4 STATES OF MATTER + topic 2b [must REVIEW; only online] - Ideal and Real Gases Qualitative Analysis Chemical Energetics Enthalpy Change Hess’s Law

Question 1

gases in container … (2)

Answer 1

-exert a pressure

• as the gas molecules are constantly colliding with the wall of the container

Question 2

changing VOLUME (3)

Answer 2

-decreasing volume (at constant temp) => the molecules are squashed together => more frequent collisions with the container wall

• so pressure of the gas increases

Question 3

relationship & graph

btwn volume & pressure.

Answer 3

-so volume is inversely proportional to pressure (at constant temperature)

• graph of the volume of gas plotted against 1/pressure gives a straight line
  ([1 ÷ pressure] directly proportional to volume)

Question 4

changing temp (3)

Answer 4

-increasing temp (at constant volume) causes molecules to gain more kinetic energy => particles will move faster and collide with container walls more frequently

• pressure of the gas increases

Question 5

relationship btwn temp & volume of gas

Answer 5

-so temperature is directly proportional to the pressure (at constant volume)

• graph of the temperature of gas plotted against pressure gives a straight line

Question 6

Kinetic theory of gases states what?

Answer 6

molecules in gases are constantly moving

Question 7

what assumptions does the kinetic theory of gases make? [5]

Answer 7

• gas molecules are MOVING very FAST and RANDOMLY

-molecules HARDLY have any VOLUME

-gas molecules do NOT attract or repel each other (no intermolecular forces, no IMF)

-No kinetic energy is lost when the gas molecules collide with each other (ELASTIC collisions)

-temperature of the gas is related to the average kinetic energy of the molecules

Question 8

so what are ideal gases? (3)

Answer 8

they have zero particle volume and no intermolecular forces of attraction (OR repulsion),

so they follow the kinetic theory of gases

Question 9

& what are real gases?

Answer 9

in reality, gases do not fit this description exactly but real gases may come very close

Question 10

changing TEMP,
constant pressure

INCREASES the volume.

Answer 10

-gas is heated (at constant pressure), particles gain more kinetic energy and undergo more frequent collisions with the container wall

-to keep pressure constant, molecules must get further apart and so the volume increases

-so volume is directly proportional to the temperature (at constant pressure)

Question 11

Limitations of the ideal gas law,

under what conditions do these limitations arise?

Answer 11

very high pressures and low temperatures,

real gases do not obey the kinetic theory.

Question 12

why high pressure, low temp limits ideal gas law?

Answer 12

-Molecules close to each other

-there are instantaneous dipole- induced dipole or permanent dipole- permanent dipole forces between molecules

-attractive forces pull molecules away from the container wall

-volume of the molecules is not negligible

Question 13

so what assumptions are not followed by real gases?

Answer 13

at high temperatures and pressures,

-there is zero attraction between molecules (due to attractive forces, the pressure is lower than expected for an ideal gas)

-the volume of the gas molecules can be ignored (volume of the gas is smaller than expected for an ideal gas)

Question 14

Ideal gas equation
(can also be used to calc molar mass, Mr, of a gas)

Answer 14

pV = nRT

p = pressure (pascals)
V = volume (m3)
n = number of moles of gas (mol)
R = gas constant (8.31 J K-1 mol-1)
T = temp (kelvin)

Question 15

e.g v=?, n = 0.781mol, p = 220kPa = 220 x 10^3 Pa, temp = 21 Celsius = 294.15K

Answer 15

V = nRT/p = 0.781 x 8.31 x 294.15 / 220 x 10^3 =0.00868 or 0.00867 m^3 to dm^3 will be x1000

= 8.67 dm^3

Question 16

celsius to k

Answer 16

celsius + 273 = kelvin

Question 17

relationships for ideal gas + graph

vol = temp
vol = 1/p

Answer 17

an ideal gas will have a volume that is directly proportional to the TEMPERATURE and inversely proportional to the pressure.

Question 18

IONIC

Answer 18

ionic compounds are arranged in giant ionic lattices (also called giant ionic structures)

The type of lattice formed depends on the sizes of the positive and negative ions which are arranged in an alternating fashion

The ionic lattice of MgO and NaCl are cubic

Question 19

COVALENT lattices (simple molecular OR giant molecular lattices)

examples

Answer 19

Simple molecular lattices: Iodine, buckminsterfullerene (C60) and ice

Giant molecular: silicon(IV) oxide, graphite and diamond

Question 20

metallic lattices

Answer 20

metal ions are often packed in hexagonal layers or in a cubic arrangement

Question 21

Ionic bonding & giant ionic lattice structures - structure & properties

Answer 21

• Ionic compounds are strong. Strong electrostatic forces in ionic compounds keep the ions strongly together

-are brittle as ionic crystals can split apart

• Ionic compounds have high melting and boiling points. Strong electrostatic forces between the ions in the lattice act in all directions and keep them strongly together. Melting and boiling points increase with charge density of the ions due to the greater electrostatic attraction of charges
• e.g, Mg2+O2- has a higher melting point Na+Cl-

📍Ionic compounds are soluble in water as they can form ion - dipole bonds

• Ionic compounds only conduct electricity when molten or in solution; ions can freely move around and conduct electricity. In the solid state they’re in a fixed position and unable to move around

Question 22

Metallic bonding & giant metallic lattice

Answer 22

📍Metallic compounds are malleable. When a force is applied, the metal layers can slide. The attractive forces between the metal ions and electrons act in all directions. So when the layers slide, the metallic bonds are re-formed

-The lattice is not broken and has changed shape

📍Metallic compounds are strong and hard, due to the strong attractive forces between the metal ions and delocalised electrons.

📍Metals have high melting and boiling points
Due to the strong attractive forces between the metal ions and delocalised electrons

The greater the number of delocalised electrons and the smaller the cation, the greater the attractive force between them resulting in a higher melting / boiling point

📍Pure metals are insoluble in water

📍Metals can conduct electricity when in a solid or liquid state; mobile electrons which can freely move around and conduct electricity

Question 23

SIMPLE covalent lattice

Answer 23

📍Simple covalent lattices have low melting and boiling points; compounds have weak intermolecular forces between the molecules. Only little energy is required to break the lattice.

📍Most compounds are insoluble with water, unless they are polar (such as HCl) or can form hydrogen bonds (such as NH3)

📍They do not conduct electricity in the solid or liquid state as there are no charged particles. BUT Some simple covalent compounds conduct electricity in solution such as HCl which forms H+ and Cl- ions

Question 24

GIANT covalent lattice

Answer 24

📍Giant covalent lattices have high melting and boiling points; compounds have a large number of covalent bonds linking the whole structure and intermolecular forces between the molecules. A lot of energy is required to break the lattice

📍The compounds can be hard or soft. Graphite is soft as the forces between the carbon layers are weak. Diamond and silicon(IV) oxide are hard as it is difficult to break their 3D network of strong covalent bonds

📍Most compounds are insoluble in water

📍Most compounds do not conduct electricity however some do.

-Graphite has delocalised electrons between the carbon layers which can move along the layers when a voltage is applied

-Diamond and silicon(IV) oxide do not conduct electricity as all four outer electrons on every carbon atom are involved in a covalent bond so there are no free electrons available

Question 25

hess's law enthalpy change from A to B = enthalpy changes from A to X1 and X1 to Y, and Y to B ENTHALPY INDEPENDENT FINAL INITIAL DON'T CHANGE

Answer 25

📍📍📍the ENTHALPY change is INDEPENDENT 📍📍📍 of the path taken as long as the initial and final conditions do not change.📍📍📍

Question 26

qualitative analysis 1. ANIONS

Answer 26

carbonate: ADD DILUTE ACID, TEST FOR CO2, EFFERVESCENCE, CO2 PRODUCED halides - Cl, Br, I: acidify with DILUTE NITRIC ACID, add (aq) SILVER NITRATE; cl white, br cream, I yellow nitrate: add aqueous sodium hydroxide (OH- (aq)) then aluminium foil, warm/heat carefully, ammonia produced/"liberated" nitrite: same as nitrate + decolourises acidified aqueous potassium manganate (VII) sulfate: ACIDIFY with DILUTE NITRIC ACID, add aqueous barium nitrate; WHITE PPT (insoluble in excess dilute strong acids] gives white ppt with high Ca2+ (aq) sulfite: ADD DILUTE HCL, WARM GENTLY, add small volume of acidified aqueous potassium manganate (VII); purple to colourless, decolourises + white ppt with barium nitrate thiosulfate: gives OFF-WHITE / PALE YELLOW ppt slowly with H+ (HCl)

Question 27

2. CATIONS zinc so so so in, AL CHROM in so, COPPER

Answer 27

- NaOH (aq) THEN NH3 (aq) - aluminium: white ppt, soluble in excess, giving colourless solution. 📍 white ppt, insoluble in excess ammonium: ammonia produced on warming 📍 - barium: faint white ppt observed unless barium conc is very low 📍 no ppt calcium: white ppt, insoluble in excess 📍 no ppt/very slight white ppt chromium: grey green ppt, soluble in excess giving DARK GREEN solution📍 grey-green ppt, insoluble in excess copper: pale blue ppt, insoluble in excess 📍pale blue ppt, soluble in excess giving dark blue solution iron 2: green ppt, turning brown on contact with air, insoluble in excess 📍 green ppt, turning brown on contact with air, insoluble in excess iron 3: red brown ppt, insoluble in excess. 📍red brown ppt, insoluble in excess. magnesium: WHITE PPT INSOLUBLE in excess 📍WHITE PPT INSOLUBLE in excess manganese: OFF-WHITE PPT rapidly turning brown on contact with air, insoluble in excess 📍OFF-WHITE PPT rapidly turning brown on contact with air, insoluble in excess zinc: white ppt., soluble in excess 📍 white ppt, soluble in excess

Question 28

gas ammonia carbon dioxide hydrogen oxygen

Answer 28

ammonia: turns damp red litmus paper blue co2: gives white ppt with limewater; bubbles/effervescence/fizzing h2: pops with a lighted splint o2: relights glowing splint

Question 29

ideal gas exam q's 1. state three basic assumptions that scientists make about the properties of ideal gases 2. why CHF3 deviates from the properties of an ideal gas at pressures greater than 300atm. 3. two conditions necessary for these two gases to approach ideal gas behaviour. 4. State two assumptions of the kinetic theory as applied to an ideal gas. 5. State and explain the conditions at which krypton behaves most like an ideal gas 6. Explain, in terms of activation energy, Ea, and the collision of particles, how an increase in temperature affects the rate of a chemical reaction. RATE, ENERGY A GREATER/EQUAL TO, FREQ OF SUCC COLLISIONS,

Answer 29

1. particles have zero/negligible volume (compared to total volume of gas / container) * no (imf of attraction) forces / interactions between particles * collision between particles are elastic 2. particles / molecules are (so) close [1]. particle size/volume becomes significant 3. high temperature AND low pressure 4. 2 assumptions: no forces of attraction between particles * (ideal gas) particles have no / negligible volume (compared to container) * collisions between (ideal gas) particles / walls of container are perfectly elastic 5. 📍📍low pressure AND high temperature * volume of particles is negligible (compared to volume of container) * VdW forces/imf of attraction are insignificant (owing to high kinetic energy of particles) 6. 📌 -RATE INCREASES -(increase in temperature means) more particles have energy greater than or equal to activation energy * frequency of successful collisions increases

Question 30

chemical energetics, enthalpy change, hess' law [savemyexams] 1. ENTHALPY; why; exothermic

Answer 30

📍Enthalpy/heat content: total chemical energy inside a substance 📍When chemical reactions take place, changes in chemical energy take place and therefore the enthalpy changes 📍exothermic: more reactants; Heat energy is given off by the reaction to the surroundings. The temperature of the ENVIRONMENT increases - this can be measured on a thermometer The temperature of the SYSTEM decreases 📍an enthalpy decrease during the reaction so ΔH is negative 📍thermodynamically possible (because the enthalpy of the reactants is higher than that of the products) 📍rate may be too slow to observe any appreciable reaction 📍 reaction is kinetically controlled 📍means the reaction could have a high activation energy which is preventing the reaction from taking place.

Question 31

endothermic

Answer 31

📍products have more energy than the reactants 📍Heat energy is absorbed by the reaction from the surroundings 📍temperature of the environment decreases - this can be measured with a thermometer The temperature of the system increases 📍an enthalpy increase during the reaction so ΔH is positive

Question 32

tip important to specify the physical states of each species in an equation when dealing with enthalpy changes as any changes in state can cause very large changes of enthalpy. USE STATE SYMBOLS eg

Answer 32

Na+Cl- (s) → Na+ (aq) + Cl- (aq) ΔH = +4 kJ mol-1 Na+Cl- (g) → Na+ (g) + Cl- (g) ΔH = + 500 kJ mol-1

Question 33

system=? environment? meaning?

Answer 33

system is the molecules that are reacting (i.e. the reaction itself) and the surroundings are everything else (eg. the flask the reaction is taking place in).

Question 34

Reaction Pathway Diagrams/energy profile

Answer 34

shows the energies of the reactants, the transition state(s) and the products of the reaction with time

Question 35

transition state?

Answer 35

a stage during the reaction at which chemical bonds are partially broken and formed -transition state is very unstable – it cannot be isolated and is higher in energy than the reactants and products -activation energy (Ea) is the energy needed to reach the transition state

Question 36

DEFINE activation energy

Answer 36

the minimum amount of energy needed for reactant molecules to have a successful collision and start the reaction

Question 37

exo

Answer 37

-reactants are higher in energy than the products - reactants are therefore closer in energy to the transition state -means that exothermic reactions have a lower activation energy compared to endothermic reactions EXO LOWER EA ++ endo higher Ea

Question 38

Ea

Answer 38

-energy difference from the energy level of the reactants to the top of the ‘hump’ --Question is asking for the reverse reaction the Ea is the energy difference from the energy level of the products to the ‘hump’ - +20

Question 39

tips

Answer 39

-The activation energy is the energy difference from reactants to the transition state. The enthalpy change of the reaction is the energy difference from reactants to products. Remember to label the axis of the reaction pathway diagrams! axis - on y-axis Energy (kjmol-1) and "Extent of Reaction"

Question 40

To fairly compare the changes in enthalpy between reactions, all reactions should be carried out under standard conditions which are: STANDARD CONDTIONS =

Answer 40

A pressure of 101 kPa A temperature of 298 K (25 oC) Each substance involved in the reaction is in its normal physical state (solid, gas or liquid)

Question 41

To show that a reaction has been carried out under standard conditions, the symbol θ is used ΔHθ = the standard enthalpy change ++

Answer 41

+ table STANDARD ENTHALPY CHANGE OF... 1. reaction 2. formation 3. combustion 4. neutralisation atomisation dissociation - ch4 => c + 4h

Question 42

enthalpy change of atomisation (ΔHθat) EG - ΔHθat [H2] relates to the equation: ½H2 (g) → H (g) ENDOTHERMIC

Answer 42

enthalpy change when one mole of gaseous atoms is formed from its elements under standard conditions.

Question 43

Standard Enthalpy Change of reaction ΔHθr both endo exo

Answer 43

The enthalpy change when the reactants in the stoichiometric equation react to form the products under standard conditions in the molar quantities elements in standard state under standard conditions

Question 44

Standard Enthalpy Change of formation ΔHθf both endo exo +"cannot be ΔHθf as there is more than one compound being formed" TIP ;; The ΔHθf of an element in its standard state is zero. For example, ΔHθf of O2 (g) is 0 kJ mol-1

Answer 44

The enthalpy change when one mole of a compound is formed from its elements under standard conditions

Question 45

Standard Enthalpy Change of combustion ΔHθc exothermic

Answer 45

enthalpy change when one mole of a substance is burnt in excess oxygen under standard conditions

Question 46

Standard Enthalpy Change of neutralisation ΔHθneut exothermic

Answer 46

enthalpy change when one mole of water is formed by reacting an acid and an alkali under standard conditions

Question 47

Enthalpy & Bond Energies

Answer 47

📍During a reaction, enthalpy changes take place because bonds are being broken and formed 📍Energy (in the form of heat) is needed to overcome attractive forces between atoms Bond breaking is therefore endothermic Energy is released from the reaction to the surroundings (in the form of heat) when new bonds are formed Bond forming is therefore exothermic 📍If more energy is required to break bonds than energy is released when new bonds are formed, the reaction is endothermic 📍If more energy is released when new bonds are formed than energy is required to break bonds, the reaction is exothermic 📍In reality, only some bonds in the reactants are broken and then new ones are formed

Question 48

exact bond energy. what is bond dissociation energy/exact bond energy?

Answer 48

bond dissocation energy: 📍The amount of energy required to break one mole of a specific covalent bond in the gas phase -also known as "exact bond energy" or "bond enthalpy" -type of bond broken is put in brackets after E Eg. EE(H-H) is the bond energy of a mole of single bonds between two hydrogen atoms

Question 49

Average bond energy

Answer 49

📍Bond energies are affected by other atoms in the molecule (the environment) 📍an average of a number of the same type of bond but in different environments is calculated 📍 bond energy is known as the average bond energy 📍since bond energies cannot be determined directly, enthalpy cycles are used to calculate the average bond energy

Question 50

Calculating enthalpy change from bond energies

Answer 50

-Bond energies are used to find the ΔHrꝋ of a reaction when this cannot be done experimentally -equation to calculate the standard enthalpy change of reaction using bond energies is: ΔHθr = enthalpy change for bonds broken + enthalpy change for bonds formed eg 2NH3 => 6 bonds here, so 6x391

Question 51

eg may also have to change eqn -The enthalpy of combustion is the enthalpy change when one mole of a substance reacts in excess oxygen to produce water and carbon dioxide The chemical reaction should be therefore simplified such that only one mole of ethyne reacts in excess oxygen

Answer 51

-2H–C=C–H + 5O=O → 2H–O–H + 4O=C=O -ONE MOLE OF ETHYNE. so divide by 2 => H-C=C-H + 2 ½ O=O → H-O-H + 2O=C=O 2912-4142= -1230.

Question 52

Measuring enthalpy changes CALORIMETRY

Answer 52

-Calorimetry is a technique used to measure changes in enthalpy of chemical reactions A calorimeter can be made up of a polystyrene drinking cup, a vacuum flask or metal can ++

Question 53

specific heat capacity (c) of water = 4.18 J g-1 oC-1 + q = m x c x ΔT - q = the heat transferred, J & m = the mass of water, g J = g x temp deg cels x c

Answer 53

energy needed to increase the temperature of 1 g of a substance by 1 oC

Question 54

eg In a calorimetric experiment, 2.50 g of methane is burnt in excess oxygen. 30% of the energy released during the combustion is absorbed by 500 g of water, causing the temperature to rise from 25 oC to 68 oC. The specific heat capacity of water is 4.18 J g-1 K-1. Calculate the total energy released per gram of methane burnt. ANS

Answer 54

Step 1: Gather the necessary values for the q = m x c x ΔT equation: m (of water) = 500 g c (of water) = 4.18 J g-1 K-1 ΔT (of water) = 68 oC - 25 oC = 43 oC = 43 K The change in temperature in oC is equal to the change in temperature in K Step 2: Complete the calculation: q = 500 x 4.18 x 43 q = 89 870 J Step 3: Conversion from 30% to 100% The value of q calculated is only 30% of the total energy released by 2.50 g of methane Total energy x 0.3 = 89 870 J Total energy = 299 567 J Step 4: Calculate the energy released by 1.00 g of methane The energy calculated is released by 2.50 g of methane Energy released by 1.00 g of methane = fraction numerator 299567 over denominator 2.50 end fraction Energy released by 1.00 g of methane = 119 827 J Step 5: Convert the answer from J to kJ: 119827 over 1000 = 120 kJ (to 3 significant figures)

Question 55

tips -reverse reac & flip sign for same enthalpy -acid, alkali, salt (aq) => use m and c values of water - remember that ΔT is the same in oC and K!

Answer 55

📍When new bonds are formed the amount of energy released is equal to the amount of energy absorbed when the same bonds are broken 📍eg., O2 (g) → 2O (g) E (O=O) = +498 kJ mol-1 2O (g) → O2 (g) E (O=O) –498 kJ mol-1 📍reverse reaction = flip signs. 📍Aqueous solutions of acid, alkalis and salts are assumed to be largely water so you can just use the m and c values of water when calculating the energy transferred + remember that ΔT is the same in oC and K!

Question 56

HESS'S LAW savemyexams .

Answer 56

📍"The total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place as long as the initial and final conditions are the same." 📍Hess’ Law is used to calculate enthalpy changes which can’t be found experimentally using calorimetry, e.g.: 3C (s) + 4H2 (g) → C3H8(g) ΔHf (propane) can’t be found experimentally as hydrogen and carbon don’t react under standard conditions

Question 57

Calculating ΔHr from ΔHf using Hess’s Law energy cycles

Answer 57

-products can be directly formed from the elements = ΔH2 -OR The products can be indirectly formed from the elements = ΔH1 + ΔHr ++The enthalpy change from elements to products (direct route) is equal to the enthalpy change of elements forming reactants and then products (indirect route) ΔH2 = ΔH1 + ΔHr

Question 58

📌📌📌📌📌📌 eqns

Answer 58

-reaction: ΔHr = ΔH2 – ΔH1 -combustion: ΔHf = ΔH1 – ΔH2

Question 59

eg ΔHr = ΔH2 (elements to products, on right) – ΔH1 (elements to reactants, left H1)

Answer 59

Step 4: Apply Hess’s Law: ΔHr = ΔH2 – ΔH1 ΔHr = ( ΔHf [Na2CO3 (s)] + ΔHf [CO2 (g)] + ΔHf [H2O (l)] ) – 2ΔHf [NaHCO3 (s)] ΔHr = ( (–1130.7) + (–393.5) + (–285.8) ) – ( 2 x –950.8) ΔHr = +91.6 kJ mol-1

Question 60

Calculating ΔHf from ΔHc using Hess’s Law energy cycles

Answer 60

- combustion products can be formed directly from elements to combustion products = ΔH1 OR The combustion products can be formed indirectly from elements to compounds to combustion products = ΔHf + ΔH2

Question 61

Calculating average bond energies using Hess's cycles

Answer 61

Bond energies cannot be found directly so enthalpy cycles are used to find the average bond energy This can be done using enthalpy changes of atomisation and combustion or formation eg atomisation - formation/combu

Question 62

TIPS -moles - if 2, x2

Answer 62

Remember to take into account the number of moles of each reactant and product. For example, there are two moles of NaHCO3 (s) so the ΔHf value is multiplied by 2.