m3 and m5 - enthalpy and rates of reaction Flashcards
enthalpy def
(H)
the thermal energy stored in a chemical reaction
Ea def
(activation energy)
minimum energy required to start a reaction
by breaking bonds in the reactants
enthalpy diagrams
exo: —__ (giving energy to surroundings so atoms lose it)
endo: __— (taking in energy from surroundings)
add: curvy arrow for activation energy, arrow for delta H
standard conditions and standard states
conditions:
temp in K usually 298K
pressure 100kPa
solutions 1mol/dm3
states: physical state they would be at the standard conditions
enthalpy change of reaction (ΔHr)
enthalpy change between molar quantities of reactants, under standard conditions to give products in their standard states
enthalpy change of formation (ΔHf)
enthalpy change when one mole of a compound is formed from its elements under standard conditions, everything is in their standard states
enthalpy change of combustion ΔHc
enthalpy change when one mole of an element/compound reacts by complete combustion with excess oxygen under standard conditions
enthalpy change of neutralisation ΔHn
enthalpy change when 1 mole of water is formed from a neutralisation reaction
q = mcΔT
m = mass of water usually
c = specific heat capacity 4.18
ΔT = change in temp of water usually
average bond enthalpy def
breaking of 1 mol of bonds in gaseous atoms
exothermic reactions in terms of bond energies (bendomexo)
more energy required to form the new bonds, than to break existing bonds
is breaking bonds endo or exo
endo
BENDOMEXO
bond enthalpy changes
sum of reactant bond enthalpy - sum of product bond enthalpy
(left hand side - right hand side)
Hess’ law def
the total enthalpy change of a reaction is independent of the route taken
(same enthalpy change of a reaction even if it’s done in different steps basically)
hess law in a triangle diagram
A —ΔH1—> B
\ /
ΔH2 ΔH3
\ /
C
ΔH1 = ΔH2 + ΔH3
USING enthalpies of combustion: calculate ΔHf C3H8 using ΔHc data
3C (s) + 4H2 (g) -> C3H8 (g)
\ /
3xΔΗc of C | 4xΔΗc of H2 ΔΗc of C3H8
\ (arrow down) / (arrow down)
CO2 (g) + H2O (l)
ΔΗf = 3ΔΗcC + 4ΔΗcH2 - ΔΗcC3H8
USING enthalpies of formation: calculate ΔΗr using ΔΗf data
ZnCO3 —ΔΗr—> ZnO + CO2
\ (arrow up) / /
ΔHf ZnCO3 ΔHf ZnO | ΔΗf CO2
\ / /
[ Zn (s) + C (s) + O2 (g) ]
ΔΗr = -ΔΗf of ZnCO3 + ΔΗf of ZnO + ΔHf of CO2
= +812 - 348 - 393 = +71kJmol-1 (endothermic)
collision theory
for a reaction to occur particles must:
- collide with enough energy to react (above Ea)
- be in the correct orientation
how does temp effect ROR
increased kinetic energy so increased movement, more successful collisions per second, more likely to have sufficient activation energy
how does concentration/pressure affect ROR
increases, particles are closer together, so frequency of collisions increase
how does surface area affect ROR
increases, more points of contact so frequency of collisions increase
how do catalysts affect ROR
offers alternative pathway of lower activation energy
used to reduce energy demand from combustion of fossil fuels, therefore reduced CO2 emissions
boltzmann distribution graphs
at end: line will come close to the x axis but never touch it
what is the boltzmann distribution (def)
the distribution of energies of molecules at a particular temperature
effect on graph of higher temp and catalysts
higher temp: peak is lower and line moves to the right
catalysts: Ea moves to the left
homogenous vs heterogenous catalysts
homo: catalysts in same PHASE as reactants
hetero: different phases
why do people use catalysts
lower energy demands, reducing costs and environmental impact
burning less fossil fuels, producing less CO2
rate of reaction def
change in conc of reactant or product / unit time
rate constant =
k
half life def
time taken for the conc of a reactant to reduce by half
half life time intervals: zero order, first order, second order (inc or dec)
zero order => half life decreases
first order => half life remains constant
second order => half life increases
first order reaction we can determine rate constant ‘k’ from the constant half life (t1/2) using:
k = ln2
—— s-1
t1/2
ln = calculator button (natural log)
rate determining step is
the slowest step in a multi stage reaction
lattice enthalpy def
enthalpy change when 1 MOLE of a solid ionic lattice is formed from its GASEOUS ions under STANDARD conditions (1atm, 298K)
lattice enthalpy: -500 is bigger than -100 so the bond strength is
stronger
arrhenius equation compared to y = mx + c
y = ln k
mx = -Ea/RT
x = 1/T so to find activation energy you do: - gradient x R (always needs to be a positive value)
c = + ln A
what 2 things affect lattice enthalpy and how
charge and ionic raidius
same charge but smaller the atomic radius, then higher electron density so more exothermic
same size but higher charge, then higher charge density so more exothermic
enthalpy of atomisation def
1 mole of GASEOUS atoms forms from the ELEMENTS in their standard states
first electron affinity def
enthalpy change when 1 electron is added to each atom in 1 MOLE of gaseous atoms to form 1 mole of gaseous 1- ions
(opposite to first ionisation energy)
born haber cycle info
use bendomexo, if endothermic so bond breaking (DeltaH atomisation, ionisation energy) arrow up.
exothermic bond making (e- affinity, lattice enthalpy) arrow down.
equations: DeltaH formation = all the other numbers
is SECOND electron affinity endo or exo
endo, as the repulsion between -ve ion and -ve electron has to be overcome
enthalpy of solution def
enthalpy change when 1MOLE of a solute is COMPLETELY dissolved in water under standard conditions
enthalpy of hydration def
enthalpy change when 1MOLE of aqueous ions are formed from their GASEOUS ions under standard conditions
entropy def
a measure of dispersal of energy in a system, greater the more disordered a system is
ΔS
entropy units
J K-1 mol-1
entropy increases as temp increases because
particles gain energy and move faster apart (less ordered)
change in entropy when there’s an increase in number of gaseous molecules
entropy increases (gas is most disordered)
entropy calculation
ΔS = ΣΔ products — ΣΔ reactants
when disorder increases, ΔS…
increases and is positive
free energy topic - we say a reaction is feasible or spontaneous if
chemical system becomes more stable and overall energy decreases
overall energy is Gibbs free-energy ΔG
ΔG equation free energy
ΔG = ΔH - TΔS
t = temp in Kelvin
kelvin’s cancel out, and have to divide ΔS by 1000 so it’s also in kJmol-1
for a reaction to be feasible G must be…
negative (as overall energy decreases so more stable)
if they ask you to calculate temp at which a reaction is feasible
make ΔG = 0 so…
ΔG = ΔH - TΔS
0 = ΔH - TΔS
TΔS = ΔH
T = ΔH / ΔS
if asked to find minimum temp a reaction is feasible…
make ΔG = 0
in the ΔG = ΔH - TΔS
rearranged =
T = ΔH / ΔS
