m3 and m5 - enthalpy and rates of reaction Flashcards by Jemima Dixon

enthalpy def
(H)

the thermal energy stored in a chemical reaction

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Ea def
(activation energy)

minimum energy required to start a reaction
by breaking bonds in the reactants

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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

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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

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enthalpy change of reaction (ΔHr)

enthalpy change between molar quantities of reactants, under standard conditions to give products in their standard states

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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

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enthalpy change of combustion ΔHc

enthalpy change when one mole of an element/compound reacts by complete combustion with excess oxygen under standard conditions

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enthalpy change of neutralisation ΔHn

enthalpy change when 1 mole of water is formed from a neutralisation reaction

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q = mcΔT

m = mass of water usually
c = specific heat capacity 4.18
ΔT = change in temp of water usually

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average bond enthalpy def

breaking of 1 mol of bonds in gaseous atoms

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exothermic reactions in terms of bond energies (bendomexo)

more energy required to form the new bonds, than to break existing bonds

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is breaking bonds endo or exo

endo
BENDOMEXO

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bond enthalpy changes

sum of reactant bond enthalpy - sum of product bond enthalpy

(left hand side - right hand side)

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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)

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hess law in a triangle diagram

A —ΔH1—> B
\ /
ΔH2 ΔH3
\ /
C

ΔH1 = ΔH2 + ΔH3

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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

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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)

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collision theory

for a reaction to occur particles must:
- collide with enough energy to react (above Ea)
- be in the correct orientation

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how does temp effect ROR

increased kinetic energy so increased movement, more successful collisions per second, more likely to have sufficient activation energy

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how does concentration/pressure affect ROR

increases, particles are closer together, so frequency of collisions increase

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how does surface area affect ROR

increases, more points of contact so frequency of collisions increase

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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 =

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