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Chemistry IB HL > Topic 5: Energetics & Thermochemistry > Flashcards

Topic 5: Energetics & Thermochemistry Flashcards

1
Q

heat

A
  • a form of energy
  • transferred from a warmer body to cooler body
  • due to temp equilibrium
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2
Q

absolute zero

A

-273 degrees Celsius

the temp at which all particle movements cease completely

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3
Q

1st law of thermodynamics

A

AKA: law of conservation of energy

  • energy can neither be created nor destroyed
  • it can only be converted from one form to another
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4
Q

System

A
  • a specified part of the universe

- under observation/where a chem rxn is taking place

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5
Q

surroundings

A
  • the remaining portion of the universe

- NOT part of the system

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6
Q

enthalpy

A
  • heat content of the system

- denoted by H

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7
Q

enthalpy change

A
  • heat absorbed/evolved during a rxn
  • at constant temp/pressure
  • is denoted by ΔH
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8
Q

enthalpy change equation

A 
ΔH = Hp - Hr
ΔH = enthalpy change
Hp = enthalpy of products
Hr = enthalpy of reactants
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9
Q

enthalpy change of exothermic reactions

A

ΔH = negative

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10
Q

exothermic reaction

A
  • heat energy is released

- energy released in bond formation on product side is greater than energy consumed in bond breaking on reactant side

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11
Q

enthalpy change of endothermic reactions

A

ΔH = positive

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12
Q

endothermic rxn

A
  • heat energy is absorbed

- energy released in bond formation on product side is less than energy consumed in bond breaking on reactant side

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13
Q

Specific heat capacity

A 
heat required to raise the temp of 1g of a substance by 1 degree Celsius/Kelvin
unit: J/kg
j = joules
k = kelvin
g = grams
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14
Q

heat capacity

A

heat needed to increase the temp of the object by 1 degree K/C

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15
Q

Hess’ Law

A
  • the total enthalpy change for a chem rxn doesn’t depend on the pathway it takes
  • only considers initial and final states
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16
Q

Standard enthalpy change

A

enthalpy change when 1 mol of the gaseous bond is broken or formed

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17
Q

calorimetry

A

technique of measuring enthalpy change

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18
Q

what is a calorimeter?

A
  • a well-insulated container (e.g. polystyrene cup)
  • in which temp change of a liquid is measured
  • before and after the change
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19
Q

assumptions in the calculations of heat transferred to water

A
  • there is no heat transfer between the soln, the thermometer, the surrounding air, and the calorimeter itself
  • the solution is dilute enough so that its density and specific heat capacity are equal to water’s
  • the rxn is assumed to have occurred sufficiently rapidly for max temp to be achieved before rxn mixture begins cooling to room temp
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20
Q

problems with calorimetry

A
  • not having the desired rxn occur (e.g. incomplete combustion)
  • loss of heat to surroundings in exothermic reactions
  • absorption of heat from surroundings in endothermic reactions
  • using an incorrect specific heat capacity in enthalpy calculations
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21
Q

what is a cooling graph?

A
  • for slow rxns (e.g. metal ion displacement), the results will be less accurate
  • due to heat loss over time
  • an allowance can be made by plotting a temp-time graph (cooling graph)
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22
Q

enthalpy of formation

A
  • denoted by ΔHf
  • enthalpy change when 1 mole of a substance is formed from its elements
  • all substances being in standard states
  • enthalpy of formation of every element in its standard state is assumed to be 0!
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23
Q

what happens to enthalpy change when you reverse a rxn

A

signs are reversed

e.g. negative value turns positive and vice versa

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24
Q

cycle of ozone depletion and formation

A
  1. Strong covalent double bonds in normal O2 is broken by high-energy UV radiation with λ O•(g) + O•(g)
  2. The oxygen atoms have unpaired electrons, so they are reactive, and will react to form ozone.
    O•(g) + O2 (g) -> O3 (g)
    The rxn is exothermic and the energy given raises the temp of the stratosphere
  3. Bonds in ozone are weaker than the double bond in oxygen, so lower energy UV rays can break them (λ O•(g) + O2(g)
  4. The free atomic oxygen reacts with another ozone molecule.
    O3 (g) + O•(g) -> 2O2 (g)
    This is another exothermic reaction, which produces heat that maintains the relatively high temp of the stratosphere.
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25
Energy
measure of the ability to do work
26
conditions of standard enthalpy change
- pressure: 100kPa - conc.: 1 mol/dm3 for all solutions - temp.: 298K (usually) - all substances in standard state
27
temperature
measure of average kinetic energy of the particles of a substance
28
how to calculate reaction enthalpies from temp changes
ΔH (reaction) = - ΔH (water) = - m (H2O) x c(H2O) x ΔT(H2O) - it's negative bc water gains heat (positive enthalpy change) from the reaction - so if water's ΔH is positive then the reaction's ΔH is negative
28
assumptions made in calculating enthalpy change of reactions in a solution
- no heat loss from system - all heat goes from rxn to water - the soln is dilute - water's density is 1 g/cm
29
how to calculate enthalpy change using Hess' Law
ΔH1 = ΔH2 + ΔH3 ΔHf (products) = ΔH (reaction) + ΔHf (reactants) ``` ΔH1 = enthalpy change of formation (of products) ΔH2 = enthalpy change of formation (of reactants) ΔH3 = enthalpy change of reaction ```
30
how are bonds in oxygen and ozone broken
- upon absorption of UV radiation of sufficient energy | - as bond-breaking is endothermic
31
difference in UV wavelength required to break Oxygen bonds vs Ozone bonds
- Oxygen exists as diatomic molecules (with double bonds) - Ozone exists as molecules with 1 double bond and 1 single bond - Oxygen's bond is stronger than Ozone's - so it's only broken by radiation of higher energy and shorter wavelengths
32
UV wavelength required to break oxygen bonds
wavelength
33
UV wavelength required to break oxygen bonds
wavelength
34
first ionisation energy
ΔHi - energy needed to form the positive ion of the atom - cations (positive ions) only
35
first electron affinity
ΔHe - enthalpy change when 1 mol of gaseous atoms attracts 1 mol of electrons - anions (negative ions) only
36
is the formation of ionic compounds exothermic or endothermic? what are the implications?
- endothermic - can be found by adding ΔHi to ΔHe - thus energetically unfavourable despite forming ions with stable noble gas config
37
lattice enthalpy
ΔHlat - defined as the enthalpy change when 1 mol of gaseous ions is formed from 1 mol of solid crystal - enthalpy change in terms of the reverse process of lattice formation - ionic lattice formation is highly exothermic due to strong attraction between oppositely-charged ions
38
Born-Haber cycle
- used to determine ΔHlat - based on Hess' law - ΔH for the overall formation of the solid = sum of all ΔH of individual steps
39
Born-Haber cycle steps
1. ΔHatom of cation e.g. Na (s) -> Na (g) ΔHatom (Na) = +107 2. ΔHatom of anion e.g. 1/2 Cl2 (g) -> Cl (g) 1/2 E (Cl-Cl) = 1/2 (+242) [Cl's bond enthalpy] 3. ΔHi of cation e.g. Na (g) -> Na+ (g) + e- ΔHi (Na) = +496 4. ΔHe of anion e.g. Cl (g) + e- -> Cl- (g) ΔHe (Cl) = -349 5. - ΔHlat (negative lattice enthalpy of the molecule) e.g. Na+ (g) + Cl- (g) -> NaCl (s)
40
enthalpy change of atomisation
enthalpy change occurring when a solid is atomised to form 1 mol of gaseous ions
41
factors affecting lattice enthalpy
- size of ions: as size decreases, ΔHlat increases | - charge of ions: as charge increases, ΔHlat increases
42
assumptions used in calculating lattice enthalpies with ionic model
- interactions are due only to electrostatic forces between ions - increase in ionic radius will decrease attraction between ions - increase in ionic charge will increase attraction between ions - the crystal is made up of perfectly spherical ions
44
Relationship between enthalpies of solution, lattice, and hydration
ΔHsol = ΔHlatt + ΔHhyd
45
enthalpy change of hydration
enthalpy change when 1 mol of gaseous ions dissolve in sufficient water to form a soln
46
bond dissociation enthalpy
energy change when 1 mol of gaseous ions have their bonds broken
47
conditions for hydrogenation
- nickel/platinum/palladium | - 150-200 deg Celsius
48
enthalpy change of hydration
ΔHhyd - enthalpy change when 1 mol of a specific gaseous ion dissolves in sufficient water to form a dilute soln - exothermic process - negative enthalpy change
49
entropy
- measure of randomness/disorder of a system | - represented by S
50
driving force
force responsible for the spontaneity of a process
51
tendencies of driving force
- tendency for minimum energy rxns - tendency to favor exothermic rxns (ΔH = negative) - tendency to favor rxns with max randomness - however, as rxns like liquification of gas and solidification of liquid are feasible, a rxn's spontaneity doesn't fully depend on randomness
52
factors affecting entropy
- temp: as temp increases, entropy increases - change in state: S -> L -> G, entropy increases - no. of particles: as no. of particles increases, entropy increases - mixing of particles: as disorder increases, entropy increases
53
predicting whether ΔS is positive or negative
- if rxn is s -> l -> g, ΔS = positive | - if rxn is g -> l -> s, ΔS = negative
54
gibbs free energy equilibrium
- ΔG describes the spontaneity & temp dependence of a rxn - always opposite to ΔS
55
difference between ΔH and ΔS
``` ΔH = enthalpy change of rxn ΔS = enthalpy change of system ```
56
meaning of ΔG = positive
non-spontaneous rxn, reactants favored
57
meaning of ΔG = negative
spontaneous rxn, products favored
58
meaning of ΔG = 0
equilibrium (neither products nor reactants favored)
59
what is k?
k = conc. of products / conc. of reactants - value of k indicates whether products or reactants are favored in equilibrium - if k > 1, ΔG 0
60
gibbs free energy formula
ΔG = ΔH - TΔS | T must be in Kelvin (in standard conditions, T is 298K)
61
calculating entropy of surroundings
ΔS (surrounding) = - ΔH (system) / T (in Kelvin) ΔS (total) = ΔS (system) + ΔS (surrounding) OR ΔG = - TΔS (total)
62
ΔG value if ΔS = positive and ΔH = positive
ΔG = depends on temp | ΔG will be negative at high temp (meaning rxn is spontaneous at high temps)
63
ΔG value if ΔS = positive and ΔH = negative
ΔG = negative | spontaneous at all temps
64
ΔG value if ΔS = negative and ΔH = negative
ΔG = depends on temp | ΔG will be negative at low temp (meaning rxn is spontaneous at low temps)
65
ΔG value if ΔS = negative and ΔH = positive
ΔG = positive | non-spontaneous at all temps
66
spontaneous
- when a rxn can give gibbs free energy that can do work - ΔG = negative - rxn will occur without added energy

Chemistry IB HL (11 decks)

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