer:
Strong acid/Strong base:
• Starts at low pH
• Slowly increases to near equivalence
• pH rapidly changes near equivalence pt from pH of about 3-11
• equivalence pt neutral
Weak acid/Strong base:
• similar to strong acid/strong base, but starts at higher pH
• pH changes slowly at first
• pH change occurs rapidly near equivalence pt b/w pH range about 7-11 (shorter than strong acid/strong base)
• equivalence pt on basic side
Weak base/Strong acid:
• curve drops in pH rather than rising (therefore starts w/high pH, basic)
• pH slowly declines at first
• abruptly falls from about 7-3 pH

Front 38

Write chemical equations for the acid ionizations of each of the following weak acids, in terms of H3O+

Back 38

HCO2H (formic acid)
HCO2H (aq) +H2O(l) HCO2- (aq) +H3O+ (aq)
HF (hydrofluoric acid)
HF (aq) + H2O(l) F--(aq) + H3O+(aq)
HN3 (hydrazoic acid)
HN3(aq) + H2O(l) N3-(aq) + H3O+(aq)
HOCN (cyanic acid)
HOCN (aq) + H2O(l) OCN--(aq) + H3O+(aq)

Front 39

A chemist wanted to determine the concentration of a solution of lactic acid. HC3H5O3. She found that the pH of the solution was 2.51. What was the concentration of the solution? The Ka of lactic acid is 1.4 x10-3.

Back 39

HLa + H2O H3O+ + La-
X 0 0
(X - .0031) .0031 .0031
[.0031][.0031] / [X - .0031] = 1.4 x 10-4
9.61 x 10-6 = [1.4 x 10-4][X - .0031]
9.61 x 10-6 / 1.4 x 10-4 = X - .0031
6.86 x 10-2 = x - .0031
(6.86 x 10-2) + (3.1 x 10-3) = X
X = .0717 or 7.17 x 10-2

Front 40

Write the equation for the acid ionization of the Cu(H2O)62+ ion

Back 40

Cu(H2O)62+ (aq) + H2O (l) H3O+ (aq) + Cu(H2O)5 (OH)+(aq)

Front 41

Explain why barium fluoride dissolves in dilute hydrochloric acid but is insoluble in water

Back 41

BaF2(s) + 2Cl-(aq)  Ba2+(aq) +2Cl- (aq) + 2F-(aq)
Solubility is affected by the pH if the compound supplies an anion conjugate to a weak acid. In this case, the Fluoride is an anion to the H+ provided by the HCl. In the case of water, it is less likely to provide that H+ than would the HCl, and the Barium Fluoride is therefore soluble in HCl solution but not H2O.

Front 42

Lead chloride at first precipitates when sodium chloride is added to a solution of lead nitrate. Later, when the solutions is made more concentrated in chloride ion, the precipitate dissolves. Explain what is happening. What equilibira are involved? Note that lead ion forms the complex ion PbCl42-

Back 42

The chloride anion acts as a ligand solution because the complex ion PbCl42- is created so as more Cl- is added it lowers the concentration of the Pb making it more soluble. So at first when more lead is added it’s ion product exceeds Ksp and precipitates but once Cl- is added it lowers the ion product so it dissolves

Front 43

You add dilute hydrochloric acid to a solution containing a metal ion. No precipitate forms. After the acidity is adjusted to .3 M hydronium ion, you bubble hydrogen sulfide into the solution. Again no precipitate forms. Is it possible that the original solution contained silver ion? Could it have contained copper (II) ion?

Back 43

No, it could not have contained either a silver ion or a copper (II) ion. If it had contained a silver ion, it would have precipitated in the first step (adding HCL) as a chloride. If it had contained a copper (II) ion, it would have precipitated in the second step (bubbling in Hydrogen Sulfide) as a sulfide.

Front 44

Use the solubility rules (table 4.1) to decide which of the following compounds are expected to be soluble and which insoluble. It

Back 44

a. Ca(NO3)2 - Soluble, b/c nitrates are soluble (Rule 2)

b. AgBr - Insoluble, b/c this is an exception to the solubility of bromides (Rule 3)

c. MgI2 – Soluble, b/c most iodides are soluble and this is not an exception (Rule 3)

d. PbSO4 - Insoluble, b/c this is an exception to the solubility of sufates (Rule 4)

Front 45

Describe how you could separate the following mixture of metal ions: Na+ Hg2+ Ca2+

Back 45

Using Qualitative Analysis, by adding sulfide, Mercury Sulfide would precipitate.
(Filtrate now contains Ca2+ and Na+)
Add (NH4)2CO3
Ca2+(aq) + CO32-(aq) → CaCO3(s)
(What’s left is Sodium)

Front 46

A student was asked to identify a compound. In an effort to-do so, he first dissolved the compound in water. He found that no precipitate formed when hydrochloric acid was added, but when H2S was bubbled into this acidic solution, a precipitate formed. Which one of the following could be the precipitate?

Back 46

a. PbCl2 – incorrect b/c Pb2+ would get precipitated as a sulfide (PbS2) in Group II if H2S was bubbled in

b. CdS - incorrect b/c most sulfides are insoluble

c. MnS - correct b/c Mn2+ gets precipitated by H2S as a sulfide in analytical group III

d. Ag2S – incorrect b/c Ag+ will not precipitate as a sulfide but will as a chloride

Front 47

What is a spontaneous process? Give three examples of spontaneous processes. Give three examples of non-spontaneous?

Back 47

Physical or chemical change that occurs by itself.
Spontaneous Ex:
1. rock, at the top of a hill rolls down
2. heat flows from a hot object to a cold one
3. an iron object rusts in moist air
Nonspontaneous Ex:
1. rock could be moved to the top of the hill, but work would have to be expended
2. heat can be made to flow from a cold to hot object, but a heat pump or refrigerator is needed.
3. rust can be converted to iron but the process requires chemical reactions used in the manufacture of iron form its ore (iron oxide)

Front 48

Which contains greater entropy, a quantity of frozen benzene or the same quantity of liquid benzene at the same temperature? Explain in terms of dispersal energy in the substance.

Back 48

Liquid Benzene has greater Entropy. It includes the phase changes between frozen, liquid and sublimated benzene and its attendant molecular disorder.

Front 49

State the second law of thermodynamics

Back 49

the total entropy of a system and its surroundings always increases for a spontaneous process. Also, for a spontaneous process at a given temperature, T, the change in entropy of the system is greater than the heat divided by the absolute temperature, q/T

Front 50

Which of the following is spontaneous?

Back 50

a. cube of sugar dissolves in a cup of hot tea
b. rusty crowbar turns shiny (requires chemical reaction)
c. butane from lighter burns in air (required a spark)
d. clock pendulum, initially stopped, begins swinging (requires a force)
e. hydrogen and oxygen gases bubble out from a glass of pure water (requires hat or electric current)

Front 51

Predict the sign of entropy change for each of following

Back 51

a. drop food coloring diffuses through a glass of water (+)
b. tree leafs out in spring (-)
c. flowers wilt and stems decompose in fall (+)
d. lake freezes over in the winter (+)
e. rainwater on pavement evaporates (-)

Front 52

What is delta U for the following reaction at 25 degrees C
2H2(g) + O2(g) 2H2O(l)

Back 52

0 + 0 → 2(-285.8)
∆U = Uf – U1
∆U = 2(-285.8) – 0
∆U = -571.6 kj

Front 53

Predict sign for delta S, if possible for each of the following if not possible explain why

Back 53

a. N2H4(l) N 2(g) +2H2O(g)
+
b. C2H5OH (l) + 3O2(g) 2CO2(g) + 3H2O(l)
-
c. P4(s) + 5O2 (g) P4O10(s)
-
d. 2NaHCO3(s) Na2CO3(s) + H2O(g) +CO2 (g)
+

Front 54

Calculate the delta S for the following, using standard entropy values

Back 54

a. Pb(s) + .5O2(g)  PbO(s)
ΔS° = (66.32) – ((64.78) + (1/2 x 205)
66.32 – (64.78 + 102.5)
66.32 – 167.28
ΔS° = 100.96
b. CS2(g) + 4H(g)  CH4(g) + 2H2S(g)
ΔS° = (186.1 + (2 x 205.6)) – (237.9 + (4x 130.6))
(186.1 + 411.2) – (237.9 + 522.4)
597.3 – 760.3
ΔS° = -163
c. C2H4(g) + 3O2(g)  2CO2 (g) + 2H2O(g)
ΔS° = ((2 x 188.7) + (2 x 213.7)) – (219.2 + (3 x 205))
(377.4 + 427.4 ) – (219.2 + 615)
804.8 – 834.2
ΔS° = -29.4
d. Ca(s) + 2HCL (aq)  CaCl2 (aq) + H2 (g)
ΔS° = (104.6 + 130.6) – (41.59 + (2 x 56.5))
235.2 – (41.9 + 113)
235.2 – 154.9
ΔS° = 80.3

Front 55

The combustion of acetylene, CH is a spontaneous reaction given. As expected for combustion, the reaction is exothermic. What is the sign of delta H? What do you expect the sign for delta S? Explain the spontaneity of the reaction in terms of the enthalpy and entropy changes.

Back 55

2C2H2(g) + 5O2(g)  4CO2(g) + 2H2(l)
ΔH°f = 226.7 0 -393.5 -241.8
ΔS° = 205 205 213.7 69.95
ΔH° = ((4 x -393.5) + (2 x 241.8)) – (2 x 226.7)
(-1574+ -4836) – (453.4)
-2057.6 – 453.4
ΔH° = -2511
ΔS° = ((4 x 213.7) + (2 x69.95)) – ((2 x 200.9) + (5x 205))
(854.8 + 139.9) – (401.8 + 1025)
994.7 – 1426.9
ΔS° = -432.1
The reaction produces a heat of reaction but is not spontaneous, which is shown by the negative outcome of the ΔS° equation.