Thermochemistry Flashcards
Define enthalpy change
ΔH of a rxn is amt energy absorbed or released in chem rxn wrt no of mol of rxt or pdt specified
ΔH = sum of Hpdt - sum of Hrxt
where
Hpdt is enthalpies of pdt,
Hrxt is enthalpies of rxt
Unit: kJ mol -1
Explain endothermic and exothermic reactions
- Endo
- rxn absorb energy fr surrounding
- temp of surrounding decrease
- sum Hpdt > sum Hrxt (ΔH>0)
- pdt r thermally LESS stable than rxt
- energy profile diagram shows increase - Exo
- rxn release energy to surrounding
- temp of surrounding increase
- sum Hpdt < sum Hrxt (ΔH<0)
- pdt r thermally MORE stable than rxt
- energy profile diagram show decrease
What to take note about thermochemical equations?
- MUST write state symbol (aq,s,l,g) for ALL eqn
- MUST write sign (to know if exo or endo)
- MUST write ΔH > or < 0
What are conditions affecting enthalpy change?
- amt substance (eg if twice, twice enthalpy change)
- state of rxt & pdt (change state can release/absorb energy)
- temp & Pa (usually standard condit n; if changed, enthalpy change affected)
Define standard enthalpy change of reaction, ∆Hϴ
∆Hϴ or ∆Hrϴ is amt energy absorbed or released in chem rxn when molar qty stated in chemical eqn react under standard condit n 298K, 1 bar
What does standard conditions mean?
(denoted by super script ϴ)
- temp = 298K = 25 deg C
- Pa = 1 bar = 1.0e5 Pa
- sol n at conc of 1 mol dm-3
Define standard enthalpy change of combustion, ∆Hcϴ
∆Hϴ of substance is energy RELEASED when 1 mol of substance completely burnt in oxygen under standard condit n 298K, 1 bar
NOTE:
- always exo
- water in eqn is liquid, NOT gas at standard condt n, and CH4 is gas, C8H18 is liquid
- values of ∆Hcϴ aka energy value of fuel; more exo the ∆Hcϴ, the larger amt heat liberated upon complete combust n fr given amt fuel
Define standard enthalpy change of neutralisation, ∆Hnϴ
∆Hϴ is energy RELEASED when acid & base react form ONE mol water under standard conditions 298K, 1 bar
NOTE:
∆Hnϴ always exo, same value for strong acid, strong base rxn
Why is enthalpy change of neutralisation involving weak acids or bases is lower than -57.3kJ mol-1
- weak acid & bases oni slightly dissociated in aqueous sol n
- some of energy evolved fr neutral n process is used to further dissociate weak acid/base completely
=> enthalpy change of neutral n involving weak acids, bases is less exo than that btw strong acid, strong base
Define standard enthalpy change of formation, ∆Hfϴ
∆Hfϴ of substance is energy CHANGE when 1 mol of substance is formed fr its element under standard condit n 298K, 1 bar
NOTE:
- ∆Hfϴ of elements is ZERO
- Graphite is used as basis for carbon since it is the most stable form of C vs other form eg diamond
- ∆Hfϴ is measure of stability relative to element (the more exo value, greater stability relative to element - less energy content)
Define standard enthalpy change of atomisation of element, ∆Hatϴ
∆Hatϴ of element is energy ABSORBED when one mole gaseous atoms is formed fr element in its standard state under standard condit n 298K, 1 bar
Define standard enthalpy change of atomisation of compound, ∆Hatϴ
∆Hatϴ of compound is energy ABSORBED when one mole of compound in a given state is being broken down into its constituent gaseous atoms under standard condit n 298K, 1 bar
Define bond energy (BE)
of covalent bond is avg energy absorbed to break 1 mol of COVALENT bond in gas phase into constituent gaseous atom under standard condit n 298K, 1 bar
NOTE:
BE of homonuclear diatomic molecule (ie X-X) is twice of ΔHatθ of constituent element X2 (so, usually BE found in databooklet)
Eg
BE (Cl-Cl) = 2ΔHatθ(Cl)
Define First electron affinity (1st EA)
of element is energy released when 1 mol gaseous atom gain 1 mol e- form 1 mol singly charged negative gaseous ions
Note:
Always exo, unlike 2nd EA
What to take note about First and Second Ionisation Energy?
- always endo
Define Second electron affinity (2nd EA)
of element is energy ABSORBED when 1 mol singly charged negative gaseous ion gain 1 mol e- form 1 mol doubly charged negative gaseous ions
Note:
Always endo bcos e- added to -ve ion (energy needed overcome repuls n)
Define lattice energy ΔHlattθ
ΔHlattθ of ionic cpd is energy released when 1 mol of ionic cpd is formed fr constituent gaseous ion under standard condit n 298K, 1 bar
Define Standard Enthalpy change of hydration ΔHhydtθ
ΔHhydθ of an ion is energy released when 1 mol gaseous ions is hydrated under standard condit n 298K, 1 bar
Define standard enthalpy change of solution ΔHsolθ
ΔHsolθ of substance is energy CHANGE when 1 mol of substance is completely dissolved in a solvent form infinitely dilute sol n under standard condit n 298K, 1 bar
Note:
Can be exo or endo depending on chemical properties
Define standard enthalpy change of fusion ΔHfusθ
ΔHfusθ of substance is energy absorbed when 1 mol of substance is converted fr solid to liquid state w/o change in temp at 1 bar
Define standard enthalpy change of vaporisation ΔHvapθ
ΔHvapθ of substance is energy absorbed when 1 mol of substance is converted fr liquid to gaseous state w/o change temp at 1 bar
Note:
Energy required to overcome intermolecular attract n in liquid state to vaporise => endo
Define standard enthalpy change of sublimation ΔHsubθ
ΔHsubθ of substance is energy absorbed when 1 mol substance is converted fr solid to gaseous state w/o change temp at 1 bar
Give formula for temperature change ΔT
ΔT = Highest/lowest temp - initial temp
ΔT>0 indicate temp increase
ΔT<0 indicate temp decrease
How to measure temperature change?
- directly recording initial and highest/lowest temp reached
- monitor temp over time, plot temp-time graph & extrapolate to find max/min temp (for exo & endo rxn respectively)
Give formula for heat absorbed/released by solution Qsol
Qsol = mcΔT = VcΔT (Only magnitude)
where
m is mass of sol n in cup (g)/V is vol of sol n (cm3) (when density assumed 1g cm-3),
c is specific heat capacity of sol n (when assumed, 4.18) (J g-1 K-1),
ΔT is temp change (K)
Give formula for heat absorbed/released by reaction Qrxn
- Qrxn = Qsol (assume no heat loss; 100% efficiency)
- Qrxn = (100/x)Qsol
where
x% is efficiency of heat transfer
Give formula for enthalpy change ΔH
ΔH = +/- (Qrxn/n) x N
where
Qrxn is heat released/absorbed,
n is amount of limiting reagent,
N is stoichiometric coeff of limiting reagent in eqn (usually 1)
*MUST HAVE SIGN
What to take note about calculating enthalpy change of reaction ΔH
- all enthalpy changes of rxn MUST hv correct sign
- reversing chem eqn oso reverse sign of ΔHθ
- multiply chem eqn by factor oso multiply ΔH by same factor
- ΔH can be calculated theoretically or any following methods (if can):
1 Hess’ Law
2. Standard enthalpy change of format n
3. Standard enthalpy change of combust n
4. Bond energy
5. Algebraic method
Define Hess’ law
Enthalpy change for chem rxn is same regardless of route by which chem change occur, provided initial state of rxt & final state of pdt r same
How to calculate enthalpy changes with Hess’ Law?
By Hess’ Law,
dH1 = dH2 + dH3 + dH4
(sum of dH in clockwise direct n = sum of dH in anti-clockwise direct n)
*enthalpy change is path-independent (regardless of route)
What are the general steps for using Hess’ Law
- construct balanced chem eqn for rxn to find unknown (aim eqn)
- identify thermochem eqn for all given ΔH info in qn
- form energy cycle (include state symbol, species along arrow, balance species, include appropriate ΔH notat n, values)
- apply Hess’ Law solve for unknown ΔH
Give formula for standard enthalpy change of formation. When can this formula be used?
ΔHr = sum ΔHθf (pdt) - sum ΔHθf (rxt)
*use oni when given values of enthalpy change of form n of ALL rxt, pdt
When using formation or combustion formulae, what to take note?
- ΔHr is not always the unknown to find
Give formula for standard enthalpy change of combustion. When can this formula be used?
ΔHr = sum ΔHθc (rxt) - sum ΔHθc (pdt)
*use oni when given enthalpy change of combust n of ALL rxt, pdt
NOTE:
- sum of RXT - PDT, not other way ard like for form n
- same eqn may sometimes represent more than 1 type of enthalpy change (but same process)
eg
C(s) + O2(g) –> CO2(g)
ΔHθf (CO2) = ΔHθc (C)
Give formula for bond energy method
ΔHr = sum BE(rxt) - Sum BE (pdt)
NOTE:
- energy absorbed in rxt => (+ve sign)
- energy released in bond form n in pdt => (-ve sign)
Enthalpy change of reaction that is calculated differs slightly from actual value. Suggest why
BE calculat n is approximat n method as BE values given in data book r avg value, not specific to cpd in rxn
What are the steps to find enthalpy change using algebraic method
- construct balanced chem eqn (aim eqn)
- write thermochem eqn for all given enthalpy change in qn (optional if u clear of eqn involved)
- manipulate given eqn so rxt in aim eqn appear on left side, pdt appear on right side (ensure balanced eqn)
- add eqn so same substance on both side cancel out
- enthalpy change of rxn = sum given enthalpy changes of rxn added
How does energy diagram show endothermic or exothermic?
- Upwards arrow mean energy absorbed (endo)
- downward arrow mean energy released (exo)
What to take note about Born-Haber Cycle
- apply ONI to ionic cpd
- use upward arrow for endo rxn (+), downward arrow for exo rxn (-)
- follow this sequence in making cycle:
F orm n (exo)
A tom n of metal then non-metal (endo)
I on fr IE (metal) then EA (non-metal) (2nd EA is endo, larger mag than 1st EA, which is exo)
L attice energy (exo) - use ruler, ensure arrows are rather proportional in mag
Why is theoretical lattice energy for eg FeO different from value obtained using Born-haber cycle?
Fe2+ cation hv high charge density, so polarise O2- anion e- cloud, induce partial covalent character. Theoretical value no consider partial covalent character of FeO
NOTE:
- calculat n using born-haber cycle involve expt values, presense of partial cov character in ionic cpd is accounted for
- for theoretical values (fr use of | ΔHlatt | ∝ (q+)(q-)/(r+ + r-) , oni pure ionic lattice structure considered, not partial cov character usually
Give formula for standard enthalpy change of solution. What can be deduced?
fr Hess’ Law,
ΔHθsol = sum ΔHθhyd - ΔHθlatt
- more exo, more soluble salt (unless data suggest otherwise)
Deduct n:
- salt likely soluble if ΔHθsol < 0, bcos enough hydrat n energy released to compensate for lattice energy required to break down ionic cpd
- salt NOT likely soluble if ΔHθsol > 0, bcos insuff hydrat n energy released compensate for lattice energy needed break down ionic cpd
NOTE: some salts still soluble w endo ΔHθsol of small mag (due to entropy)
