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Chemistry IB HL (2016+) > Topic 5: Energetics & Thermochemistry > Flashcards

Topic 5: Energetics & Thermochemistry Flashcards

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1
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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2
Q

enthalpy change of exothermic reactions

A

ΔH = - x

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3
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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4
Q

enthalpy change of endothermic reactions

A

ΔH = + x

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5
Q

system

A
  • a specified part of the universe

- under observation/where a chem rxn is taking place

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6
Q

surroundings

A
  • the remaining portion of the universe

- NOT part of the system

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7
Q

enthalpy

A

heat content of the system

symbol: H

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8
Q

enthalpy change

A
  • heat absorbed/evolved during a rxn
  • measured at a constant temp/pressure
    symbol: ΔH
    unit: kJ mol^-1
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9
Q

enthalpy change equation

A 
ΔH = Hp - Hr
ΔH = enthalpy change
Hp = enthalpy of products
Hr = enthalpy of reactants
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10
Q

absolute zero

A

-273 degrees Celsius

the temp at which all particle movements cease completely

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11
Q

heat

A
  • form of energy
  • measure of total energy in a given amount of substance
  • depends on the amount of substance present
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12
Q

temperature

A
  • measures the ‘hotness’ of a substance
  • avg kinetic energy of the substance
  • independent of the amount of substance present
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13
Q

specific heat capacity

A 
heat required to raise the temp of 1g of a substance by 1°C/K
unit: J K^-1 g^-1
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°C/K

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15
Q

standard enthalpy change

A

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

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16
Q

calorimetry

A

technique of measuring enthalpy change

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17
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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18
Q

assumptions made when making heat calculations

A
  • no heat transfer between the soln, the thermometer, the surrounding air, and the calorimeter itself
  • the solution’s density and specific heat capacity are equivalent to water’s
  • rxn occured fast enough for max temp to be achieved before cooling begins
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19
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 incorrect specific heat capacity value in enthalpy calculations
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20
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
  • this can be compensated by plotting a temp-time graph (cooling graph) to extrapolate backwards
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21
Q

how to draw a cooling graph

A
  • draw a line of extrapolation backwards from cooling curve
  • draw a vertical line where the reactants were mixed (i.e. when the curve begins to rise)
  • the y-coordinate of the point of intersection is the temp that would’ve been reached if no heat was lost to surroundings
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22
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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23
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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24
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!
25
Q

what happens to enthalpy change when you reverse a rxn

A

signs are reversed

e.g. negative value turns positive and vice versa

26
Q

Energy

A

measure of the ability to do work

27
Q

conditions of measuring standard enthalpy change

A
  • pressure: 100kPa
  • conc.: 1 mol/dm3 for all solutions
  • temp.: 298K (usually)
  • all substances in standard state
28
Q

how to calculate reaction enthalpies from temp changes

A

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
29
Q

how to calculate enthalpy change using Hess’ Law

A

ΔH1 = ΔH2 + ΔH3

ΔHf (products) = ΣΔH (reaction) + ΣΔHf (reactants)

ΔH1 = enthalpy change of formation (of products)
ΔH2 = enthalpy change of formation (of reactants)
ΔH3 = enthalpy change of reaction
30
Q

average bond enthalpy

A
  • energy needed to break one mole of a bond in a gaseous molecule
  • can be used to calculate enthalpy change
31
Q

limitations of using bond enthalpies

A
  • can only be used if all reactants and products are in gaseous state; in other states, more heat would be evolved
  • bond enthalpies are averaged values that are obtained considering a number of compounds containing the bond, but in reality the bond energy varies between compounds
32
Q

formula for specific heat capacity

A

q = mcΔT

q: heat change
m: mass
c: specific heat capacity
ΔT: change in temperature

33
Q

formula for heat capacity

A

C = q/ΔT

C: heat capacity
q: heat change
ΔT: change in temperature

34
Q

assumptions made in enthalpy change calculations for solutions

A 
ΔH(system) = ΔH(water) + ΔH(reaction) = 0
ΔH(reaction) = - ΔH(water)
35
Q

enthalpy change: does change in temperature differ depending on volume of reactants?

A

volume halved = enthalpy halved = heat required also halved = change in temperature will be constant!

36
Q

lattice enthalpy

A

AKA ΔH(lat)

  • endothermic: enthalpy change when 1 mol of gaseous ions form from 1 mol of a solid crystal
  • exothermic: enthalpy change when 1 mol of solid crystal forms from 1 mol of gaseous ions
  • sign (+/-) of ΔH(lat) denotes whether it’s endothermic/exothermic
37
Q

factors affecting lattice enthalpy/enthalpy of hydration

A

↓ size + ↑ charge = ↑ ΔH(lat)

↓ size + ↑ positivity of charge = ↓ ΔH(hyd) – because the enthalpy of hydration is always negative

38
Q

enthalpy change of hydration

A
  • enthalpy change occurring when 1 mol of gaseous ions is dissolved to form an infinitely dilute solution of 1 mol of aqueous ions
  • always exothermic so ΔH(hyd) is always negative
39
Q

entropy

A
  • i.e. randomness/disorder
  • denoted by S
  • defined as the distribution of available energy among particles
  • higher energy distribution (i.e. diffused state) = higher entropy
  • all entropy values are positive except for perfectly ordered solids at absolute zero (S = 0)
40
Q

how to predict ΔS with relation to state

A
  • conversions from s –> l/g have positive ΔS

- vice versa for g –> l/s

41
Q

formula to calculate ΔS

A

ΣΔS(system) = ΣΔS(products) - ΣΔS(reactants)

42
Q

why are exothermic reactions more common than endothermic reactions?

A
  • because ΔS(surroundings) ∝ -ΔH(system)

- exothermic reactions cause an increase in the entropy of the surroundings

43
Q

relationship between ΔS and temperature

A

ΔS(surroundings) ∝ 1/T

- i.e. entropy of surroundings is inversely proportional to absolute temperature

44
Q

formulae for ΔS

A

ΔS(total)* = ΔS(system) - ΔS(surroundings)
*total denotes total entropy of the universe.

ΔS(surroundings) = - ΔH(system) / T

45
Q

how to deduce whether a reaction is feasible

A

as long as ΔS(total) > 0, a reaction will occur spontaneously

46
Q

Gibbs free energy

A
  • criterion of feasibility of a reaction occurring
  • feasible reactions = spontaneous
  • ΔG is always negative for spontaneous processes and positive in the reverse case
  • ΔG is NOT the same as ΔS
47
Q

formula for Gibbs free energy

A

ΔG(system) = -TΔS(total)
ΔG(system) = ΔH(system) - TΔS(system)
T must be in Kelvin (in standard conditions, T is 298K)

48
Q

how can ΔG(system) predict feasibility?

A
  • as ΔS is inversely proportional to T, the T value adjusts the importance of ΔS(system) in the gibbs free energy equation
  • at low temps, ΔS is high but T is low so overall TΔS(system) ≈ 0. Thus ΔG(system) = ΔH(system)
  • at high temps ΔH is negligible so ΔG(system) = TΔS(system)
49
Q

formula for ΔG(reaction)

A

ΣΔG(reaction) = ΣΔG(products) - ΣΔG(reactants)

50
Q

Relationship between ΔG(reaction) and equilibrium

A

as ΔG(reaction) becomes more negative (i.e. as reaction feasibility increases), the equilibrium position shifts more to the right

51
Q

linking ΔS to ΔG

A

ΔS(total) = ΔS(system) - ΔS(surroundings)
Knowing ΔS(surroundings) = - ΔH(system) / T, we can substitute that:
ΔS(total) = ΔS(system) - ΔH(system) / T
-TΔS(total) = ΔH - TΔS(system)
Knowing ΔG(system) = -TΔS(total), we can substitute that:
ΔG(system) = ΔH(system) - TΔS(system)

52
Q

enthalpy of atomization

A

ΔH(at)*

  • standard enthalpy change when 1 mol of gaseous atoms are formed from the element in its standard state under standard conditions
  • for diatomic molecules the ΔH(at) value is half that of the bond dissociation enthalpy

*standard enthalpies like ΔH(atom) are always written with a theta in superscript

53
Q

difference between bond dissociation enthalpy and bond enthalpy

A
  • bond dissociation enthalpy refers to the enthalpy of a specific bond in a specific compound
  • bond enthalpy refers to the average enthalpy of bond, taking into account different compounds
  • a specific bond’s energy differs somewhat between compounds, so bond dissociation enthalpy is more specific/accurate
54
Q

enthalpy of solution

A

ΔH(sol)

- enthalpy change when 1 mol of ionic substance dissolves in water to form a solution of infinite dilution

55
Q

Hess’ Law for the solubility of salts

A

ΔH(sol) = ΔH(lat) + sum of ΔH(hyd) of ALL constituents

NOTE: ΔH(sol) and ΔH(lat) are values of compounds while ΔH(hyd) is individual values of ions

56
Q

why can’t H2O be added to SO3?

A
  • taking SO3’s non-ionic nature into account, its ΔH(lat) value will be small
  • considering ΔH(sol) = ΔH(lat) + ΔH(hyd of ALL constituents), remember the fact that ΔH(hyd) is always negative (exothermic)
  • as ΔH(lat) will be small, the overall ΔH(sol) value for H2O + SO3 will be highly exothermic
  • the evolved heat could be high enough to cause the product H2SO4 to boil
57
Q

factors affecting ΔH(hyd)

A
  • ionic radius

- charge of ion

58
Q

factors affecting ΔH(hyd): ionic radius

A

smaller ionic radius = higher ΔH(hyd)

59
Q

factors affecting ΔH(hyd): charge of ion

A

more positive charge = higher ΔH(hyd)

- due to increased attraction between the positive ion and the partially negative O atoms in water

Chemistry IB HL (2016+) (13 decks)

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