Enthalpy change (ΔH) Flashcards
Define (and apply the term) ‘enthalpy of formation’
The standard enthalpy of formation is the enthalpy change when one mole of a substance is formed from its constituent elements under standard conditions (100kPa, 298K), with all reactants and products in their standard states.
Always negative: -↓ (gives out heat; exothermic)
E.g.
H2(g) + 1/2O2(g) → H2O(l) Δ = -286 kJ mol -1
Define (and apply the term) ‘enthalpy of atomisation’
The standard enthalpy of atomisation (ΔHato) is the enthalpy change when 1 mole of gaseous atoms is formed from an element in its standard state, under standard conditions.
Always positive: +↑ (takes in heat; endothermic)
E.g:
Mg(s) → Mg(g) ΔHato = +147.7 kJ mol-1<br></br> 1/2Br2 → Br(g) ΔHato= +111.9 kJ mol-1<br></br> 1/2Cl2 → Cl(g) ΔHato= +111.9 kJ mol-1
N.B. enthalpy of atomisation is given per mole of chlorine or bromtine atoms, and not per mole of chlorine or bromine molecules.
Define (and apply the term) ‘first ionisation enthalpy/first IE’
The first ionisation enthalpy/energy (Δie1) is the enthalpy change when 1 mole of gaseous atoms is converted into 1 mole of gaseous ions each with a single positive charge (the amount of energy required to remove an electron from the atom or molecule in the gaseous state).
Always positive: +↑ (takes in heat; endothermic)
X → X+ + e-
E.g.
Na(g) → Na+(g) + e- first IE = +496 kJmol-1
N.B. The second ionisation energy (second IE) referes to the loss of a mole of electrons from a mole of positively charged ions.
X+ → X2+ + e-
E.g.
Na+(g) → Na2+(g) + e- second IE = +4563 kJ mol-1
Define (and apply the term) ‘first electron affinity’
The first electron affinity (EA) ΔHoea is the standard enthalpy change when a mole of gaseous atoms is converted into a mole of gaseous ions, each with a single negative charge.
(when 1 mole of gaseous 1- ions is made from 1 mole of gaseous atoms)
Always negative: -↓ (gives out heat; exothermic)
E.g.
O(g) + e- → O-(g) ΔHoea = -141.1 kJmol-1
N.B. this refers to single atoms, not to diatomic molecules, or Oxygen molcules, O2 etc.
Define (and apply the term) ‘second electron affinity’
The second electron affinity (EA) ΔHoea is the enthalpy change when 1 mole of electrons is added to 1 mole of gaseous ions each with a single negative to charge to form ions with two negative charges.
(enthalpy change when 1 mole of gaseous 2- ions is made from 1 mole of gaseous 1- ions)
Always negative: -↓ (gives out heat; exothermic)
E.g.
O-(g) + e- → O2-(g) ΔHoea = +798 kJ mol-1
Define (and apply the term) ‘lattice formation enthalpy’
Lattice formation enthalpy ΔHoL is the standard enthalpy change when 1 mole of solid (possibly crystalline) ionic compound is formed from its (scattered) gaseous ions.
Lattice formation enthalpies are always negative; -↓
(gives out heat; exothermic)
E.g.
Na+(g) + Cl-(g) → NaCl(s) ΔoL= -788 kJ mol-1
It’s the amount of energy released when a lattice is formed from its scattered gaseous ions.
Defien (and apply the term) ‘lattice dissociation enthalpy’
The lattice dissociation enthalpy is the standard enthalpy change when 1 mole of ionic compound (solid crystal) is converted/dissociated (completely) into its (scattered) gaseous ions.
Lattice dissociation enthalpies are always positive. +↑
(takes in heat; endothermic)
E.g.
NaCl → Na+(g) + Cl-(g) ΔoL = +788 kJ mol-1
It’s the amount of energy needed to split up a lattice into its scattered gaseous ions.
Defien (and apply the term) ‘enthalpy of hydration’
The enthalpy of hydration is the standard enthalpy change when 1 mole of aqueous ions is formed from gaseous ions (when water molecules surround 1 mole of gaseous ions).
Always negative: -↓
Na+ + aq → Na+(aq) ΔHohyd = -406 kJ mol-1
Cl- + aq → Cl-(aq) ΔHohyd = -363 kJ mol-1
It is the heat energy released when new bonds are made between the ions and water molecules.
You can think of an imaginary process where the crystal lattice is first broken up into its separate gaseous ions, and then those ions have water molecules wrapped around them. That is how they exist in the final solution.
Defien (and apply the term) ‘enthalpy of solution’
The enthalpy of solution (ΔHosol) is the standard enthalpy change when 1 mole of solute dissolves completely in sufficient solvent to form a solution in which the molecules or ions are far enough apart not to interact with each other; where no further enthalpy change occurs on further dilution.
Enthalpies of solution may be either positive or negative - in other words, some ionic substances dissolved endothermically (for example, NaCl); others dissolve exothermically (for example NaOH).
E.g.
NaCl(s) + aq → Na+(aq) + Cl-(aq) ΔHosol = +19 kJ mol-1
The enthalpy change of solution is the enthalpy change when 1 mole of an ionic substance dissolves in water to give a solution of infinite dilution.
Defien (and apply the term) ‘mean bond enthalpy’
Mean bond enthalpy is the enthalpy change when 1 mole of gaseous molecules each breaks a covalent bond to form 2 free radicals, averaged over a range of compounds.
E.g.
CH4(g) → C(g) + 4H(g) ΔHodiss = +1664 kJ mol-1
Therefore the mean (or average) C-H bond enthalpy in methane is 1664/4 = +416 kJ mol-1
How do you construct a Born-Haber cycle?
Construct a Born-Haber cycle for the formation of NaCl, and calculate the lattice enthalpy.
- (Elements in their standard states; energy zero of the diagram)
Enthalpy of atomisation; the reaction involves soild Na, not gaseous, and Cl molecules; not seperate atoms.
Na(s) → Na (g)
ΔHoat = +108 kJ mol-1
1/2Cl2(g) → Cl(g)
ΔHoat = +122 kJ mol-1
∴ Na(s) + 1/2Cl2(g)
↑ ΔHoat (Na) = +108
∴ Na(g) + 1/2Cl2(g)
↑ ΔHoat (Cl) = +122
- First IE; the gaseous Na atoms must give up an electron to form gaseous Na+ ions:
Na(g) → Na+(g) + e-
First IE = +496 kJ mol-1
∴ Na(g) + Cl(g)
↑ First IE (Na) = +496
- First electron affinity; the Cl atoms must gain an electron (e-) to form gaseous Cl- ions:
Cl(g) + e- → Cl-(g)
First EA = -349 kJ mol-1
∴ Na+(g) + e- + Cl(g) *(at the top)
↓* First EA (Cl) = -349
- Entalpy of formation; add in the enthalpy of formation of NaCl; the direct route from zero energy to the exothermic floor at the start, below the atomisation.
Na(s) + 1/2Cl2(g) → NaCl(s)
ΔHof = -411 kJ mol-1
∴ NaCl (at the bottom exothermic line, ionic lattice)
↓ ΔHof (Na+ + Cl-) = -411
- Lattice formation enthalpy; add below electron affinity; when positively charged ions come together with negatively charged ions, they form a solid crystalline lattice and energy is given out due to attraction between oppositely charged ions; NaCl exists as a solid lattice at RTP, not as seperate gaseous ions.
Na+(g) + Cl-(g) → NaCl(s)
ΔHoL = -788 kJ mol-1
∴ Na+(g) + Cl-(g)
↓ ΔHoL (NaCl) = -788
Lattice formation enthalpy of -788 kJ mol-1 calculated via Hess’ Law; going either way from the line of first electron affinity; going opposite the arrow reverses the sign.
How do Born-Haber cycles of Group 2 Elements differ?
(e.g. for MgCl2)
- Two chlorines are involved; all the quantities related to Cl are doubled, i.e.:
- 2 x Δoat *(2 lots of atomisation of Cl; 2 moles, just double the figure)
- 2 x First EA* (2 lots of first EAs; not first + second electron affinities)
- Mg exists as Mg2+ (group 2); need a First IE AND a Second IE, one on top of each other, but only one atomisation enthalpy.
How do lattice enthalpies form Born-Haber cycles differ with lattice enthalpies calculated from calculations based on a perfect ionic model, and why?
- A theoretical lattice enthalpy is worked out on calculations based on the purely ioninc model of a lattice; which assumes that all ions are spherical, and have their charge evenly distributed around them.
- Experimental lattice enthalpy from the Born-Haber cycle is usually different; this is evidence that ionic compounds usually have some covalent character.
- The positive and negative ions in a lattice aren’t usually exactly spherical; positive ions polairse nerighbouring negative ions to different extents, and the more polarisation there is, the more covalent the bond will be.
- E.g. ZnSe; the Zn2+ ion is relatively small and has a high positive charge, while Se2- is relatively large and has a high negative charge.
The small Zn2+ can approach closely to the electron colouds of the Se2- and distort them by attacting them towards it. Se2- is fairly easy to distort as its large size means the electrons are far from the nucleus, its double charge meaning there is plenty of negative charge to distort.
This distortion means there are more electrons than expected concenrated between the Zn and Se nuclei, representing a degree of covalency (electron sharing) which accounds for ther lattice enthalpy discrepancy.
The Se2- ion is said to be polarised.
- Factors increasing polarisation:
- positive ion (cation): small size, high charge.
- negative ion (anion): large size, high charge.
- The greater the difference in experimental figures, the greater the covalent character; more polarisation leading to stronger bonding than the ionic model predicts.
What occurs when a solid ionic lattice dissolves in water?
Two things occur:
- The bonds between the ions break - this is endothermic. The enthalpy change is the lattice dissociation enthalpy.
- Bonds between the ions and the water are made - this is exothermic. The enthalpy change here is the enthalpy of hydration.
- The enthalpy change of solution is the overall effect of the enthalpy of the above.
What is the enthalpy of solution for NaCl?
Lattice dissociation enthalpy = +787
Enthalpy of hydration (Na+) = -406
Enthalpy of hydration (Cl-) = -364
You need to construct an enthalpy cycle; to construct one, you need the lattice dissociation enthalpy and the enthalpy of hydration for the ions.
- NaCl(s) → Na+(aq) + Cl-(aq)
ΔHosol
↘ ΔHoL = +787 ΔHohyd Na+= -406 ↗
ΔHohyd Cl- = -364
Na<sup>+</sup>(g) + Cl<sup>-</sup>(g)
Ionic lattice (NaCl) connected to gaseous ions by lattice enthalpy of dissociation. (if a negative value of lattice enthalpy is given, that’s the lattice enthalpy of formation. You want the reverse; flip the sign)
Gaseous ions are joined to the dissolved ions by respective enthalpies of hydration for each ion.
- Use Hess’s Law to work out ΔHosol:
+787 + (-406 + -364) = +17 kJ mol-1 - The enthalpy change of solution is slightly endothermic, but this is compensated for by a small increase in entropy, so NaCl still dissolves in water.
If an enthalpy of solution is much more endothermic, there may not be a great enough increase in entropy to compensate, thus it is insoluble; e.g. AgCl.
What’s the enthalpy of solution for AgCl?
ΔoL = +905
ΔHohyd (Ag+) = -464
ΔHohyd (Cl-) = -364
ΔHosol = +905 + (-464 + -364) = +77 kJ mol-1.
How do you use mean bond enthalpies to calculate an approximate ΔH value for other reactions?
- ΔH = ∑ΔH (reactants) - ∑ΔH (products)
- Enthalpy change for a reaction = sum of enthalpies of bonds broken - sum of enthalpies of bonds formed.
- Exothermic = more energy released than taken in = negative ΔH.
- Endothermic = less energy released than taken in = positve ΔH.
Why do values of mean bond enthalpy calculations differ from those determined from enthalpy cycles?
- The bond enthalpy for a bond (e.g. N-H bond) is not exactly the same for every bond (not every N-H bond is the same value).
- A given type of bond will vary in strength for compound to compound and can even vary within a compound.
- Mean bond enthalpies are the averages of these bond enthalpies.
- Only the bond enthalpies of diatomic molecules such as H2 and HCl will always be the same.
- Hence calculations from mean bond enthalpies will never be perfectly accurate. You get much more exact results from experimental data obtained from the specific compounds.
Why is ΔH alone insufficient to explain spontaneous change (reactions)?
- Many reactions that occur of their own accord tend to be exothermic, where ΔH is negative.
- Negative ΔH is a factor in whether a reactopm is spontaenous, however it does not explain why a number of endothermic reactions are spontaneous.
- Spontaneous endothermic reactions are the reason why ΔH alone is insufficient.
So why do reactions occur? What is it? Under what circumstances will a reaction take place?
- Due to the entropy ΔS of the system; the disorder/randomness of a system.
- If the products of a reaction has particles that are more spread out and random (i.e. have a greater entropy) than the reactants, then the reaction is likely to be spontaneous.
- Hence endothermic reactions can be spontaneous due to an increase in entropy.
(exothermic reactions thought to be spontaneous as you need to supply energy for endothermic reactions to occur, but such an increase in entropy could cause it to occur)
E.g.
- Water evaportates at room temperature; it’s endothermic (needs energy to break bonds) but because of its changing state (from liquid to gas), the entropy increases.
- Dissolution; when sodium hydrogencarbonates are mixed with hydrochloric acid, there is an increase in entropy and CO2 is evolved
(a spontaneous reaction)
What factors affect entropy?
- Physical state; solid particles wobble around a fixed point, with hardly any randomness, hence having the lowest entropy.
Gas particles whizz around wherever the fuck they want; they have the most random arrangements of particles, hence the highest entropy.
- Dissolving; dissolving a solid increases entropy; dissolved particles move freely as they are no longer held in one place.
- More particles; more particles means more entropy; the more particles you have, the more ways they and their energy can be arranged; if the no. of moles of products is greater than the reactants, entropy increases.
How do you calculate entropy change, ΔS?
ΔS = ∑ΔS (products) - ∑ΔS (reactants)
What is the equation that governs the feasibility of a reaction ocurring via two main factors?
Gibbs free-energy change, ΔG
This combines the entropy and the enthalpy change to govern the feasibility of a chemical reaction.
ΔG = ΔH - TΔS
ΔG = free energy change, J mol-1
ΔH = enthalpy change, J mol-1 (remember to convert)
T = temperature, K
ΔS = entropy change, J K-1 mol-1
A reaction is deemed feasible if ΔG is negative or equal to zero.
How does temperature affect the feasibility of a reaction, ΔG?
ΔG depends on temperature due to the therm TΔS; thus some reactions may be feasible at one temperature, but not at aother.
Thus an endothermic reaction can become feasible when temperature is increased if there is a large enough positive entropy change; a positive entropy change makes ΔG more negative due to - TΔS.
How do you calculate the temperature of a reaction that is just feasible?
When ΔG = 0, a reaction is just feasible.
- ΔG = ΔH - TΔS; substitute ΔG = 0.
- Rearrange for T.
T = ΔH
ΔS
ΔH = enthalpy change, J mol-1 (remember to convert)
T = temperature, K
ΔS = entropy change, J K-1 mol-1
