Thermodynamics & Gas Law

Structure with a greater extent of intermolecular forces have ______ boiling points because more energy is required to overcoem the attractions between molecuels during ______ from a liquid state to a gas state.

Hydrogen atoms bonded with N,O, or F form polar bonds with partial charge producing dipoles which ______ boiling point

higher

phase change

increases

Branched structures experience fewer london forces comparable to straight chain molecules, which ____ boiling point

decreases

A. deposition

B. Melting

C. condensation

D. Sublimation

Answer: A

Passage A

Passage A

Which of the following conclusions can be made based on figure 1 and data in table 1

Answer: D

Passage A

Answer: C

Freezing occurs when the _____ energy of the molecules can no longer overcome the _________ forces attracting nearby molecule together. As this happens, the molecules are pulled together and held as a solid.

Pure water molecuels strongly itneract rhough hydrogen bonding, which greatly assists freezing as the temperature reaches freezing point. As pure water molecules begin to freeze, molcules transition from _____ entropy (liquid0 to ____entropy (solid).

Add a salt solute disrupts some of these intermolecular interactions between the water molecules and forms new _______ interactions with salt ions. This inclusion of a salt also ____ the entropy (disrupts the usual order)

kinetic

intermolecular

High (liquid) to lower entropy (solid)

ion dipole interactions

increases the entropy

Passage A

A. lower, because vapor pressure of briny water is higher than the vapor pressure of pure water

B. Higer, because the vapor pressure of briny water is lower than vapor pressure of pure water

C. lower, because the vapor pressure of briny water is lower than the vapor pressure of prue water

D. higher, because the vapor pressure of briny water is higher than vapor pressure of pure water.

Answer: B

equillibrium partial pressure formula

Kp= (product)^m/(reactant)^n

8= x/(0.5)²

x= 2 atm

Ideal gas vs deviation (conditions)

Ideal gas

High temp

low pressure

low molecular volume

Answer: C

Charles law

direct relationship exists between Temp (in kelvin) and volume

V1/T1=V2/T2

Boyle’s law

inverse relationship between volume and pressure

P1V1=P2V2

is it greater or less than 500ml? and which law says so

Charles’s law

V1/T1=V2/T2

500/293=V2/310

so increasing T2 increases V2, V2 is greater than 500 mL

Passage B

Passage B

A. Both metal sample and apparatus experience an increase in entropy, resulting in an overal increase in entropy within the system

B. The entropy of the metal sampel decreases, but the entropu of the apparatus icnreases by greater amount, resulting in an overal increase in entropy within the system

C. The entropy of the mettal sample icnreases but the entropy of the apparatus decreases by a greater amount, resulting in an overal decrease in entropy within the system

Answer:B

Passage B

100ml = 100 g

heat capacity

amount of heat required to raise the temp of the entire smaple by 1ºC

context formula: q= CT (q=CAT)

specific heat capcity

amount of heat required to raise the temp of 1 gram of substance by 1ºC

q=mCT (q=mCAT)

Passage B

fomula q= CT for heat capacity (not q=MCAT— which is for specific heat capacity)

Note: -qsample = qwater + qcalorimeter

rearrange: qcal= -qwater-qsample

q/T= C

-qwater-qsample/ (change in Temp of the system) = Ccal

And based on this conclusion, what is the unknown metal?

Passage B

No, because the calorimeter is greater than 0 in experiment 2,this is shown in experiment 1 when the calorimetet absorbs a significant amount of heat. This means that the Cp is wrong.

Because of this, the unknown sample is Ti, because caloeimeter is >0, more than 2100 J of heat is released by the metal, this means the specific heat capacity is greater than the 0.42

Answer: A

Dalton’s law of partial pressure

dimensional analysis to figure out the torr of other gas and add them the sum

0.4 (Co2) + 0.2 + 0.2 ( N2 and H2O)= 0.8 torr

effect of temperature on the quilibrium constant

decrease

Note: T is in kelvin

Passage C

Passage C

(just solve) + what order of reaction?

Passage C

solve for the rate question first to get (rate = k [P][RSH] (2nd order)

Passage C

solve for the rate question first to get (rate = k [P][RSH] (2nd order)

Passage C

passage D

Passage D

Passage D

in lipid, aqueous, or gaseous environemnt?

Aqueous

regarding formation, is it thermodynamically favorable under what temp?

Passage D

becomes more favorable at lower temps

Passage D

so its 1,2 and 3

regarding Keq…. what happens to it?

it will decrease

the reaction is exothermic based on the figures (change in H is negative) so heat is a product, adding heat shifts the reaction to the left (reactants)

N2

because there are mroe moles of N2(g)

Water is a constant

273+ (-196) = 77

so 75-79K

Passage E

Passage E

1,3 only

Passage E

Passage E

Volatlity

how easily a substance tansitions from liquid or solid into a vapor or gas

Volatile substances

easily evaporate or sublime

Properties: high vapro pressure, low boiling points, weak IMF

nonvolatile substances

do not readily evaporate

negligible vapor pressure (no pressure), high boiling points, strong IMF

At any given temp, there is enough energy for some molcules in a liquid to overcome the IMF between molecules and escape into gas phase as vapors that are in equilibrium with liquid phase. The pressure resulting from these is called ______ pressure. When the temp is high enough for that specific pressure to equal the external pressure exerted by surroundings, the liiquid boils.

Vapor

Passage E

Nonvolatile solutes dissolved in the liquid reduce the amount of surface area availbale for liquid molecules to occupy and escape into gas phase. Nonvolatile solute also increases the boiling temp

So removing and purification of the proteins would decrease the boiling point.

mmHg and kPA and atm equivalence

101.3 kPa= 760 mmHg = 1 atm

state function

thermodynamic system whose value depends only on the state of the system. As such, it is independent of the path taken

Passage F

Passage F

C

Passage F

2 and 3 only

1,2,3

Passage G

Passage G

Answer A

passage G

Describe what the graph would look like

Is the enthalpy formation positive or negative, bc the WX2 formation is favored with……

Negative, because its favored with decreasing tmeperature.

Is it a closed, open, or isolated system?

Closed system, because only the heat adiated form the filament is echanged with the surroundings

Open system

both heat and matter are freely exchanged with the surroundings

Closed system

only heat is exchanged with surroundigns

Isolated

neither heat nor matter can be exchanged with the surroundings.

Answer: B

Heat of formation vs bond enthalpies

Larger, smaller, or same?

Larger

Passage H

Passage H

Passage H

Endergonic or exergonic?

Exergonic

Passage H

Passage H

Which compound is kinetic and which is thermodynamic product?

2- kinetic (lower Ea), 3- thermodynamic (most stable-lowest energy)

Passage I

Passage I

state whether each entropy is less than or greater than 0

Heat of fusion = to?

Heat of solidification

Heat of vaporization =?

Heat of condensation

open system vs isolated system, vs closed vs invalid image

Answer B

33-23= 10

Fe3+ and I- (catalysts in general) do not affec thte heat of reaction

With and without catalyst image

V1/n = V1/n

V2=3V1

n2 is 3 times greater

heat capcity is given, use q=cat

note= b/c it says increase by 10ºC, it is the change in temp so it is the same as 10Kelvin