Physical Chemistry

ΔH_at^⦵

is the enthalpy change when 1 mole of gaseous atoms is formed from its elements under 298K and 1atm

Mg_(s) → Mg_(g)

ΔH_at^⦵ type of energy change and reason

Always Endothermic; energy is required to break down bonds between elements

ΔH_latt^⦵

enthalpy change when 1 mole of ionic compound is formed from its ions in the gaseous state under 298K and 1atm

X^-_(g) + Y^-_(g) → X⁺Y^-_(s)

ΔH_latt^⦵ type of energy change and reason

Always Exothermic; energy is released when new bonds are made.

(more exothermic = stronger ionic bond)

ΔH_Ea1^⦵

is the enthalpy change when 1 mole of electrons is added to 1 mole of gaseous atoms to form 1 mole of gaseous 1- ions under 298K and 1 atm

Cl_(g) + e^- → Cl^-_(g)

ΔH Ea type of energy change and reason

Ea₁ :

• Exothermic; energy is released when new bonds are made

Ea₂ onwards :

• Endothermic; energy is required to overcome repulsion between electrons and negative ions

Factors affecting the magnitude of electron affinity

1. Nuclear charge :

   Nuclear charge ↑ = attractive forces between electron and nucleus ↑

2. Distance :

   Distance between nucleus and outermost shell ↑ = force of attraction ↓

3. Shielding effect of electron :

   Number of shells ↑ = Shielding effect ↑ = force of attraction ↓

Does Cl have a more exothermic EA₁ than S? and why?

Yes :

• Positive nuclear charge ↑

• Shielding effect =

• Distance ↓

• Attractive forces between nucleus and incoming electron ↑

• therefore, Cl is more exothermic

Trend of EA in group 16 and 17

Becomes less exothermic :

• Positive nuclear charge ↑

• Shielding effect ↑ and increase in distance outweighs ↑ in nuclear charge

• Forces of attraction between nucleus and incoming electron ↓

• so its magnitude decreases

Exceptions to the group 16 and 17 EA trend

Fluorine :

• Very small atomic radius

• Electron density is high

• More repulsion between incoming electrons and electrons already present at fluorine

• Less attractive force between nucleus and incoming electron

• It is less exothermic than expected

Order for born haber cycle

• ΔH_^f (start cycle)

• ΔH_latt (end cycle)

• ΔH_at (metal)

• IE (metal)

• ΔH_at (non-metal)

• EA₁

First Ionization energy

energy needed to remove one electron from each atom in one mole of atoms of the element in the gaseous state to form one mole of gaseous ions.

M(g) →M⁺(g) +e^-

ΔH_f^⦵

is the enthalpy change when one mole of a compound is formed from its elements under 298K and 1atm

Na(s) + (1/2)Cl₂ → NaCl(s)

Factors affecting ΔH_latt^⦵ and how

1. Size of the ions
   when the size of the ion ↑:
   -the charge density (charge/size) of the ion will ↓.
   -forces of attraction between the ions ↓.
   -lattice energy becomes less exothermic
   

2. Charge on the ions
   when the charge of the ion ↑:
   -the charge density of the ion will ↑
   -forces of attraction between the ions ↑
   -lattice energy becomes more exothermic

(Note : After stating if size/charge is the one changing mention change in charge density aswell)

Na⁺ has similar size with Ca²⁺, why is the lattice energy of NaCl less exothermic than CaCl₂

this is because CaCl₂ has a greater charge since Ca2+ has a greater charge density than Na+ which means forces of attraction between ions are greater, more energy will be released when the new bonds are formed between them.

Polarising power

Ability of the cation to attract electrons and hence distorts an anion.

The more polarised an ionic bond is, the greater covalent character, so an ionic bond is perfect if there is complete separation of charge

Factors affecting ion polarisation

1. Charge density of cation:
   Greater charge density means the polarising power of the cation is higher
   

2. Size and charge of anion:
   Greater size of anion means higher polarisability
   Charge of anion becoming more negative means higher polarisability

Trend of thermal stability of group 2 carbonates and nitrates

Since :

• Atomic number increases

• Cation size increases, charge remains constant so charge density decreases

• Polarising power of the cation decreases down the group. (stronger ionic bonds)

• Thermal stability will increase down the group

ΔH_sol^⦵

Energy when 1 mole of ionic solid dissolves in very sufficient amount of water to form a very dilute solution under 298K and 1atm

NaCl (s) + aq → Na⁺(aq) +Cl^-(aq) [ which is just NaCl(aq) ]

ΔH_sol^⦵ type of enthalpy

can be exothermic and endothermic ;

the more exothermic it is, the more likely it is to be soluble in water

(if its a positive value; then entropy is considered)

ΔH_hyd^⦵

the enthalpy change when 1 mole of a gaseous ion dissolves in sufficient water to form a very dilute solution under 298K and 1atm.

Ca²⁺(g) + aq → Ca²⁺(aq)

ΔH_hyd^⦵ type of enthalpy change

always exothermic because when an ion dissolves in water; new bonds are formed between ions and water molecules

ΔH_hyd^⦵ factors

1. Size

   1. Size increases = less exothermic

2. Charge

   1. Charge increases = more exothermic

Solubility trends for group 2 sulfates

• Solubility depends on the balance between ΔH_latt^⦵ and ΔH_hyd^⦵

• ΔH_latt^⦵ and ΔH_hyd^⦵ both become less exothermic due to increase in cation size and decrease in charge density.

• ΔH_hyd^⦵ will become less exothermic faster due to SO4 large size

• Enthalpy change in solution will therefore become less exothermic

• Solubility will decrease down the group
  

Solubility trends for group 2 hydroxides

• Solubility depends on the balance between ΔH_latt^⦵ and ΔH_hyd^⦵

• ΔH_latt^⦵ and ΔH_hyd^⦵ both become less exothermic due to increase in cation size and decrease in charge density.

• ΔH_latt^⦵ magnitude will decrease exothermic faster due to OH size

• Enthalpy change in solution will therefore become more exothermic

• Solubility will increase down the group

What does solubility depend on?

The balance between ΔH_latt^⦵ and ΔH_hyd^⦵

Calculation of ΔH_sol^⦵ from lattice energy and Enthalpy change of hydration

• correction : its -delta hlatt

Entropy and its unit

measure of the degree of disorder of a system

unit = Jmol^-1k^-1

A system is more stable when its energy is more spread out in a more disordered state.

Factors that contribute to increase in entropy

• Solids melting

• Temperature increase

• Liquids boiling

• Increase in number of gas molecules

• ionic solids dissolve in water

CaCO₃ has a greater entropy than CaO despite being in the same state,explain

• more atoms present in CaCO₃ (5) than CaO (2)

• More disorder in CaCO₃

Spontaneous/Feasible Reaction

a reaction that will occur naturally without external influence

how to calculate ∆S_sys

∆S_sys= ΣS_products-ΣS_reactants

∆G equation and units for each

∆G=∆H_reaction-T∆S_sys :

(∆G<0;spontaneous)

∆G=kJmol^-1
∆Hr = kJmol^-1
T = K
∆S_sys= Jk^-1mol^-1 (conversion to kJ → divide 1000)

minimum temperature where reaction is feasible (∆G=0)

0K (or -273.15 C → convert!!)

feasibility of a reaction can be altered by

Temperature change

Generally, exothermic reactions are (Spontaneous/Non-Spontaneous) and why?

Exothermic reactions are spontaneous in general :

• energy is released to the surroundings increasing its arrangements.

• which causes increase in entropy

Generally, endothermic reactions are (Spontaneous/Non-Spontaneous) and why?

Endothermic reactions are non-spontaneous in general :

• Energy is absorbed from the surrounding, decreasing its arrangements

• which causes decrease in entropy

(Inorganic Part) decomposition of group 2 nitrates - Magnesium nitrate eg

2Mg(NO₃)₂ (s)→2MgO(s) + 4NO₂(g) + O₂(g)

standard conditions

• pressure of 1atm (or 101kPa)

• temperature 298K

• ions at a concentration of 1.0moldm^-3

Standard electrode potential (E⁰)

emf of cell when a half-cell is connected to a standard hydrogen electrode under standard conditions

Use of salt bridge

• maintain charge balance

• complete the circuit

Criteria for a salt bridge

• does not react with either solutions in the half-cells

• no precipitate will form with any ions in contact with it

(KNO3 is usually used as a salt bridge)

standard hydrogen electrode (SHE) and features

reference half-cell used for measuring standard electrode potential.

Consists of :

• H₂ gas at 1atm and 298K

• platinum electrode

• 1moldm^-3 H⁺ (aq) ion

why is finely divided platinum electrode used in SHE

• hydrogen does not conduct electricity so platinum is used to conduct and is inert so that it doesn’t react with H+ ions

• finely divided to increase surface area for a faster reaction

A more positive E⁰ value implies that the ions are..

a greater oxidising agent

Standard cell potential (E_cell)

potential difference of a cell in a circuit between two half-cells under standard conditions

electron flow goes from…

more negative E0 to more positive E0 value

A reaction with a positive E_cell value is said to be..

feasible

why does the prediction of feasibility sometimes fail on positive E_cell value

• Reaction has high activation energy

• condition of the reaction are not standard

If a change causes the position of equilibrium to be shifted to the right, the value of E0 will…

increase

Nernst equation

electrolysis

breakdown of an ionic compound by passing electric current

selective discharge of an ion is affected by :-

• position of ion in electrochemical series

• concentration of ion

Cations are selectively discharged when..

• a more positive E0 (favours reduction more)

• higher concentrtion (increasing E0 value)

why does an ionic compound have to molten during electrolysis

so that ions are free to move

why is graphite a suitable material for an electrode (two reasons)

• conducts electricity

• inert

hydrogen will be selectively discharged at the cathode unless…

the ions are gold,silver or copper

O₂ will be selectively discharged at the anode unless…

the ions are halogens (it will discharge chlorine at a higher concentration)

formula of charge (Q)

Q=IT

1 Faraday

is the charge carried by one mole of electrons (96500C)

calculating 1 Faraday using avogadro constant

F=Le

L=avogadro constant

e = charge of one electron

Bronsted acid

proton donor

Bronsted base

proton acceptor

strong acid/base

one that completely dissociates/ionizes in water

weak acid/base

one that partially dissociates/ionizes in water

strong acid has a … conjugate base

weak

conjugate acid-base pairs

pair of reactants and products that are linked to each other by the transfer of a proton

K_a (acid dissociation constant) equation for
HA ⇌ H⁺+A^-

K_a= [H⁺][A^-] / [HA]

the greater the K_a

the stronger the acid

pK_a

pK_a = -log₁₀(K_a)

the lower the pK_a

the stronger the acid

pH

pH = -log₁₀[H⁺]

[H⁺]=10^-pH

K_w value at 25^oC and equation

K_w=[H⁺][OH^-]

K_w = 1.0×10^-14mol²dm^-6 at 25⁰C

explain why pH of a neutral solution is lower at higher temperature

• H₂O⇌H⁺+ OH^-

1. forward reaction is endothermic

2. so when temperature increases, equilibrium shifts to the right increasing concentrations of both H⁺ and OH^- , since [H⁺] increases, the pH will decrease

3. but that does not mean that it is more acidic at higher temperatures since concentration of H+ and OH- are the same

strong acid [H⁺] formula

[H⁺] = [HA(aq)] (HA = strong acid)

(since the acid gets fully dissociated - it is assumed that the [H⁺] formed due to water is negligible and so is not included)

strong base [H⁺] formula

from K_w :

[H⁺] = K_w / [OH^-]

(strong alkali have small amounts of H⁺ which are formed due to ionisation of water — we are dealing with aqueous solutions)

*remember to multiply concentration by x if molar ratio is 1:x

assumption made when deriving for pH of weak acids

1) ignore concentration of hydrogen ions produced by the ionisation of water

2) assuming the dissociation of weak acids is negligible (c-x approx = c)

calculation of [H⁺] for weak acids

[H⁺]=√(K_aC)

C=concentration of acid

buffer solution

a solution which resists change in pH when small amounts of acid or alkali are added

Acid buffer

made up of a weak acid and one of its salts

Alkaline buffer

made up of a weak base and one of its salts

what happens when an acid is added to the buffer

• (Assume dissociation is HA⇌H⁺ + A^-)

• [H⁺] added means equilibrium will shift to the left

• a large supply of A^- ensures that its concentration does not change rapidly (that supply comes from the salt — assuming its dissociation is NaA⇌Na⁺+ A^- )

• a large supply of HA ensures its concentration does not change rapidly

• so pH does not change significantly

pH of buffer solution

[H+] = Ka * [Acid] / [Salt]

pH = pKa + log( [Salt] / [Acid])

solubility

number of grams/moles of compound needed to saturate unit mass of water at a given temperature

K_sp

product of concentration of ions on a saturated solution of a sparingly soluble salt at 298K, raised to the power of their relative concentrations

common ion effect

reduction in the solubility of a dissolved salt achieved by adding a solution of a compound which has an ion common with the dissolved salt. which often results in precipitation

how to predict precipitation

if the product of the concentration is :

• less than K_sp — no precipitate will be formed

• more than K_sp — precipitate is formed

• equial to K_sp — precipitate is not formed yet (limiting equilibrium ts)

* it causes equilibrium to shift to the left to the side of the solid so more solid is precipitated

K_pc

ratio of concentration of solute in two immiscible solvents at equilibrium at a particular temperature.

factors affecting K_pc

• solubilities of the solute in two solvents

  • strength of imf between solute and solvents

    • polarity of the solute and solvents molecules

rate of reaction

rate of change in the concentration of a reactant or product per unit time

unit of rate (or rate of reaction)

moldm^-3s^-1

order of reaction (with respect to a reactant)

the power to which the concentration of that reactant is raised to in the rate equation

Zero order

rate = k[A]⁰

this means that the rate of reaction is independent on the concentration of A

Zero order :

rate-time graph

concentration of reactant-time graph

concentration of product-time graph

• concentration of product just increases with +ve gradient passing through origin

First order

rate=k[A]

rate of reaction is proportional to the concentration of A

First order :

rate-time graph

concentration of reactant-time graph

concentration of product-time graph

*product concentration against time is just an increasing graph with decreasing gradient

half life

time taken for the initial concentration of reactant to decrease to half its original value

half life of a first order reaction is…

constant

relationship between half-life and rate constant is given by

K = ln(2) / t_1/2

*only for first order