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thermochemistry past exam notes

Shannon Ruscoe Chemistry exam NCEA Level 3 Englisch 5 Seiten

thermochemistry past exam notes

Inhalt der Notizen (5)

Seite 1

Resource booklet

thermochemistry past exam notes

Seite 2

2017 Question one

(a) Complete the following table

Merit question

remember that when electrons are lost (particle has a positive charge) they are removed from the 4s energy level before the 3d, despite the 4s being filled before the 3d.

use periodic table for the none electron configuration related questions

thermochemistry past exam notes

(b) (i) Define the term electronegativity.

Achievement question

Electronegativity is the ability of an atom to attract a bonding pair of electrons to itself.

(ii) Explain why the electronegativity of chlorine is greater than that of phosphorus.

Excellence question

Electronegativity increases as you go across a period.

Both Cl and P are row 3 elements and have valence electrons in their 3rd shell.

The electrons are in the same shell so experience the same shielding effect.

Chlorine has more protons in its nucleus than phosphorus so its nuclear charge is greater.

This means that chlorine will have more attraction for the bonding pair of electrons so its electronegativity is greater.

(c) The following graph shows the first ionization energies of the Group 2 elements from Be to Ba

(i) Write an equation to show the first ionization energy for the element calcium

achievement question

Ca(g) → Ca+ (g) + e–

(ii) Explain the trend shown of first ionization energies of the Group 2 elements

Excellence question

The first ionization energy is the energy required to remove one mole of the most loosely held electrons from one mole of gaseous atoms.

The trend is that the ionization energy decreases going down the group two elements.

Although the nuclear charge increases due to more protons in the atoms going down a group,

it is offset by the increasing distance of the outer electrons from the nucleus as the atomic radius increases due to more energy levels being added.

The full inner energy levels shield the outer electrons from the protons in the nucleus so the electrostatic attraction is less.

Additional energy levels result in greater shielding/repulsion between energy levels.

The further the outer electron is from the nucleus, the less energy needed to remove it.

thermochemistry past exam notes

Seite 3

2017 Question two

thermochemistry past exam notes

Use the information in the table above to compare and contrast the boiling points of the substances below. In your answers, you should:

list the types of intermolecular forces present for each substance

explain the relative strength between the particles involved.

achievement question- to correctly name all of the intermolecular forces between two of the three substances

(a) (i) Hydrazine and iodomethane.

Excellence question

Hydrazine (N2H4) hydrogen bonding, (permanent dipole attractions) temporary dipole attractions.

iodomethane (CH3I) permanent and temporary dipole attractions.

The hydrogen bonds between N2H4 molecules are stronger than the permanent dipole forces between CH3I molecules,

therefore, require more energy to break resulting in a higher boiling point.

The presence of hydrogen bonding outweighs the expected higher temporary dipole in CH3I due to the greater molar mass.

(ii) Iodomethane and decane

Excellence question

Iodomethane (CH3I) has permanent and temporary dipole attractions.

Decane (C10H22) has temporary dipole attractions.

Despite the molecules having the same molar mass, decane C10H22 has stronger intermolecular attractions.

Decane is a longer molecule when compared to the spherical shape of iodomethane CH3I,

so it has a greater surface area/electron cloud/number of electrons meaning stronger temporary dipole attractions.

Therefore, more energy is needed to break these attractions, resulting in a higher boiling point

(b) Explain why the solubility of hydrazine in water is greater than that of decane in water

Merit question

Hydrazine is a polar molecule.

Decane is non-polar

As water is a polar solvent, the hydrazine will be more soluble than the decane.

The attractive forces between the molecules of hydrazine are less than the attractive forces between the hydrazine and water molecules,

and therefore it is more soluble than decane,

where the attractive forces between the decane molecules are greater than the attractive forces between the decane and water molecules.

(c) Carbon dioxide and water are formed when decane burns completely in oxygen. The reaction is shown in the equation below.

C10H22() + 15 1 2 O2(g) → 10CO2(g) + 11H2O()

Calculate the enthalpy of combustion for decane, given the following data:

Δf H° (C10H22(ℓ)) = –301 kJ mol–1

Δc H° (C) = –393 kJ mol–1

Δc H° (H2) = –286 kJ mol–1

Excellence question

10C + 11H2 → C10H22 –301 kJ mol–1

C + O2 → CO2 –393 × 10 kJ mol–1

H2 + ½O2 → H2O –286 × 11 kJ mol–1

ΔH = +301 + (10 × –393) + (11 × –286)

ΔH= –6775 kJ mol–1

(d) The reaction for the complete combustion of hydrazine is shown in the equation below.

N2H4() + O2(g) → N2(g) + 2H2O(g)

This is an exothermic reaction.

Explain the entropy changes associated with this reaction

Excellence question

System – as the number of gaseous molecules is greater on the product side than the reactant side, then there is an increase in disorder / the dispersal of matter/degree of randomness/dispersal of energy, thus the entropy of the system increases.

Surroundings – as the reaction is exothermic the entropy of the surroundings increases, as there is an increase in disorder / the dispersal of matter/degree of randomness/dispersal of energy.

Seite 4

2017 Question three

Chlorine, Cl2, bromine, Br2, and iodine, I 2, are all halogens.

Bromine is a liquid at room temperature.

(a) (i) In the box below, tick the type(s) of intermolecular attractions in liquid bromine

(ii) Explain why bromine is a liquid at room temperature, whereas chlorine is a gas.

Merit question

The first box ticked (temporary dipole-dipole attractions)

Bromine is a larger molecule than chlorine so the temporary dipole intermolecular attractions are greater.

Thus, more energy is required to separate the molecules and so Br2 has a higher boiling point and is a liquid at room temperature

thermochemistry past exam notes

(b) (i) Write an equation for the sublimation of iodine in the box below

(ii) Define the enthalpy of sublimation for iodine.

achievement question

I2(s) → I2(g)

This is the heat energy required to change one mole of a substance from a solid state to gaseous state (at a given combination of temperature and pressure).

(iii) Explain why the sublimation of iodine is spontaneous, even though the enthalpy of sublimation is a positive value.

Excellence question

Spontaneity is determined by the total entropy change (system + surroundings).

The entropy of the system increases as the solid becomes a gas because the gas particles are more disordered.

positive enthalpy due to the endothermic process of breaking bonds is offset by entropy changes in the system

(c) Iodine forms a linear I3 – ion.

(i) Draw the Lewis structure for the I 3 – ion in the box below

Achievement question

thermochemistry past exam notes

(ii) Explain why the I 3 – ion has a linear shape.

Merit question

the I3 ion has five regions of negative charge around the central I atom

These areas all repel each other for maximum seperation

Therefore the arrangement of areas of electron density is a trigonal bipyramidal geometry

As there are three non-bonding pairs (in the equatorial area) and two bonded atoms,

the shape is linear.

(iii) IF5 has a square pyramidal shape.

Indicate whether the molecule IF5 is polar or non-polar. Circle your choice.

polar OR non-polar

Justify your choice

Excellence question

Polar.

The I-F bond is polar due to a difference in electronegativity.

There are six regions of negative charge giving IF5 an octahedral geometry.

The five bonded and one lone pair around the central iodine atom gives it the square pyramid shape.

This means the molecule is asymmetric

so the bond polarities dipoles don’t cancel

causing the molecule to be polar.

Seite 5

2017 summary

Question one

For achievement

writing some electron configurations

Defining periodic trends

writing the equation for first ionization energies

being able to identify what the actual trend is as a result of the periodic trend

For merit

writing all correct electron configurations

link resulting effects such as an increase/decrease in another related periodic trend, to the observed trend.

For excellence

full and correct explanations explaining the periodic trend, that must answer all aspects of the asked question

Question two

For achievement

Naming present intermolecular forces for substances

State relative strength for particles involved

state polarity

state solubility

state an increase/decrease in entropy

state an increase/decrease in enthalpy

For merit

link relative strengths of intermolecular forces to the properties of substances

link the relative strengths of intermolecular forces to the surface area of a substance

fully explain solubility in terms of polarity

partially explain entropy changes in a system and surroundings

For excellence

justifies properties in terms of intermolecular forces

Able to calculate the enthalpy of a given reaction and give the answer with correct significant figures, with units, and appropriate sign

fully explain entropy changes in a system and surroundings.

Question three

For achievement

Naming present intermolecular forces for substances*

state relative strength for particles involved*

Define different types of enthalpy changes

Define spontaneity

Draw Lewis structure

state polarity

state shape

For merit

link relative strengths of intermolecular forces to the properties of substances*

partially explain entropy changes in a system and surroundings

correctly explain the shape

link shape to the polarity

For excellence

the complete explanation of entropy and enthalpy linked to the spontaneity of the sublimation process.

write a full explanation of the shape and polarity

2017 past exam summary

Achieved

writing some electron configurations

Defining periodic trends

writing the equation for first ionization energies

being able to identify what the actual trend is as a result of the periodic trend

Naming present intermolecular forces for substances

State relative strength for particles involved

state polarity

state solubility

state an increase/decrease in entropy

state an increase/decrease in enthalpy

Define different types of enthalpy changes

Define spontaneity

Draw Lewis structure

state shape

Merit

writing all correct electron configurations

link resulting effects such as an increase/decrease in another related periodic trend, to the observed trend.

link relative strengths of intermolecular forces to the properties of substances

link the relative strengths of intermolecular forces to the surface area of a substance

fully explain solubility in terms of polarity

partially explain entropy changes in a system and surroundings

correctly explain the shape

link shape to the polarity

Excellence

full and correct explanations explaining the periodic trend, that must answer all aspects of the asked question

the complete explanation of entropy and enthalpy linked to the spontaneity of the sublimation process.

write a full explanation of the shape and polarity

justifies properties in terms of intermolecular forces

Able to calculate the enthalpy of a given reaction and give the answer with correct significant figures, with units, and appropriate sign

fully explain entropy changes in a system and surroundings.

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NCEA Level 3 Chemistry exam Notiz am thermochemistry past exam notes, erstellt von Shannon Ruscoe am 09/10/2018.
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Shannon Ruscoe
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