Physical Chemistry Flashcards

Question

a sample of tungsten is 0.1% 180W, 26.5% 182W, 14.3% 183W, 30.7% 184W and 28.4%186W. Calculate the Ar of tungsten

Answer 1

a sample of tungsten is 0.1% 180W, 26.5% 182W, 14.3% 183W, 30.7% 184W and 28.4%186W. Calculate the Ar of tungsten ( 0.1 x 180 ) + ( 26.5 x 182 ) + ( 14.3 x 183 ) + ( 30.7 x 184 ) + ( 28.4 x 186 ) =183.891 18389.1/100 =183.9 (to 1 d.p)

Answer 2

what is relative atomic mass? (Ar) The average mass of an atom of an element what is relative molecular mass? (Mr) the average mass of a molecule what is relative formula mass? the average mass of a formula unit

Answer 3

Two techniques to ionise the sample: 1 Electrospray ionisation 2 Electron impact ionisation

Answer 4

Electrospray ionisation is a gentler technique and prevents fragmentation. It’s typically used for polymers and biological materials like DNA. The sample is dissolved in a solvent and a high voltage is applied. The high voltage rips a proton off the solvent and attaches it to the sample molecules. The sample molecules are now positively charged ions.

Answer 5

In electron impact ionisation, the sample is first vaporised and then hit with electrons from an electron gun. The electrons knock off electrons from the molecule The molecules are now positively charged ions. This method often causes the sample to fragment.

Answer 6

2) Acceleration Molecules are accelerated to all have the same kinetic energy. Kinetic energy = 1/2mv^2 All the molecules have the same kinetic energy, so the speed is dependent on the mass of the molecule. Lighter particles move faster and are detected before heavier particles.

Answer 7

3) Ion drift The time of flight is given by: Time of flight = distance / velocity This leads to an equation for the time travelled that depends on mass: Time = distance × the square root of this -->mass / 2KE Lighter ions take less time as the time is dependent on the square root of the mass.

Answer 8

4) Detection The ions hit a negatively charged plate. This causes a current and the size of this current gives a measure of the number of molecules hitting the plate. This gives the abundance of the molecule.

Answer 9

Analysis to Identify Molecules Once a sample has passed through the mass spectrometer, we can analyse the data to identify the molecule. Spectrum produced When the sample has passed through the mass spectrometer, a spectrum is produced by the spectrometer. On this spectrum: The x-axis is mass/charge ratio. The y-axis is % abundance.

Answer 10

Main peak The spectrum produces lots of peaks, but the most important is the molecular ion peak. This is the peak of the greatest mass/charge ratio. This represents the mass/charge value of the molecule we are analysing.

Answer 11

Isotopes Smaller peaks will cluster around the molecular ion peak. These are from the same molecules but with different isotopes in them. The isotopic molecules have different masses and so different mass/charge ratio values.

Answer 12

Any smaller and significantly lighter peaks in the spectrum are because of fragmentation. The molecule can fragment in the spectrometer.

Answer 13

Analysis to Calculate Once we have the mass spectrum, we can calculate the relative atomic masss Example - boron This is the mass spectrum of a sample of elemental boron. Boron has two isotopes, 10B and 11B. You can see from the spectrum that approximately 20% of the boron is 10B and 80% is 11B. You can use this to work out the relative atomic mass: (80% × 11)+ (20% × 10) = 10.8

Answer 14

Electron shells electrons are arranged in shells. These shells are defined by the principal quantum number, given the symbol 'n'. The lowest energy shell has n = 1. Other shells have higher energy and higher n. The higher the n of an electron, the further from the nucleus it orbits.

Answer 15

Sub-shells For atoms with more than one electron, shells are split into sub-shells that have slightly different energies. The difference in energy between sub-shells is much less than the difference in energy between shells. A shell with a given n will have n sub-shells. E.g. the n = 3 electron shell has three sub-shells.

Answer 16

Orbitals Sub-shells are composed of orbitals. Orbitals in the same sub-shell have the same energy. Each orbital can hold a maximum of two electrons.

Answer 17

Bringing it all together Electrons occupy orbitals. An orbital can hold a maximum of two electrons. A set of orbitals of the same energy is called a sub-shell. Sub-shells make up different shells of electrons. Different sub-shells do not have the same energy.

Answer 18

electronic structure electron shells sub-shells orbitals electrons

Answer 19

Types of orbitals Orbitals are labelled by letters. The first three orbitals are called s, p and d: An a sub-shell is made up of one s orbital. A p sub-shell is made of three p orbitals. A d sub-shell is made of five d orbitals. Orbitals of exactly the same energy are called 'degenerate'.

Answer 20

Shells and sub-shells The n = 1 shell has one sub-shell. It is an s sub-shell. The n = 2 shell has two sub-shells. They are an s sub-shell and a p sub-shell. The n = 3 shell has three sub-shells. It has an s sub-shell, a p sub-shell and a d sub-shell.

Answer 21

Filling orbitals Orbitals are filled from the lowest energy to the highest energy. The usual energy order is from s ⇒ p ⇒ d. The transition metals are an anomaly. The 4s sub-shell is of a lower energy than the 3d sub-shell, and so here the filling order is: 3s ⇒ 3p ⇒ 4s ⇒ 3d.

Answer 22

Electron shells n=1 s sub-shell 1s orbital ``` n=2 s sub-shell p sub-shell 2s orbital 3 × 2p orbitals ``` ``` n=3 s sub-shell p sub-shell d sub-shell 3s orbital 3 × 3p orbitals 5 × 3d orbitals ```

Answer 23

An atom has 16 electrons to hold. How are they arranged within the sub-shells? ``` 1 1s orbital contains 2 electrons 2 2s orbital contains 2 electrons 3 2p orbital contains 6 electrons 4 3s orbital contains 2 electrons 5 3p orbital contains 4 electrons ```

Answer 24

the ionisation energy equation. This is best shown by a few examples. The first ionisation energy equation of Na: Na(g) → e− + Na +(g) The second ionisation energy equation of Mg: Mg(g) → e− + Mg 2+(g)

Answer 25

The electrons Ionisation is the reaction where one electron is removed from an atom. There will be one electron in the product of the equation. So we will always have an e− term on the right-hand side of the equation.

Answer 26

The charges We must understand the charges on the atom during the reaction. During ionisation, we remove one electron from an atom. As the negative electron is removed, the atom's charge will increase by 1. E.g. the second ionisation of Na converts Na+ → Na2+.

Answer 27

The state symbols Ionisation energies are always defined as being in the gas phase. So atomic species of any charge, on either side of the equation, will have the state symbol (g). We do not define the state of the electron - it has no state symbol.

Answer 28

Ionisation Energies A specific amount of energy is needed to remove an electron from an atom or ion.

Answer 29

First ionisation energy The first ionisation energy is the energy required to remove an electron from every atom in a mole of atomic gas, to produce a mole of unipositive gaseous ions.

Answer 30

Second ionisation energy The second ionisation energy is the energy required to remove an electron from every ion in a mole of unipositive gaseous ions, to produce a mole of dipositive gaseous ions.

Answer 31

What do clustered small peaks around the molecular ion peak represent Isotopic molecules

Answer 32

Shielding The greater the number of electrons between the nucleus and the outer electrons, the lower the effective nuclear charge. This is because the positive charge felt by the electron is reduced by the electrons in between. This means that the greater the number of electrons, the lower the ionisation energy.

Answer 33

Nuclear charge The greater the number of protons in the nucleus, the greater the attraction of the electron to the nucleus. A greater attraction of the electron means more energy is needed to remove the electron. This means that ionisation energy is greater.

Answer 34

Distance from the nucleus Electrostatic attraction decreases sharply with distance. This means that less energy is needed to remove electrons which are further away. This means that as distance increases, ionisation energy decreases In practice, this means that the higher the principal quantum number of an electron, the lower its ionisation energy.

Answer 35

The three main factors affecting ionisation energies: ``` 1 The nuclear charge 2 The distance of the electron from the nucleus 3 The electron shielding ```

Answer 36

Breaking down the trend Consider the sodium atom of electron configuration 1s2 2s2 2p6 3s1 Its first ionisation energy corresponds to the removal of an electron from the n = 3 shell. Its second ionisation energy corresponds to the removal of an electron from the n = 2 shell. Between the third and ninth ionisation energies, an electron is being removed from the n = 2 shell of an increasingly positive ion. Its tenth ionisation energy corresponds to the removal of an electron from the n = 1 shell.

Answer 37

Explaining the patterns Ionisation energy increases with proximity to the nucleus. The n = 1 shell will have a higher ionisation energy than n = 2 shell. The n = 2 shell will have a higher ionisation energy than n = 3 shell. Within the n = 2 shell, successive ionisation energies will rise as the ion becomes more positively charged.

Answer 38

Trend down a group First ionisation energy decreases down a group. This is because the atomic radius increases. The electron is further from the nucleus, and so experiences lower attraction. There are also more electrons between the nucleus and the outer electrons. This means the outer electrons experience greater shielding from the nucleus. This also provides evidence for the existence of shells.

Answer 39

Trend across a period First ionisation energy increases across a period. This is because the atomic radius decreases. Proton number increases across a period, so electrons are more attracted to the nucleus. The electron is closer to the nucleus, and so experiences a greater attraction. Nuclear charge increases across a period.

Answer 40

Why does ionisation energy increase across a period? Atomic radius is smaller and so there is a greater attraction

Answer 41

How does ionisation energy vary within the periodic table? Decreases down a group Increases across a period

Answer 42

Example of sub-shell impact An example of the impact of the sub-shell structure is the difference in ionisation energies between magnesium and aluminium within Period 3. They are in the same period and so we would expect aluminium's ionisation energy to be greater due to a greater nuclear charge.

Answer 43

They are in the same period and so we would expect aluminium's ionisation energy to be greater due to a greater nuclear charge But the ionisation energies are: Mg = 738 kJ/mol. Al = 578 kJ/mol. These values do not align with the general trend we predicted. We need to look at the sub shell structures of these atoms to understand what is going on.

Answer 44

Mg vs Al sub-shell structure Mg = 1s2 2s2 2p6 3s2 So the first ionisation removes an electron from the 3s orbital. Al = 1s2 2s2 2p6 3s2 3p1 So the first ionisation removes an electron from the 3p orbital.

Answer 45

Sub-shell energies The 3p orbital is further from the nucleus on average than the 3s orbital. So this orbital has a higher energy and its electrons have a lower electrostatic attraction to the positive nucleus. It is also slightly shielded by the 3s2 electrons. So it's easier to ionise an electron in the 3p orbital than in the 3s orbital. This explains why it is easier to ionise Al than Mg. This effect outweighs the general trend we looked at first.

Answer 46

Why is it easier to ionise a 3p electron than a 3s electron? 1 3p orbital is higher in energy 2 3p electron has a lower electrostatic attraction to the positive nucleus 3 3p electrons are shielded by 3s electrons

Answer 47

The impact of electron repulsion can be seen in the ionisation energies of Group 5 and Group 6. Phosphorus and sulfur Both P and S have their outer electron in a 3p orbital. ``` P = 1s2 2s2 2p6 3s2 3p3 S = 1s2 2s2 2p6 3s2 3p4 ``` The shielding for each atom's outer electron is identical. S has a higher nuclear charge and so you might expect S to have a higher ionisation energy. This is NOT the case.

Answer 48

Singly vs doubly occupied orbitals The P electron is removed from a singly occupied orbital but the S electron is removed from a doubly occupied orbital. S has a lower ionisation energy because electrons in the same orbital repel each other. This means that an electron is easier to remove from a doubly occupied orbital. This gives more evidence for the electronic structure model we are using.

Answer 49

It’s easier to remove an electron from a doubly occupied orbital because the electrons repel each other

Answer 50

Concentration Concentration is defined as moles / unit volume. The usual units of concentration are moles per litre. Litre is often written as dm3. So moles per litre is mol ÷ dm3. This is often written as moldm-3.

Answer 51

Three moles of NaCl are dissolved in half a litre of water. Workout the concentration Concentration = number of moles ÷ volume Concentration = 3 mol ÷ 0.5 dm3 Concentration = 6 moldm-3

Answer 52

Mole calculations We can calculate the number of moles present in a sample if we know its mass, and its Mr: Moles = mass ÷ Mr Remember mr mole!

Answer 53

Benzene has an Mr of 78. How many moles of benzene are in 7.8 g of pure benzene? ``` Moles = mass ÷ Mr Moles = 7.8 g ÷ 78 Moles = 0.1 mol ```

Answer 54

Calculating masses 1) . Balance the equation 2) . Work out how many miles of the reactant you have 3) . Use the molar ratio from the balanced equation to work out the number of miles of product that will be formed 4) . Calculate mass of that many miles of product

Answer 55

Find the mass of iron oxide produced when 28.0 g of iron is burnt in air 1). Balance equation: 2Fe + 1 1/2 O2 —-> Fe2O3 2). Work out number of moles of iron you have - Ar of Fe = 55.8 Moles = mass / Ar = 28.0/55.8 =0.502 3). The molar ratio of Fe: Fe2O3 is 2:1 For every 2 moles of Fe you will produce 1 mole of Fe2O3 But you only have 0.502 / 2 = 0.251 4). Now find the mass of 0.251 moles of Fe2O3 Mr of Fe2O3 = (2x55.8)+(3x16) = 159.6 Mass = moles x Mr 0.25x159.6 = 40.0 g of iron oxide

Answer 56

Definition of enthalpy change An enthalpy change is a measure of the heat given out or taken in during a process. When objects are heated, they use energy to expand. Enthalpy takes into account the energy used in the expansion.

Answer 57

Enthalpy vs energy changes We use enthalpy instead of energy because we cannot easily measure energy changes. This is because objects expand when heated. To measure an energy change, we would have to fix the volume of the object. Enthalpy is much easier to use as it allows for expansion. Enthalpy changes are instead measured under constant pressure. The atmosphere is at a constant pressure, so we require no extra equipment.

Answer 58

Endothermic vs exothermic reactions An exothermic reaction is one which gives out heat energy. An exothermic reaction has a negative enthalpy change. An endothermic reaction is one which takes in heat energy. An endothermic reaction has a positive enthalpy change.

Answer 59

Combustion is exothermic Combustion is an exothermic process as it gives out heat! E.g. Burning methane: CH4 + 2O2 → CO2 + 2H2O ΔH = −882.00kJmol-1

Answer 60

Bond enthalpies During a reaction, some chemical bonds must be broken and made. The energy to break or make a bond is known as the bond enthalpy. Energy is needed to break a bond so it is an endothermic process. Energy is given off when a bond is made so it is an exothermic process. The enthalpy change of a reaction is a sum of the individual bond enthalpies being broken and made during the reaction.

Answer 61

The dominating term During a chemical reaction, we must determine whether the reaction requires more energy to break bonds or to make bonds. This will give an overall reaction enthalpy that is either positive or negative.

Answer 62

Endothermic vs exothermic Since bond breaking is endothermic and bond making is exothermic: More energy required to break bonds in a reaction will lead to an overall endothermic reaction. More energy released making bonds in a reaction will lead to an overall exothermic reaction.

Answer 63

When calculating reaction enthalpies, we must add the bond enthalpies of all the bonds that are being broken and then minus the bond enthalpies of all the bonds that are being made.

Answer 64

Bond breaking and making: 1 Bond breaking is an endothermic process 2 Bond making is an exothermic process

Answer 65

``` Enthalpy change ΔH = + Endothermic E.g. Thermal decomposition of calcium carbonate ΔH = − Exothermic E.g. Combustion of methane ```

Answer 66

Standard conditions: For many measurements, we can record them under standard conditions. This is true for enthalpy changes - we call it the standard enthalpy change. ``` 1 Pressure of 1bar 2 Temperature of 298K 3 Substances in their most stable state under the above conditions ```

Answer 67

The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under the standard conditions of 298K and 100 kPa (1bar).

Answer 68

Bomb calorimetry Bomb calorimetry uses a machine called a bomb calorimeter to measure enthalpy changes of combustion. This process involves burning a sample of a compound in a sealed vessel and measuring the temperature change. Often the calorimeter will just determine the temperature change in the vessel and you will have to calculate the enthalpy change of combustion.

Answer 69

Inaccuracy Bomb calorimetry can be inaccurate due to: Heat lost to the surroundings. Any incomplete combustion that may take place. Loss of some reactant that evaporates before it combusts.

Answer 70

Calculating enthalpy changes ``` The equation to calculate enthalpy changes from temperature changes is: q = m × c × ΔT q is the heat energy. m is the mass of the sample. c is the specific heat capacity. ΔT is the temperature change. If the pressure is constant, q = ΔcH ```

Answer 71

Calculating enthalpy changes ``` We have calculated q, the energy given off to the surroundings (the enthalpy change). The units of q are Joules. To determine the enthalpy change of combustion, we must convert Joules into the unit of enthalpy change, Joules per mole. Calculate the number of moles: Moles = mass ÷ Mr ΔH = \frac{q}{moles} moles q ```

Answer 72

Calorimetry is used to physically measure changes in enthalpy.

Answer 73

A student carried out a simple laboratory experiment to measure the enthalpy change for Process 3. The student showed that the temperature of 200 g of water increased by 8.0 °C when 0.46 g of pure ethanol was burned in air and the heat produced was used to warm the water. Use these results to calculate the value, in kJ mol–1, obtained by the student for this enthalpy change. (The specific heat capacity of water is 4.18 J K–1 g–1) Give one reason, other than heat loss, why the value obtained from the student’s results is less exothermic than a data book value. q = m x c x delta temp = 200 x 4.18 x 8.0 = 6688 moles = mass / Mr 0.46 / 46 = 0.01 6.688 / 0.01 kj/mol =668.8 kjmol^-1 (to convert an number of moles to one is to divide it b themselves 0.01 / 0.01 = 1) the reaction is exothermic so we add a negative

Answer 74

For best accuracy, we must plot the temperature loss after the reaction is complete. Then we can extrapolate the line to find the most accurate temperature change value.

Answer 75

Why are the initial and final temperatures in a calorimeter inaccurate? heat is always being lost to surroundings

Answer 76

Combustion of cyclohexane Cyclohexane fuel is burned completely in a calorimeter. There are 200 g of water in the calorimeter. There are 0.5 moles of cyclohexane burnt. The temperature of the water was raised from 298 K to 368 K. q = 200 x 4.18 x 70 = 58,520 j enthalpy change of combustion ΔH = q / moles ΔH = −58520 J ÷ 0.5 moles Note the minus sign added. This is because we know the reaction is exothermic since the water's temperature was increased. ΔH = −117040 Jmol-1 ΔH = −117.04 kJmol-1 Note the final units of kJmol-1 as this is more standard.

Answer 77

Neutralisation reaction Calculate the heat lost/gained during the reaction between H2SO4(aq) and NaOH(aq): 20 cm3 of the acid is added to an insulate container. 30 cm3 of the base is then added. The temperature change is recorded to be 40 K. Assume the density of the solutions to be the same as water, 1 gcm-3. Assume the specific heat capacity is the same as water's, 4.18 Jg-1K-1. The calculation ``` Because we have assumed that the density is the same as water, we can calculate the mass of the solution as: 1 cm3 = 1 g (20 + 30) cm3 = 50 g The heat change: q = mcΔT q = (50) g x 4.18 Jg-1K-1 x 40 K q = 8360 Joules q = 8.36 kJ ```

Answer 78

Two things that change if calculating temperature changes in reaction mixtures: 1 The density is no longer equal to the density of water (1 cmg-1) 2 The specific heat capacity may not be equal to 4.18 Jg-1K-1

Answer 79

What is the only case where we can state that a reaction mixture has the same density as water or the same specific heat capacity as water? if the q lets us assume so

Answer 80

The units of q are Joules. | The units of enthalpy change are Joules per mole.

Answer 81

IIt is NOT accurate to just measure the initial and final temperatures on the calorimeter.

Answer 82

``` The enthalpy change of X to Y is 10 Jmol-1. The enthalpy change of Y to Z is 30 Jmol-1. What is the enthalpy change of X to Z? 1 By Hess' Law: ΔH(X → Z) = ΔH(X → Y) + ΔH(Y → Z) 2 Substitute the values in: ΔH(X → Z) = 10 Jmol-1 + 30 Jmol-1 ΔH(X → Z) = 40 Jmol-1 ```

Answer 83

The enthalpy of formation for any element is zero.

Answer 84

hess' law allows us to indirectly calculate enthalpy change of a complex reaction by considering simpler reactions

Answer 85

Hess' Law states that the enthalpy change of a reaction is independent of the pathway.

Answer 86

Bond enthalpy is the enthalpy change when one mole of bonds is broken in the gas phase. This varies from molecule to molecule and from bond to bond.

Answer 87

Bond enthalpy Bond enthalpy is the enthalpy change when one mole of bonds is broken in the gas phase. This varies from molecule to molecule and from bond to bond.

Answer 88

Mean bond enthalpies ``` Some molecules (e.g. methane - CH4) have multiple identical bonds. The energy to break the second CH bond in methane will be greater than for the first bond because the molecule will rearrange after the first bond is broken. The CH bond enthalpy value for methane is the average of all four bond enthalpies. ```

Answer 89

Example Looking at the previous example of methane: First C-H broken: 401 kJmol-1 Second C-H broken: 413 kJmol-1 Third C-H broken: 428 kJmol-1 Fourth C-H broken: 439 kJmol-1 So overall, the enthalpy change will be an average: (439 + 428 + 413 + 401) ÷ 4 = 420.25 kJmol-1

Answer 90

Inaccuracies There are two main reasons mean bond enthalpies are inaccurate: Mean bond enthalpies are only valid in the gas phase. You might not be doing your reaction in the gas phase. Bond enthalpies depend on the particular molecule, whereas mean bond enthalpies are quoted generally. So we can't expect a perfect answer - they are only good as a guide.

Answer 91

Which of the following best describes bond enthalpies? often mean values

Answer 92

In an endothermic reaction, is the energy released by the formation of bonds GREATER or LESS THAN the energy needed to break bonds? Less than

Answer 93

Which of the following is the correct definition for ΔfHθ? the enthalpy

Answer 94

Conditions for a collision to result in a reaction Activation energy The collision must have enough energy for a reaction to happen Steric factor A collision must take place at the correct location within the molecule for a reaction to happen

Answer 95

Enthalpy profiles show the progress of reaction on the x-axis and energy level on the y-axis. They contain the following information: The relative enthalpies of reactants and products. The activation energy of a reaction. The overall enthalpy change of a reaction. This diagram shows an exothermic reaction.

Answer 96

Types of enthalpy profiles: Exothermic enthalpy profile Endothermic enthalpy profile

Answer 97

The axes in enthalpy profile diagrams: x -axis Progress of reaction. y -axis Enthalpy.

Answer 98

"Steric factor" is just a complex phrase to describe how the location of a collision is important.

Answer 99

how would u form carbon dioxide (give equation) ΔH C(s) + O2(g) ---> CO2(g)

Answer 100

how would u form methane (give equation) ΔH C(s) + 2H2(g) --->CH4(g)

Answer 101

calculate the enthalpy chnage for this reaction ΔH=? C2H2(g) + 2H2(g) ---> C2H6(g) enthalpies of formation: ΔHf (C2H2) = +228 Kj mol ^-1 ΔHf (C2H6) = -85 kj mol^-1 ΔHf = -85 2C(s) + 3H2(g) --------------> C2H6(g) ΔHf = +228 2C(s) + H2(g) -------------------> C2H2(g) ΔH = -228 + (-85) = 313 kj mol^-1 (see elliot rintoul energetics pt 3 10.46 minns)

Answer 102

ΔHc = enthalpy chnage of combustion

Answer 103

# define ΔHc when one mole of a compound is completely burned in oxygen with all reactants and products in their standard states and standard conditions

Answer 104

complete combustion products CO2 + H2O

Answer 105

incomplete combustion products CO + H20

Answer 106

complete combustion of methane CH4 + 2O2 ---> CO2 + 2H20

Answer 107

ENTHALPY CHANGE OF COMBUSTION ΔHc = ? C2H2(g) + 2H2(g) ----------------> C2H6(g) ΔHc (C2H2(g)) = - 1301 ΔHc (2H2(g)) = - 285 ΔHc (C2H6(g)) = - 1560

Answer 108

when ΔHc arrows go down when ΔHf arrows go up

Answer 109

for a reaction to occur particles must have energy greater than the activation energy

Answer 110

The Maxwell-Boltzmann distribution of energies is a handy little graph showing the energy distribution of all the molecules in a gas. The graph The number of molecules is on the y-axis. The kinetic energy of the molecules is on the x-axis.

Answer 111

The origin No molecules have zero energy, so the curve passes through the origin at (0,0).

Answer 112

The peak There is a peak in the middle. This represents the most likely energy of any molecule. There are more molecules with this energy than with any other energy.

Answer 113

The area under the graph The area under the graph gives the total number of molecules. You can think of this as like adding up the number of molecules with every particular energy.

Answer 114

Molecule energies The average energy is to the right of the maximum (peak). The tail of the graph is asymptotic to the x-axis. This means it tends to zero, but never touches the x-axis. This reflects that there is a small possibility of having a very high energy molecule.

Answer 115

Activation energies We can mark the activation energy on the Maxwell-Boltzmann distribution. This allows us to view the number of molecules with the energy to react. On the left-hand side of the activation energy mark, molecules have less energy than the activation energy and so they can't react. On the right-hand side, molecules have more energy than the activation energy and so they can react.