Use LEFT and RIGHT arrow keys to navigate between flashcards;
Use UP and DOWN arrow keys to flip the card;
H to show hint;
A reads text to speech;
66 Cards in this Set
- Front
- Back
|
chemical reactions that occur inside a battery |
Electrical work |
|
|
Chemical reac4ons are based on changes in _________ as _________________________ |
ENERGY bonds are broken or formed |
|
|
Just like there is a xed amount of mass in the universe there is also a xed amount of energy Energy can be transferred from one body to another Energy cannot be created or destroyed This is called: |
CONSERVATION OF ENERGY |
|
|
System: |
only the chemicals(reactant and products) |
|
|
Surroundings: |
everything else (solvent, glassware and air) |
|
|
ENDOTHERMIC: Energy transferred ______ system PE of the system ____________ Beaker feels ______ Heat is a __________ |
INTO INCREASES COLD reactant |
|
|
EXOTHERMIC: Energy transferred ______ system PE of the system ____________ Beaker feels ______ Heat is a __________ |
OUT OF DECREASES HOT product |
|
|
The heat of a chemical reaction (q) is due to the _________________________. This is due to the making and breaking of bonds |
the change in potential energy Heat = q = ΔH |
|
|
If energy is released it goes ______ the system ΔH < 0 = If energy is absorbed it goes ______ the system ΔH > 0 = |
out of exothermic reaction into endothermic reaction |
|
|
H + H --> H2 When bonds are formed energy is ________ by the system H2 --> H + H When bonds are broken energy is ________ by the system |
released absorbed |
|
|
ΔHRXN = ΔHFormed + ΔHBroken |
ΔHRXN = ΔHBonds Formed ΔH = - 436 kJ/mol of H2 |
|
|
Breaking Bonds ΔH ____ Making Bonds ΔH ____ |
> 0 < 0 |
|
|
|
The NET ΔHRXN depends on the strength of the bonds that are broken and the bonds that are formed. For most reactions we can’t predict ΔHRXN simply bylooking at the reaction.
|
|
|
lΔHformed l > ΔHbroken _________ reaction ΔHbroken > lΔHformed l __________ reaction |
Exothermic Endothermic |
|
|
First Law of Thermodynamics |
The only way a system can experiencechanges in its energy is through theprocess of energy transfer to or from thesurroundings as either HEAT or WORK ΔESYSTEM = q + w |
|
|
INTERNAL energy of the system |
ESYSTEM = KEAVE + PEAVE for all the molecules that comprise the system For the Universe: ΔE = 0 For a chosen system: ΔE = 0 |
|
|
Energy: The capacity to do ______ or to produce _____ |
WORK- Motion against an opposing force HEAT - Motion in random directions |
|
|
Heat and Work are: Examples of work: ex of heat: |
processes for the transfer of energy • Radiation from the sun • A candle burning • Hot and cold packs • Baseball bat hitting a baseball • Picking up a book or walking up the stairs • A battery running an electrical device |
|
|
work - ____________________ Work done ON the system w _ 0 Energy of the system _____________ Work done BY the system w _ 0 Energy of the system _____________ |
force acting over a distance Work done ON the system w > 0 Energy of the system INCREASES Work done BY the system w < 0 Energy of the system DECREASES |
|
|
heat - ___________________ Heat INTO the system q _ 0 Energy of the system ____________ Heat OUT OF the system q _ 0 Energy of the system ____________ |
Process of energy transfer in a random direction Heat INTO the system q > 0 Energy of the system INCREASES Heat OUT OF the system q < 0 Energy of the system DECREASES |
|
|
Consider the reaction for the evaporation of water: H2O(l) → H2O(g) 1. Is this reaction endothermic or exothermic? HINT: Think about IMF. Are bonds being broken or formed? 2. What is the system and what are the surroundings? 3. Is heat posi5ve or nega5ve for this reac5on? 4. Is the change in energy of the surroundings posi5ve or nega5ve? |
1. IMF are being broken so this is an endothermic reacton Energy required to break the Hydrogen bonds. 2. System: liquid and gaseous water Surroundings: everything else (beaker, stove, bench etc.) 3. Endothermic energy is moving into the system q > 0 4. Energy is moving out of the surroundings into the system q < 0 |
|
|
The Ideal Gas model consists of four postulates that explain the behavior of an ideal gas: |
1. Gas molecules have no volume of their own. 2. Gas molecules have perfectly elastic collisions and are in constant linear motion. 3. Gas molecules experience no intermolecula forces acting between one another. 4. The KE of an ideal gas is directly proportional to it temperature measured in degrees K. |
|
|
There are 4 variables that indicate the state of a gas |
Pressure (P) Measured in atmospheres (atm): This indicates the number of collisions Volume (V) Measured in liters (L): This indicates the space the gas occupies Number of moles (n indicates amount) measured in moles (mol): This indicates how MUCH gas there is Temperature (T) Measured in Kelvin (K): This indicates the internal kinetic energy of the gas |
|
|
The KE of a gaseous system is related to both theamount of gas and the internal temperature of the gas
The molar KE (KEMOLAR) of a gas is the KE per mole Any collec4on of gases at the sametemperature will have the SAME KEMOLAR KEMOLAR = KE/n = [(3/2)nRT]/n = (3/2)RT KEMOLAR = (3/2)RT |
Kinetic Molecular Theory |
|
|
wPV = - PatmΔV |
work is force exerted over a distance |
|
|
When a gas expands (more moles of gas on product side)its volume _________. It pushes against an external pressure and its energy goes ____. __________, and so w _ 0. (is work done or does system do work?) When a gas is compressed (fewer moles of gas onproduct side) its volume _________. An external forcepushes on the gas and its energy goes ____. _________, and w > 0. (is work done or does system do work?) |
increases down the system does work so w < 0 decreases up work is done on the system so w > 0 |
|
|
1. Consider the reac5on for the formation of water: H2(g) + ½ O2(g) → H2O(g) ΔHf° = -‐‐242 kJ/mol Does this systemdo pressure-volume work, or is work done on the system? |
Work is done on the system w > 0 # moles of gas is decreasing(less moles on the product side) Atmosphere does work on the system, w > 0 |
|
|
Temperature is propor4onal to the internalkine4c energy of a system. Temperature is thus all about theRANDOM MOTION of par4cles. In calorimetry temperature is usuallymeasured in units of degrees Celsius (°C) |
ya 3ami |
|
|
Constant Pressure Calorimetry |
q = mcΔT |
|
|
Specic Heat Capacity (c or s) |
Substances withsmall heat capacityvalues heat upvery quickly. Substances withlarge heat capacityvalues will need a lotof heat to raise theirtemperatures |
|
|
If q is postive, then energy was ____________. |
absorbed by the system as heat |
|
|
Overtime the energy from the hot water will be transferred to the cold water until both systems reach the same nal temperature. |
Thermal Equilibrium |
|
|
Energy lost = Energy gained |
I ΔElost I = ΔEgained ‐ q1 = + q2 ‐(m1c1ΔT1) = +(m2c2ΔT2) q = mcΔT |
|
|
Consider a Styrofoam cup that contains 50.0 g of water at 10.0oC. If you add a 50.0 g iron ball at 90.0oC to the water, what is thees5mate for the nal temperature of the water? (cH2O = 4.18 J/oC-‐‐g and cFe = 0.45 J/oC-‐‐g) A. Above 90oC B. Between 50oC and 90oC C. 50oC D. Between 10oCand 50oC E. Below 10oC |
The two systems end at the SAME temperature. The amounts ofwater and iron are the same, but the heat capacity of iron is muchlower so it will cool down more quickly than the water will heat up D. Between 10oCand 50oC |
|
|
Latent Heat: heat of ___________ |
phase change There is no change in temperature during a phase change:The energy associated with the phase change is called LATENT HEAT The energy going in isused to overcome theintermolecular forces holding the molecules together |
|
|
LIQUIDto GAS Latent heat of _______________ SOLID to LIQUID Latent heat of _______________ qphase transition = mΔH |
VAPORIZATION ΔHvap = 2257 J/g q = mΔHv FUSION ΔHf = 334 J/g q = mΔHf The reverse processes would use the negative ΔH values! |
|
|
Process Quantity Heat and Work Variables depend ______________. ________ path dependent State Function Energy and Enthalpy Variables depend ______________ ________ path dependent |
Process Quantity Heat and Work Variables depend on how we conduct a process. ARE path dependent State Function Energy and Enthalpy Variables depend only on the current state of the system. are NOT path dependent |
|
|
Enthalpy is the heat of a reaction |
H = E + PV ΔE = q + w Expanding gas: w = -PΔV
For constant pressure (ALL bench top chemistry) and PV work only (No electrical work is done (ALL gases)): ΔH = qrxn |
|
|
Extensive Variable Intensive variables |
Value depends on the extent of the stu i.e. length Heat (q) always has units of kJ Enthalpy (ΔH) can have units of kJ Amount of stu doesn't aect the value i.e. pressure
Enthalpy (ΔH) can have units of kJ/mol (molar enthalpy) |
|
|
2 ways to calculate the heat of reaction |
1. Hess Law: ΔH overall RXN = ΣΔH individual RXNs 2. Enthalpy of Formation: ΔHRXN = ΣΔHf(products) - ΣΔHf(reactants) |
|
|
Formation reactions |
A chemical reaction in which 1 mole of a compound is made from its pure elements in their most stable states. H2(g) + ½O2(g) → H2O(l) ½H2(g) + ½Cl2(g) → HCl(g) C(graph) + O2(g) → CO2(g) The heats of these reactions are called heats of formation(ΔHf). These have been measured and are tabulated. They can be used to compute heats of other compounds. The natural state of an element is its most stable form at room temperature and pressure. |
|
|
ΔH of : naught (o) = standard states Solids, Liquids, Gases: Pure form at 1 atm pressure Aqueous soluEons: concentraEon of 1 M |
:] |
|
|
Find the enthalpy of reacEon for the combusEon of ethane (C2H6) to yield carbon dioxide and water vapor. (what are the steps) |
step 1: write the balanced chemical rxn step 2: write ΔHf val with stoichiometry below the species step 3: ΔHRXN = ΣΔHf(products) - ΣΔHf(reactants) |
|
|
What does spontaneous mean? |
A spontaneous process is one that occurs without the help of an outside agent. A spontaneous process will occur irreversibly in the absence of an external force! |
|
|
Spontaneous processes can be reversed |
A process will be spontaneous in ONLY one direction. This does not mean that it is irreversible.The reverse reaction will NOT be spontaneous,but it can be made to occur with the help of an outside force. |
|
|
2nd Law of Thermodynamics: |
For any spontaneous process the entropy of the universe always increases (ΔSUNIV > 0) ΔSUNIV > 0ΔSUNIV = ΔSSYS + ΔSSURR |
|
|
Remember from the 1st Law of Thermodynamics: |
ΔEUNIV = 0ΔEUNIV = ΔESYS + ΔESURR |
|
|
How many of the following processes are considered spontaneous? A. Ice melting in warm water B. Sugar dissolving in water C. Candle wax burning (once the candle has been lit) D. A gas expanding E. Heat transfer from a hot object to a cooler object |
all of them are spontaneous |
|
|
According to the 2nd law of thermodynamics which of the following statements is true? Choose one statement from each set of answers. A. For any spontaneous process, ΔHSYS < 0 B. For any spontaneous process, ΔHUNIV < 0 C. For any spontaneous process, ΔSSYS > 0 D. For any spontaneous process, ΔSUNIV > 0 A. For any spontaneous process ΔSSYS must never decrease. B. The entropy of the universe is always constant. C. For any spontaneous process ΔSSYS must be positive. D. For any spontaneous process, the entropy of the universe must increase. |
D. For any spontaneous process, ΔSUNIV > 0 D. For any spontaneous process, the entropy of the universe must increase. |
|
|
2nd Law Thermo
ΔSUNIV > 0: _____________________________ ΔSUNIV < 0: _____________________________ ΔSUNIV = 0: _____________________________ |
Process is spontaneous in the forwards direction Process is spontaneous in the backwards direction Spontaneous process is at equilibrium ΔSUNIV = ΔSSYS + ΔSSURR |
|
|
MODEL the surroundings as a heat reservoir: |
Modeling the surroundings When a reaction occurs, how do the surroundings know? the surroundings can lose or absorb small amounts of heat without experiencing any changes in temperature. |
|
|
ΔSSURR =qSURR/T =‐qSYST
ΔESURR = q
qSURR = -qSYS ΔSUNIV = ΔSSYST + qSURRT ΔSUNIV = ΔSSYST ‐ qSYST |
The entropy change of the surroundings is equal tothe heat transferred to the surroundings from thesystem divided by the temperature in KELVIN |
|
|
There are 4 characteristics we will use to help predict entropy changes within a given system: |
1. Expansion of a Gas 2. Change in temperature 3. Physical State 4. Number of moles |
|
|
1. Expansion of a gas |
When a gaseous system expands SSYSTEM increases As particles move to new positions: increasing positional entropy |
|
|
2. Change in temperature |
When the temperate increases SSYSTEM increases As vibrational motion increases: increasing entropy |
|
|
3. Physical state |
SSYSTEM is highest in a gas and lowest in a solid As particles spread throughout container: increasing entropy |
|
|
4. Number of moles |
When the # moles increases SSYSTEM increases As number of particles increases: increasing entropy |
|
|
2CO(g) + O2(g) → 2CO2(g)
Less moles of gas on the product side → Gas ___________ (expansion or compression) (3 moles of gas to 2 moles of gas) Entropy of the system goes _______ |
compression down |
|
|
K2CO3(aq) + 2HCl(aq) → CO2(g) + H2O(l) + 2KCl(aq) More moles of product than of reactant (3 moles to 4 moles) All aqueous reactants → 1 mole of gaseous product → Gas _____________ (expansion or compression) In both cases the Entropy of the system goes ______ |
expansion up |
|
|
If 2 phenomenon give opposing predictions for the same system then you don’t necessarily know the absolute entropy change |
cool |
|
|
Which of the following processes is most likely to have a positive ΔS°SYS at 298 K? A. Cu2+(aq) + 4Cl-‐‐(aq) → CuCl42‐(aq) B. SiH4(g) + 2O2 (g) → SiO2(s) + 2H2O(aq) C. SiF4(g) + 2H2O(aq) →SiO2(s) + 4HF(g) D. 2H2(g) + O2(g) → 2H2O(g) |
C. SiF4(g) + 2H2O(aq) → SiO2(s) + 4HF(g) 1. Increase in # moles: 3 moles of reactants to 5 moles of products 2. Gas expansion: 1 mole of gas to 4 → Increase in entropy |
|
|
Calculating changes in entropy |
For any Spontaneous process ΔSUNIV > 0
Remember that values of ΔH of are tabulated at 298K ΔHRXN = ΣnΔH°f(products) -‐‐ ΣnΔH°f(reactants) Values of S° are also tabulated. (J/mol-‐‐K) As before, the superscript (So) means standard states: 1M solutions, 1 atm for gases and pure liquids or solids Standard states do NOT include Temperature
ΔSRXN = ΣnS°(products) -‐‐ ΣnS°(reactants) |
|
|
3rd Law of Thermodynamics |
We call the tabulated S◦ values absolute entropies For entropy the zero point is NOT arbitrary:It is the point at which we have only one possible configuration: PURE CRYSTAL 3rd law of Thermodynamics: The entropy of a perfect crystal at 0 K is zero. |
|
|
The units of Entropy are: |
Joules/mol-‐‐K This is an energy per temperature unit. Always remember that unlike enthalpy which is measured in kilojoules (kJ) entropy is measures in joules/kelvin (J/K).(1 kJ = 1000 J) |
|
|
ΔS° = 4S°(CO2) + 2S°(H2O) -‐‐ 2S°(C2H2) ‐ 5S°(O2) = 4(213.8) + 2(188.8) ‐ 2(200.9) ‐ 5(205.2) = ‐195.0 J/mol‐K ΔSUNIV _ 0
ΔSUNIV = ΔSSYS + ΔSSURR Because this reaction is spontaneous: ΔSSURR _ 0 |
> > |
|
|
1. Using tabulated values of So find ΔSo for the reaction: 2CO(g) + O2(g) → 2CO2(g) -173 J/mol‐K |
ΔSo < 0 makes sense: 3 moles gas → 2 moles of gas |
